id
stringlengths 36
36
| text
stringlengths 13
4.79k
| idx
int64 0
46
| class
stringclasses 1
value | subject
stringclasses 3
values | chapter
stringclasses 31
values |
|---|---|---|---|---|---|
bfe5f43c-ccf3-4fba-9f53-adbe33ebd80e | # PHYSICS
# SUMMARY
1. The motion that repeats itself is called periodic motion.
2. The period T is the time required for one complete oscillation, or cycle. It is related to the frequency v by,
T = 1
v
The frequency ν of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz:
1 hertz = 1 Hz = 1 oscillation per second = 1s–1
3. In simple harmonic motion (SHM), the displacement x(t) of a particle from its equilibrium position is given by,
x(t) = A cos(ωt + φ) (displacement),
in which A is the amplitude of the displacement, the quantity (ωt + φ) is the phase of the motion, and φ is the phase constant. The angular frequency ω is related to the period and frequency of the motion by,
ω = 2π
T = 2
πν (angular frequency).
4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs.
5. The particle velocity and acceleration during SHM as functions of time are given by,
v(t) = –ωA sin(ωt + φ) (velocity),
a(t) = –ω²A cos(ωt + φ) = –ω²x(t) (acceleration),
Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude vₘ = ωA and acceleration amplitude aₘ = ω²A, respectively.
6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.
7. A particle executing simple harmonic motion has, at any time, kinetic energy K = ½ mv² and potential energy U = ½ kx². If no friction is present the mechanical energy of the system, E = K + U always remains constant even though K and U change with time.
8. A particle of mass m oscillating under the influence of Hooke’s law restoring force given by F = –kx exhibits simple harmonic motion with
ω = k
m (angular frequency)
T = 2π
m
k (period)
Such a system is also called a linear oscillator.
9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by,
T = 2π
L
g | 13 | 11 | Physics | 206 |
d2b5610f-17d7-4070-b9e3-7ad04c5e005a | # OSCILLATIONS
# POINTS TO PONDER
1. The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer.
2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = – k x is simple harmonic.
3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: –mω²r. In the latter case, the phases of motion, in two perpendicular directions (x and y) must differ by π/2. Thus, for example, a particle subject to a force –mω²r with initial position (0, A) and velocity (ωA, 0) will move uniformly in a circle of radius A.
4. For linear simple harmonic motion with a given ω, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase.
5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity.
6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes.
7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law).
8. The motion of a simple pendulum is simple harmonic for small angular displacement.
9. For motion of a particle to be simple harmonic, its displacement x must be expressible in either of the following forms:
- x = A cos ωt + B sin ωt
- x = A cos (ωt + α ), x = B sin (ωt + β )
The three forms are completely equivalent (any one can be expressed in terms of any other two forms). Thus, damped simple harmonic motion is not strictly simple harmonic. It is approximately so only for time intervals much less than 2m/b where b is the damping constant.
2024-25 | 14 | 11 | Physics | 206 |
4dc1c339-2996-444b-8ea7-88a338b1b49c | # PHYSICS
# Exercises
# 13.1
Which of the following examples represent periodic motion?
- (a) A swimmer completing one (return) trip from one bank of a river to the other and back.
- (b) A freely suspended bar magnet displaced from its N-S direction and released.
- (c) A hydrogen molecule rotating about its centre of mass.
- (d) An arrow released from a bow.
# 13.2
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
- (a) the rotation of earth about its axis.
- (b) motion of an oscillating mercury column in a U-tube.
- (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
- (d) general vibrations of a polyatomic molecule about its equilibrium position.
# 13.3
Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?
Fig. 18.18
2024-25 | 15 | 11 | Physics | 206 |
f285ad98-19b2-4e21-8744-93197607f643 | # OSCILLATIONS
# 13.4
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):
- (a) sin ωt – cos ωt
- (b) sin³ ωt
- (c) 3 cos (π/4 – 2ωt)
- (d) cos ωt + cos 3ωt + cos 5ωt
- (e) exp (–ω²t²)
- (f) 1 + ωt + ω²t²
# 13.5
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is:
- (a) at the end A,
- (b) at the end B,
- (c) at the mid-point of AB going towards A,
- (d) at 2 cm away from B going towards A,
- (e) at 3 cm away from A going towards B, and
- (f) at 4 cm away from B going towards A.
# 13.6
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
- (a) a = 0.7x
- (b) a = –200x²
- (c) a = –10x
- (d) a = 100x³
# 13.7
The motion of a particle executing simple harmonic motion is described by the displacement function,
x(t) = A cos (ωt + φ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
# 13.8
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
# 13.9
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
Fig. 13.19 | 16 | 11 | Physics | 206 |
01dab5d6-655d-4a18-b605-f6657e3fd4e0 | # PHYSICS
# 13.10
In Exercise 13.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
- (a) at the mean position,
- (b) at the maximum stretched position, and
- (c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
# 13.11
Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
# 13.12
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
- (a) x = –2 sin (3t + π/3)
- (b) x = cos (π/6 – t)
- (c) x = 3 sin (2πt + π/4)
- (d) x = 2 cos πt
# 13.13
Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F.
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
2024-25 | 17 | 11 | Physics | 206 |
53848fc9-3131-4c5b-aecc-be7a31dbdf18 | # OSCILLATIONS
# 13.14
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
# 13.15
The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s–2)
# 13.16
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
# 13.17
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρₗ. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T = 2π hρ
ρ₁g
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
# 13.18
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
2024-25 | 18 | 11 | Physics | 206 |
daa19eb6-5f9b-41e4-a16e-b78dabf24b29 | # CHAPTER SIX
# SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
# 6.1 INTRODUCTION
In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a point mass having no size.) We applied the results of our study even to the motion of bodies of finite size, assuming that motion of such bodies can be described in terms of the motion of a particle.
Any real body which we encounter in daily life has a finite size. In dealing with the motion of extended bodies (bodies of finite size) often the idealised model of a particle is inadequate. In this chapter we shall try to go beyond this inadequacy. We shall attempt to build an understanding of the motion of extended bodies. An extended body, in the first place, is a system of particles. We shall begin with the consideration of motion of the system as a whole. The centre of mass of a system of particles will be a key concept here. We shall discuss the motion of the centre of mass of a system of particles and usefulness of this concept in understanding the motion of extended bodies.
A large class of problems with extended bodies can be solved by considering them to be rigid bodies. Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change. It is evident from this definition of a rigid body that no real body is truly rigid, since real bodies deform under the influence of forces. But in many situations the deformations are negligible. In a number of situations involving bodies such as wheels, tops, steel beams, molecules and planets on the other hand, we can ignore that they warp (twist out of shape), bend or vibrate and treat them as rigid.
# 6.1.1 What kind of motion can a rigid body have?
Let us try to explore this question by taking some examples of the motion of rigid bodies. Let us begin with a rectangular | 0 | 11 | Physics | 106 |
454324ce-e9e3-4999-a5ee-03ed30a2f896 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
The most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on (Fig 6.3(a) and (b)).
Fig 6.1 Translational (sliding) motion of a block down an inclined plane.
Any point like P1 or P2 of the block moves with the same velocity at any instant of time. The block is taken as a rigid body. Its motion down the plane is such that all the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion (Fig. 6.1).
In pure translational motion at any instant of time, all particles of the body have the same velocity.
Consider now the rolling motion of a solid metallic or wooden cylinder down the same inclined plane (Fig. 6.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, seems to have translational motion. But as Fig. 6.2 shows, all its particles are not moving with the same velocity at any instant. The body, therefore, is not in pure translational motion. Its motion is translational plus ‘something else.’
Fig. 6.2 Rolling motion of a cylinder. It is not pure translational motion. Points P1, P2, P3, and P4 have different velocities (shown by arrows) at any instant of time.
In fact, the velocity of the point of contact P3 is zero at any instant, if the cylinder rolls without slipping.
In order to understand what this ‘something else’ is, let us take a rigid body so constrained that it cannot have translational motion. The 2024-25
Fig. 6.3 Rotation about a fixed axis (a) A ceiling fan (b) A potter’s wheel.
Let us try to understand what rotation is, what characterises rotation. You may notice that in rotation of a rigid body about a fixed axis, | 1 | 11 | Physics | 106 |
51a0e2ef-0e22-4b34-8cbc-f2c30854cba4 | # PHYSICS
Fig. 6.5 (a) A spinning top (The point of contact of the top with the ground, its tip O, is fixed.)
Fig. 6.4 A rigid body rotation about the z-axis from blades (Each point of the body such as P1 or P2 describes a circle with its centre (C1 or C2) on the axis of rotation. The radius of the circle (r1 or r2) is the perpendicular distance of the point (P1 or P2) from the axis. A point on the axis like P3 remains stationary).
Every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. Fig. 6.4 shows the rotational motion of a rigid body about a fixed axis (the z-axis of the frame of reference). Let P1 be a particle of the rigid body, arbitrarily chosen and at a distance r1 from fixed axis. The particle describes a circle of radius r1 with its centre C1 on the fixed axis. The circle lies in a plane perpendicular to the axis. The figure also shows another particle P2 of the rigid body, P2 is at a distance r2 from the fixed axis. The particle P2 moves in a circle of radius r2 and with centre C2 on the axis. This circle, too, lies in a plane perpendicular to the axis. Note that the circles described by P1 and P2 may lie in different planes; both these planes, however, are perpendicular to the fixed axis. For any particle on the axis like P3, r = 0. Any such particle remains stationary while the body rotates. This is expected since the axis of rotation is fixed.
This is another simple example of this kind of rotation is the oscillating table fan or a pedestal fan [Fig. 6.5(b)]. You may have observed that the axis of such a spinning top moves around the vertical through its point of contact with the ground, sweeping out a cone as shown in Fig. 6.5(a). (This movement of the axis of the top around the vertical is termed precession.) Note, the point of contact of the top with the ground is fixed. The axis of rotation of the top at any instant passes through the point of contact.
Fig. 6.5 (b) An oscillating table fan with rotating blades. The pivot of the fan, point O, is fixed. The blades of the fan are under rotational motion, whereas, the axis of rotation of the fan blades is oscillating. | 2 | 11 | Physics | 106 |
78d9c138-ffd8-4eee-b332-c0cbffdb9d14 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
axis of rotation of such a fan has an oscillating (sidewise) movement in a horizontal plane about the vertical through the point at which the axis is pivoted (point O in Fig. 6.5(b)). Thus, for us rotation will be about a fixed axis only unless stated otherwise.
The rolling motion of a cylinder down an inclined plane is a combination of rotation about a fixed axis and translation. Thus, the ‘something else’ in the case of rolling motion which we referred to earlier is rotational motion.
You will find Fig. 6.6(a) and (b) instructive from this point of view. Both these figures show motion of the same body along identical translational trajectory. In one case, Fig. 6.6(a), the motion is a pure translation; in the other case [Fig. 6.6(b)] it is a combination of translation and rotation. (You may try to reproduce the two types of motion shown, using a rigid object like a heavy book.)
We now recapitulate the most important observations of the present section: The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. The rotation may be about an axis that is fixed (e.g. a ceiling fan) or moving (e.g. an oscillating table fan [Fig.6.5(b)]).
# 6.2 CENTRE OF MASS
We shall first see what the centre of mass of a system of particles is and then discuss its significance. For simplicity we shall start with a two particle system. We shall take the line joining the two particles to be the x-axis.
Fig. 6.6(a) Motion of a rigid body which is pure translation.
Fig. 6.6(b) Motion of a rigid body which is a combination of translation and rotation.
Fig 6.6 (a) and 6.6 (b) illustrate different motions of the same body. Note P is an arbitrary point of the body; O is the centre of mass of the body, which is defined in the next section. Suffice to say here that the trajectories of O are the translational trajectories Tr₁ and Tr₂ of the body. The positions O and P at three different instants of time are shown by O₁, O₂, and O₃, and P₁, P₂ and P₃, respectively, in both Figs. 6.6 (a) and (b). As seen from Fig. 6.6(a), at any instant the velocities of any particles like O and P of the body are the same in pure translation. Notice, in this case the orientation of OP, i.e. the angle OP makes with a fixed direction, say the horizontal, remains the same, i.e. α₁ = α₂ = α₃. Fig. 6.6 (b) illustrates a case of combination of translation and rotation. In this case, at any instant the velocities of O and P differ. Also, α₁, α₂ and α₃ may all be different.
Let the distances of the two particles be x and x respectively from some origin O. Let m₁ and m₂ be respectively the masses of the two. | 3 | 11 | Physics | 106 |
d95e03af-30e0-45f9-8b53-6c9b156cfe50 | # PHYSICS
particles. The centre of mass of the system is that point C which is at a distance X from O, where X is given by
X = m x₁ + m x₂ / (m₁ + m₂) (6.1)
Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.
Results of Eqs. (6.3a) and (6.3b) are generalised easily to a system of n particles, not necessarily lying in a plane, but distributed in space. The centre of mass of such a system is at (X, Y, Z), where
X = ∑mixi / ∑mi (6.4a)
Y = ∑miyi / ∑mi (6.4b)
Z = ∑mizi / ∑mi (6.4c)
Here M = ∑mi is the total mass of the system. The index i runs from 1 to n; mi is the mass of the ith particle and the position of the ith particle is given by (xi, yi, zi).
Eqs. (6.4a), (6.4b) and (6.4c) can be combined into one equation using the notation of position vectors. Let ri be the position vector of the ith particle and R be the position vector of the centre of mass:
r = xi ɵ + yi j + zi k
and R = X ɵ + Y j + Z k
Then R = ∑miri / M (6.4d)
The sum on the right hand side is a vector sum.
Note the economy of expressions we achieve by use of vectors. If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then ∑miri = 0 for the given system of particles.
A rigid body, such as a metre stick or a flywheel, is a system of closely packed particles; Eqs. (6.4a), (6.4b), (6.4c) and (6.4d) are therefore, applicable to a rigid body. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in these equations. Since the spacing of the particles is | 4 | 11 | Physics | 106 |
5c99d45b-f481-4ea2-adf4-3caf6b788584 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
Let us consider a thin rod, whose width and breath (in case the cross section of the rod is rectangular) or radius (in case the cross section of the rod is cylindrical) is much smaller than its length. Taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element dm of the rod at x, there is an element of the same mass dm located at –x (Fig. 6.8).
The net contribution of every such pair to the integral and hence the integral itself is zero. From Eq. (6.6), the point for which the integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry.
The same symmetry argument will apply to homogeneous rings, discs, spheres, or even thick rods of circular or rectangular cross section. For all such bodies you will realise that for every element dm at a point (x, y, z) one can always take an element of the same mass at the point (–x, –y, –z). (In other words, the origin is a point of reflection symmetry for these bodies.) As a result, the integrals in Eq. (6.5 a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre.
If we choose the centre of mass as the origin of our coordinate system, i.e., ∫ r dm = 0 or ∫ x dm = ∫ y dm = ∫ z dm = 0 (6.6)
Often we have to calculate the centre of mass of homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres.
Example 6.1 Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long.
Answer
Fig. 6.8 Determining the CM of a thin rod. Fig. 6.9 | 5 | 11 | Physics | 106 |
7c1038dd-772d-4cba-bc2a-bd78c36ff2c8 | # PHYSICS
With the x –and y–axes chosen as shown in Fig. 6.9, the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0), (0.5,0), (0.25,0.25√3). Let the masses 100 g, 150 g and 200 g be located at O, A and B respectively. Then,
m₁x₁ + m₂x₂ + m₃x₃
X = ---------------
m₁ + m₂ + m₃
100(0) + 150(0.5) + 200(0.25√3) g m
= --------------------- g
100 + 150 + 200
= 75 + 50√3 m = 125√3 m = 5√3 m
100(0) + 150(0) + 200(0.25√3) g m
Y = ---------------------
450 g
= 50√3/450 m = 9√3 m = 3√3 m
The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB. Why?
# Example 6.3
Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg.
Answer: Choosing the X and Y axes as shown in Fig. 6.11 we have the coordinates of the vertices of the L-shaped lamina as given in the figure. We can think of the L-shape to consist of 3 squares each of length 1 m. The mass of each square is 1 kg, since the lamina is uniform. The centres of mass C₁, C₂ and C₃ of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/2,1/2), (1/2,3/2) respectively. We take the masses of the squares to be concentrated at these points. The centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points.
The centre of mass of the L-shape lies on the line OD. We could have guessed this without calculations. Can you tell why? Suppose, the three squares that make up the L shaped lamina...
# Example 6.2
Find the centre of mass of a triangular lamina.
Answer: The lamina (ΔLMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 6.10.
Hence
X = (1(1/2) + 1(3/2) + ...)
= (1 + 1 + ...) (1/2) kg m = 5 m
1 kg
Y = (1(1/2) + 1(1/2) + 1(3/2)) kg m
= (1 + 1 + ...) = m
1 kg
The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of... | 6 | 11 | Physics | 106 |
3990045a-68fc-45a7-b973-a575b8334b9f | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
of Fig. 6.11 had different masses. How will you then determine the centre of mass of the lamina? Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
# 6.3 MOTION OF CENTRE OF MASS
Equipped with the definition of the centre of mass, we are now in a position to discuss its physical importance for a system of n particles. We may rewrite Eq.(6.4d) as
MR = ∑m r = m r1 + m r2 + ... + m rn (6.7)
Differentiating the two sides of the equation with respect to time we get
M dR = m1 dr1 + m2 dr2 + ... + mn drn dt
or
M V = m1v1 + m2v2 + ... + mnvn (6.8)
where v1 (= dr1/dt) is the velocity of the first particle, v2 (= dr2/dt) is the velocity of the second particle etc. and V = dR/dt is the velocity of the centre of mass. Note that we assumed the masses m1, m2, ... etc. do not change in time. We have therefore, treated them as constants in differentiating the equations with respect to time.
Differentiating Eq.(6.8) with respect to time, we obtain
dV/dt = m1 dv1 + m2 dv2 + ... + mn dvn
or
MA = m1a1 + m2a2 + ... + mnan (6.9)
where a1 (= dv1/dt) is the acceleration of the first particle, a2 (= dv2/dt) is the acceleration of the second particle etc. and A (= dV/dt) is the acceleration of the centre of mass of the system of particles.
Now, from Newton’s second law, the force acting on the first particle is given by F1 = m1a1. The force acting on the second particle is given by F2 = m2a2 and so on. Eq. (6.9) may be written as
MA = F1 + F2 + ... + Fn (6.10)
2024-25 | 7 | 11 | Physics | 106 |
c2524d10-cdd7-42c6-962c-01593dc17e10 | # PHYSICS
Problems without explicitly outlining and justifying the procedure. We now realise that in earlier studies we assumed, without saying so, that rotational motion and/or internal motion of the particles were either absent or negligible. We no longer need to do this. We have not only found the justification of the procedure we followed earlier; but we also have found how to describe and separate the translational motion of (1) a rigid body which may be rotating as well, or (2) a system of particles with all kinds of internal motion.
Let us consider a system of n particles with masses m₁, m₂,..., mₙ respectively and velocities v₁, v₂,.......vₙ respectively. The particles may be interacting and have external forces acting on them. The linear momentum of the first particle is p₁ = m₁v₁, of the second particle is p₂ = m₂v₂ and so on. For the system of n particles, the linear momentum of the system is defined to be the vector sum of all individual particles of the system:
P = p₁ + p₂ +... + pₙ = m₁v₁ + m₂v₂ +... + mₙvₙ (6.14)
Comparing this with Eq. (6.8):
P = MV (6.15)
Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (6.15) with respect to time:
dP = M dV = MA (6.16)
Comparing Eq. (6.16) and Eq. (6.11):
dP = Fext (6.17)
This is the statement of Newton’s second law of motion extended to a system of particles. A projectile, following the usual parabolic trajectory, explodes into fragments midway in air. The forces leading to the explosion are internal forces. They contribute nothing to the motion of the centre of mass. The total external force, namely, the force of gravity acting on the body, is the same before and after the explosion. The centre of mass under the influence of the external force continues, therefore, along the same parabolic trajectory as it would have followed if there were no explosion.
Suppose now, that the sum of external forces acting on a system of particles is zero. Then from Eq. (6.17):
dP/dt = 0 or P = Constant (6.18a)
Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum of a system of particles. Because of Eq. (6.15), this also means that when the total external force on the system is zero, the velocity of the centre of mass remains constant. (We assume throughout the discussion on systems of particles in this chapter that the total mass of the system remains constant.)
Note that on account of the internal forces, i.e. the forces exerted by the particles on one another, the individual particles may have. | 8 | 11 | Physics | 106 |
76f8e757-45cf-4fa6-8d19-f65c92df15a9 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle.
The vector Eq. (6.18a) is equivalent to three scalar equations,
Pₓ = c₁, Py = c₂ and Pz = c₃ (6.18 b)
Here Pₓ, Py and Pz are the components of the total linear momentum vector P along the x–, y– and z–axes respectively; c₁, c₂ and c₃ are constants.
move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b).
In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference.
In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star moves like a free particle, as shown in Fig.6.14 (a). The trajectories of the two stars of equal mass are also shown in the figure; they look complicated. If we go to the centre of mass frame, then we find that there the two stars are moving in a circle, about the centre of mass, which is at rest. Note that the position of the stars have to be diametrically opposite to each other [Fig. 6.14(b)]. Thus in our frame of reference, the trajectories of the stars are a combination of (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.
As can be seen from the two examples, separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.
# 6.5 VECTOR PRODUCT OF TWO VECTORS
We are already familiar with vectors and their use in physics. In chapter 5 (Work, Energy, Power) we defined the scalar product of two vectors. An important physical quantity, work, is defined as a scalar product of two vector quantities, force and displacement. | 9 | 11 | Physics | 106 |
d206d672-7e57-4f5b-8088-80eb42d17c39 | # PHYSICS
We shall now define another product of two vectors. This product is a vector. Two important quantities in the study of rotational motion, namely, moment of a force and angular momentum, are defined as vector products.
# Definition of Vector Product
A vector product of two vectors a and b is a vector c such that:
1. magnitude of c = c = ab sinθ where a and b are magnitudes of a and b and θ is the angle between the two vectors.
2. c is perpendicular to the plane containing a and b.
3. If we take a right handed screw with its head lying in the plane of a and b and the screw perpendicular to this plane, and if we turn the head in the direction from a to b, then the tip of the screw advances in the direction of c.
This right handed screw rule is illustrated in Fig. 6.15a. Alternately, if one curls up the fingers of the right hand around a line perpendicular to the plane of the vectors a and b and if the fingers are curled up in the direction from a to b, then the stretched thumb points in the direction of c, as shown in Fig. 6.15b.
It should be remembered that there are two angles between any two vectors a and b. In Fig. 6.15 (a) or (b) they correspond to θ and (360°– θ). While applying either of the above rules, the rotation should be taken through the smaller angle (<180°) between a and b.
Because of the cross (×) used to denote the vector product, it is also referred to as cross product.
Note that scalar product of two vectors is commutative as said earlier, a.b = b.a. The vector product, however, is not commutative, i.e. a × b ≠ b × a. The magnitude of both a × b and b × a is the same (ab sin θ); also, both of them are perpendicular to the plane of a and b. But the rotation of the right-handed screw in case of a × b is from a to b, whereas in case of b × a it is from b to a. This means the two vectors are in opposite directions. We have a × b = −b × a.
Another interesting property of a vector product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have:
x → −x, y → −y, and z → −z. As a result, all the components of a vector change sign and thus a → −a, b → −b. What happens to a × b under reflection?
(a) → −(a) × −(b) = a × b. Thus, a × b does not change sign under reflection.
Both scalar and vector products are distributive with respect to vector addition. Thus:
a.(b + c) = ab + ac
a × (b + c) = a × b + a × c
# Elementary Cross Products
We may write c = a × b in the component form. For this we first need to obtain some elementary cross products:
- a × a = 0 (0 is a null vector, i.e. a vector with zero magnitude). This follows since magnitude of a × a is a² sin 0° = 0.
# Figures
(a) Rule of the right handed screw for defining the direction of the vector product.
(b) Rule of the right hand for defining the direction of the vector product. | 10 | 11 | Physics | 106 |
53ff16bc-b8ea-444d-9c2a-131080d26680 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
From this follow the results
(i) i × i = 0, j × j = 0, k × k = 0
(ii) i × j = k
Note that the magnitude of i × j is sin90 or 1, since i and j both have unit magnitude and the angle between them is 90⁰.
Thus, i × j is a unit vector perpendicular to the plane of i and j and related to them by the right hand screw rule is k. Hence, the above result. You may verify similarly, j × k = i and k × i = j.
From the rule for commutation of the cross product, it follows:
j × i = −k, k × j = −i, i × k = −j
Note if i, j, k occur cyclically in the above vector product relation, the vector product is positive. If i, j, k do not occur in cyclic order, the vector product is negative.
Now,
a × b = (aₓ i + ay j + azk) × (bₓ i + by j + bzk)
= aₓ bₓ ˆk − aₓ bᵧ j − aᵧ bₓ k + aᵧ bₓ i + aᵧ bᵧ j − aᵧ bₓ i
=(aᵧ bᵧ − aᵧ bₓ)i + (aₓ bᵧ − aᵧ bₓ)j + (aₓ bₓ − aᵧ bᵧ)k
We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember.
a × b =
|i|j|k|
|---|---|---|
|aₓ|ay|az|
|bₓ|by|bz|
Example 6.4 Find the scalar and vector products of two vectors. a = 3i − 4j + 5k and b = −6 − 4 − 15i − 2i + j − 3k
Answer
a · b = (3i − 4j + 5k) (−6 − 4 − 15i − 2i + j − 3k) = −25
Fig. 6.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed (z-) axis moves in a circle with centre (C) on the axis.)
which lies in a plane perpendicular to the axis and has its centre on the axis. In Fig. 6.16 we redraw Fig. 6.4, showing a typical particle (at a point P) of the rigid body rotating about a fixed axis (taken as the z-axis). The particle describes | 11 | 11 | Physics | 106 |
b71dea26-e680-46e0-a2ae-21692e24751a | # PHYSICS
A circle with a centre C on the axis. The radius of the circle is r, the perpendicular distance of the point P from the axis. We also show the linear velocity vector v of the particle at P. It is along the tangent at P to the circle. (See Fig. 6.17a).
Let P′ be the position of the particle after an interval of time ∆t (Fig. 6.16). The angle PCP′ describes the angular displacement ∆θ of the particle in time ∆t. The average angular velocity of the particle over the interval ∆t is ∆θ/∆t. As ∆t tends to zero (i.e. takes smaller and smaller values), the ratio ∆θ/∆t approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocity by ω (the Greek letter omega). We know from our study of circular motion that the magnitude of linear velocity v of a particle moving in a circle is related to the angular velocity of the particle ω by the simple relation v = ωr, where r is the radius of the circle.
We observe that at any given instant the relation v = ωr applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by vi = ωri (6.19). The index i runs from 1 to n, where n is the total number of particles of the body.
For particles on the axis, r = 0, and hence v = ωr = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed.
Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body.
We have characterised pure translation of a body by all parts of the body having the same velocity at any instant of time. Similarly, we may characterise pure rotation by all parts of the body having the same angular velocity at any instant of time. Note that this characterisation of the rotation of a rigid body about a fixed axis is just another way of saying as in Sec. 6.1 that each particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has the centre on the axis.
In our discussion so far the angular velocity appears to be a scalar. In fact, it is a vector. We shall not justify this fact, but we shall accept it. For rotation about a fixed axis, the angular velocity vector lies along the axis of rotation, be on the axis of rotation. | 12 | 11 | Physics | 106 |
a610f4cb-7d82-474e-9d1e-03fad65ea4e3 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
Now ω
If the axis of rotation is fixed, the direction of ω and hence, that of α is fixed. In this case
But ω
Hence ω is along OC the vector equation reduces to a scalar equation
ω × r = ω × CP
α = dω / dt (6.22)
The vector ω CP is perpendicular to ω, i.e. to the z-axis and also to CP, the radius of the circle described by the particle at P. It is therefore, along the tangent to the circle at P. Also, the magnitude of ω × CP is ω (CP) since ω and CP are perpendicular to each other. We shall denote CP by r⊥ and not by r, as we did earlier.
Thus, ω × r is a vector of magnitude r⊥ and is along the tangent to the circle described by the particle at P. The linear velocity vector v at P has the same magnitude and direction. Thus,
v = ω × r (6.20)
In fact, the relation, Eq. (6.20), holds good even for rotation of a rigid body with one point fixed, such as the rotation of the top [Fig. 6.6(a)]. In this case r represents the position vector of the particle with respect to the fixed point taken as the origin.
We note that for rotation about a fixed axis, the direction of the vector ω change with time. Its magnitude may, however, change from instant to instant. For the more general rotation, both the magnitude and the direction of ω may change from instant to instant.
# 6.7 TORQUE AND ANGULAR MOMENTUM
In this section, we shall acquaint ourselves with two physical quantities (torque and angular momentum) which are defined as vector products of two vectors. These as we shall see, are especially important in the discussion of motion of systems of particles, particularly rigid bodies.
# 6.7.1 Moment of force (Torque)
We have learnt that the motion of a rigid body, in general, is a combination of rotation and translation. If the body is fixed at a point or along a line, it has only rotational motion. We know that force is needed to change the translational state of a body, i.e. to produce linear acceleration. We may then ask, what is the analogue of force in the case of rotational motion? To look into the question in a concrete situation let us take the example of opening or closing of a door. A door is a rigid body which can rotate about a fixed vertical axis passing through the hinges. What makes the door rotate? It is clear that unless a force is applied the door does not rotate. But any force does not do the job. A force applied to the hinge line cannot produce any rotation at all, whereas a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing rotation. It is not the force alone, but how and where the force is applied is important in rotational motion.
The rotational analogue of force in linear motion is moment of force. It is also referred to as torque or couple. (We shall use the words moment of force and torque interchangeably.) We shall first define the moment of force for the special case of a single particle. Later on we shall extend the concept to systems of particles including rigid bodies. We shall also relate it to a change in the state of rotational motion, i.e. is angular acceleration of a rigid body.
6.6.1 Angular acceleration
You may have noticed that we are developing the study of rotational motion along the lines of the study of translational motion with which we are already familiar. Analogous to the kinetic variables of linear displacement (s) and velocity (v) in translational motion, we have angular displacement (θ) and angular velocity (ω) in rotational motion. It is then natural to define in rotational motion the concept of angular acceleration in analogy with linear acceleration defined as the time rate of change of velocity in translational motion. We define angular acceleration α as the time rate of change of angular velocity. Thus,
α = dω / dt (6.21) | 13 | 11 | Physics | 106 |
032aa350-1326-4e0b-b193-fb0f3cfd5385 | # PHYSICS
of the line of action of F from the origin and F⊥(= F sinθ ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = 0⁰ or 180⁰. Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.
One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same.
# 6.7.2 Angular momentum of a particle
Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue of linear momentum. We shall first define angular momentum for the special case of a single particle and look at its usefulness in the context of single particle motion. We shall then extend the definition of angular momentum to systems of particles including rigid bodies.
Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. From this term one could guess how angular momentum is defined.
Consider a particle of mass m and linear momentum p at a position r relative to the origin O. The angular momentum l of the particle with respect to the origin O is defined to be
l = r × p (6.25a)
The magnitude of the angular momentum vector is
l = r p sinθ (6.26a)
where p is the magnitude of p and θ is the angle between r and p. We may write
l = r p or r p (6.26b)
Moment of force has dimensions M L² T-2. Its dimensions are the same as those of work or energy. It is, however, a very different physical quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is newton metre (N m). The magnitude of the moment of force may be written
τ = (r sinθ)F = r F (6.24b)
or τ = r F sinθ = rF⊥ (6.24c)
where r⊥ (= r sinθ) is the perpendicular distance of the directional line of p from the origin and p⊥ (= p sinθ) is the component of p in a direction perpendicular to r. We expect the angular momentum to be zero (l = 0), if the linear momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p passes through the origin θ = 0⁰ or 180⁰. | 14 | 11 | Physics | 106 |
4bf77316-2782-484b-a99d-417c6577f64e | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate l = r × p with respect to time,
dl d
dt = dt (r × p)
Applying the product rule for differentiation to the right hand side,
d dr dp
dt (r × p) = dt × p + r × dt
Now, the velocity of the particle is v = dr/dt and p = mv
Because of this × p = v × mv = 0,
dt
as the vector product of two parallel vectors vanishes. Further, since dp / dt = F,
r × dp = r × F = τ
dt t
Hence d (r × p) = τ
dt
Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation F = dp/dt, which expresses Newton’s second law for the translational motion of a single particle.
# Torque and angular momentum for a system of particles
To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles,
The angular momentum of the iᵗʰ particle is given by
li = ri × pi
where ri is the position vector of the iᵗʰ particle with respect to a given origin and p = (mivi) is the linear momentum of the particle. (The particle has mass mi and velocity vi) We may write the total angular momentum of a system of particles as
This is a generalisation of the definition of angular momentum (Eq. 6.25a) for a single particle to a system of particles.
Using Eqs. (6.23) and (6.25b), we get | 15 | 11 | Physics | 106 |
40ce97bf-b335-400e-9fda-b3d006a5f1b0 | # PHYSICS
dL = d(l) = ∑ dl = ∑ τ
Note that like Eq.(6.17), Eq.(6.28b) holds good for any system of particles, whether it is a rigid body or its individual particles have all kinds of internal motion.
# Conservation of angular momentum
If τext = 0, Eq. (6.28b) reduces to
dL/dt = 0
or L = constant. (6.29a)
Thus, if the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved, i.e. remains constant. Eq. (6.29a) is equivalent to three scalar equations,
L = K1, L = K2 and L = K3 (6.29b)
Here K1, K2 and K3 are constants; Lx, Ly and Lz are the components of the total angular momentum vector L along the x, y and z axes respectively. The statement that the total angular momentum is conserved means that each of these three components is conserved.
Eq. (6.29a) is the rotational analogue of Eq. (6.18a), i.e. the conservation law of the total linear momentum for a system of particles. Like Eq. (6.18a), it has applications in many practical situations. We shall look at a few of the interesting applications later on in this chapter.
# Example 6.5
Find the torque of a force + – about the origin. The force acts on a particle whose position vector is r = i - j + k and F = 7i + 3j - 5k.
We shall use the determinant rule to find the torque τ = r × F.
# Example 6.6
Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion. | 16 | 11 | Physics | 106 |
2412a1ab-56e8-45a2-b331-10c648d863ae | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
Answer Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.
(1) The total force, i.e. the vector sum of the forces, on the rigid body is zero;
F1 + F2 + ... + Fn = ∑ F = 0
If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body.
(2) The total torque, i.e. the vector sum of the torques on the rigid body is zero,
τ1 + τ2 + ... + τn = ∑ τ = 0
The angular momentum is l = r × mv. Its magnitude is mvr sinθ, where θ is the angle between r and v as shown in Fig. 6.19. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ is a constant.
Further, the direction of l is perpendicular to the plane of r and v. It is into the page of the figure. This direction does not change with time.
Thus, l remains the same in magnitude and direction and is therefore conserved. Is there any external torque on the particle?
One may raise a question, whether the rotational equilibrium condition [Eq. 6.30(b)] remains valid, if the origin with respect to which the torques are taken is shifted. One can show that if the translational equilibrium condition [Eq. 6.30(a)] holds for a rigid body, then such a shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the location of the origin about which the torques are taken. Example 6.7 gives a proof of this result in a special case of a couple, i.e. two forces acting on a rigid body in translational equilibrium.
We shall recapitulate what effect the external forces have on a rigid body. (Henceforth we shall omit the adjective ‘external’ because unless stated otherwise, we shall deal with only external forces and torques.) The forces change the translational state of the motion of the rigid body, i.e. they change its total linear momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The total torque on the body may not vanish. Such a torque changes the rotational state of motion of the rigid body, i.e. it changes the total angular momentum of the body in accordance with Eq. (6.28 b).
A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear nor angular acceleration.
Eq. (6.30a) and Eq. (6.30b), both, are vector equations. They are equivalent to three scalar equations each. Eq. (6.30a) corresponds to
∑ Fix = 0
∑ Fiy = 0
∑ Fiz = 0
where Fix, Fiy and Fiz are respectively the x, y and z components of the forces Fi. Similarly, Eq. (6.30b) is equivalent to three scalar equations
∑ τix = 0
∑ τiy = 0
∑ τiz = 0
where τix, τiy and τiz are respectively the x, y and z components of the torque τi.
Eq. (6.31a) and (6.31b) give six independent conditions to be satisfied for mechanical equilibrium. | 17 | 11 | Physics | 106 |
993b5c0a-7e6a-4f9a-a1e5-0946f79818aa | # PHYSICS
Equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis perpendicular to the plane of the forces must be zero.
The conditions of equilibrium of a rigid body may be compared with those for a particle, which we considered in earlier chapters. Since consideration of rotational motion does not apply to a particle, only the conditions for translational equilibrium (Eq. 6.30 a) apply to a particle. Thus, for equilibrium of a particle the vector sum of all the forces on it must be zero. Since all these forces act on the single particle, they must be concurrent. Equilibrium under concurrent forces was discussed in the earlier chapters.
A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium. A pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque. A couple produces rotation without translation.
When we open the lid of a bottle by turning it, our fingers are applying a couple to the lid . Another known example is a compass needle in the earth’s magnetic field as shown in the . The earth’s magnetic field exerts equal forces on the north and south poles. The force on the North Pole is towards the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field.
Let C be the midpoint of AB, CA = CB = a. The moment of the forces at A and B will both be equal in magnitude (aF), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational equilibrium; ∑F ≠ 0.
Our fingers apply a couple to turn the lid. | 18 | 11 | Physics | 106 |
8b16c02c-5d9e-4439-98db-9ae94e0b1a78 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
length. This point is called the fulcrum. A see-saw on the children’s playground is a typical example of a lever. Two forces F₁ and F₂, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances d₁ and d₂ respectively from the fulcrum as shown in Fig. 6.23.
Fig. 6.23
The Earth’s magnetic field exerts equal and opposite forces on the poles of a compass needle. These two forces form a couple.
# Example 6.7
Show that moment of a couple does not depend on the point about which you take the moments.
# Answer
R is the reaction of the support at the fulcrum; R is directed opposite to the forces F₁ and F₂. For translational equilibrium,
R – F₁ – F₂ = 0 (i)
For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero,
d₁F₁– d₂F₂ = 0 (ii)
Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum.
Fig. 6.22
Consider a couple as shown in Fig. 6.22 acting on a rigid body. The forces F and -F act respectively at points B and A. These points have position vectors r₁ and r₂ with respect to origin O. Let us take the moments of the forces about the origin.
The moment of the couple = sum of the moments of the two forces making the couple
= r₁ × (–F) + r₂ × F
= r₂ × F – r₁ × F
= (r₂–r₁) × F
But r₁ + AB = r₂, and hence AB = r₂ – r₁. The moment of the couple, therefore, is AB × F.
Clearly this is independent of the origin, the point about which we took the moments of the forces.
The moment of the couple is
M.A. = F₁/F₂ = d₂/d₁ (6.32b)
Clearly this is independent of the origin, the point about which we took the moments of the forces.
If the effort arm d₂ is larger than the load arm, the mechanical advantage is greater than one. Mechanical advantage greater than one means that a small effort can be used to lift a large load. There are several examples of a lever around you besides the see-saw. The beam of a balance is a lever. Try to find more such.
2024-25 | 19 | 11 | Physics | 106 |
92f2b68f-f174-4116-b497-f932d594be12 | # PHYSICS
examples and identify the fulcrum, the effort and effort arm, and the load and the load arm of the lever in each case. You may easily show that the principle of moment holds even when the parallel forces F₁ and F₂ are not perpendicular, but act at some angle, to the lever.
If ri is the position vector of the ith particle of an extended body with respect to its CG, then the torque about the CG, due to the force of gravity on the particle is τi = ri × mig. The total gravitational torque about the CG is zero, i.e.
τg = ∑ τi = 0
Many of you may have the experience of balancing your notebook on the tip of a finger. Figure 6.24 illustrates a similar experiment that you can easily perform. Take an irregular-shaped cardboard having mass M and a narrow tipped object like a pencil. You can locate by trial and error a point G on the cardboard where it can be balanced on the tip of the pencil. (The cardboard remains horizontal in this position.) This point of balance is the centre of gravity (CG) of the cardboard. The tip of the pencil provides a vertically upward force due to which the cardboard is in mechanical equilibrium. As shown in the Fig. 6.24, the reaction of the tip is equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the cardboard due to the forces of gravity like m1g, m2g …. etc, acting on the individual particles that make up the cardboard.
Fig. 6.24 Balancing a cardboard on the tip of a pencil. The point of support, G, is the centre of gravity.
Fig. 6.25 Determining the centre of gravity of a body of irregular shape. The centre of gravity G lies on the vertical AA₁ through the point of suspension of the body A.
2024-25 | 20 | 11 | Physics | 106 |
0a31b8dc-12b2-41d5-9145-1a74b3d78b53 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
free space. We note that this is true because = 30 cm, PG = 5 cm, AK₁= BK₂= 10 cm and K₁G = the body being small, g does not vary from one point of the body to the other. If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Basically, the two are different concepts. The centre of mass has nothing to do with gravity. It depends only on the distribution of mass of the body.
In Sec. 6.2 we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.
Figure 6.25 illustrates another way of determining the CG of an irregular shaped body like a cardboard. If you suspend the body from some point like A, the vertical line through A passes through the CG. We mark the vertical AA₁. We then suspend the body through other points like B and C. The intersection of the verticals gives the CG. Explain why the method works. Since the body is small enough, the method allows us to determine also its centre of mass.
For translational equilibrium of the rod,
R₁ + R₂ – W₁ – W = 0 (i)
Note W₁ and W act vertically down and R₁ and R₂ act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of R₂ and W₁ are anticlockwise (+ve), whereas the moment of R₁ is clockwise (-ve).
For rotational equilibrium,
–R₁ (K₁G) + W₁ (PG) + R₂ (K₂G) = 0 (ii)
It is given that W = 4.00g N and W₁ = 6.00g N, where g = acceleration due to gravity. We take g = 9.8 m/s².
With numerical values inserted, from (i) R₁ + R₂ – 4.00g – 6.00g = 0 or R₁ + R₂ = 10.00g N (iii) = 98.00 N
From (ii), – 0.25 R₁ + 0.05 W₁ + 0.25 R₂ = 0 or R₁ – R₂ = 1.2g N = 11.76 N (iv)
Example 6.8 A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)
Answer
Figure 6.26 shows the rod AB, the positions of the knife edges K₁ and K₂, the centre of gravity of the rod at G and the suspended load at P.
Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP
Example 6.9 A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.6.27. Find the reaction forces of the wall and the floor.
Answer
2024-25 | 21 | 11 | Physics | 106 |
d2e7230f-b0f7-4ac3-9fe5-b0d53b2db71b | # PHYSICS
The ladder AB is 3 m long, its foot A is at distance AC = 1 m from the wall. From Pythagoras theorem, BC = 2√2 m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces F₁ and F₂ of the wall and the floor respectively. Force F₁ is perpendicular to the wall, since the wall is frictionless. Force F₂ is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction,
N – W = 0 (i)
Taking the forces in the horizontal direction,
F – F₁ = 0 (ii)
For rotational equilibrium, taking the moments of the forces about A,
F₁ − (1/2) W = 0 (iii)
Now W = 20 g = 20 × 9.8 N = 196.0 N
From (i) N = 196.0 N
From (iii) F₁ = W/4√2 = 196.0 / 4√2 = 34.6 N
From (ii) F = F₁ = 34.6 N
F₂ = F₁² + N² = 199.0 N
The force F₂ makes an angle α with the horizontal,
tanα = N/F = 4√2, α = tan⁻¹(4√2) ≈ 80°
# 6.9 MOMENT OF INERTIA
We have already mentioned that we are developing the study of rotational motion parallel to the study of translational motion with which we are familiar. We have yet to answer one major question in this connection. What is the analogue of mass in rotational motion? We shall attempt to answer this question in the present section. To keep the discussion simple, we shall consider rotation about a fixed axis only. Let us try to get an expression for the kinetic energy of a rotating body. We know that for a body rotating about a fixed axis, each particle of the body moves in a circle with linear velocity given by Eq. (6.19). (Refer to Fig. 6.16). For a particle at a distance r from the axis, the linear velocity is υi = riω. The kinetic energy of motion of this particle is k = 1/2 m υ² = 1/2 m r²ω² where m is the mass of the particle. The total kinetic energy K of the body is then given by the sum of the kinetic energies of individual particles,
K = ∑ki = ∑(1/2 m r²ω²) (6.34)
Here n is the number of particles in the body. Note ω is the same for all particles. Hence, taking ω out of the sum,
K = (1/2) ω²(∑m r²)
We define a new parameter characterising the rigid body, called the moment of inertia I, given by
I = ∑m r² (6.34)
With this definition, K = (1/2) Iω² (6.35)
Note that the parameter I is independent of the magnitude of the angular velocity. It is a characteristic of the rigid body and the axis about which it rotates.
Compare Eq. (6.35) for the kinetic energy of a rotating body with the expression for the kinetic energy of a body in linear (translational) motion,
K = (1/2) m υ²
Here, m is the mass of the body and v is its velocity. We have already noted the analogy between angular velocity ω (in respect of rotational motion about a fixed axis) and linear velocity v (in respect of linear motion). It is then evident that the parameter, moment of inertia I, is the desired rotational analogue of mass in linear motion. In rotation (about a fixed axis), the moment of inertia plays a similar role as mass does in linear motion.
We now apply the definition Eq. (6.34), to calculate the moment of inertia in two simple cases. | 22 | 11 | Physics | 106 |
43783da9-34ea-49ad-8b78-da2d2447c1fb | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
(a) Consider a thin ring of radius R and mass M, rotating in its own plane around its centre with angular velocity ω. Each mass element of the ring is at a distance R from the axis, and moves with a speed Rω. The kinetic energy is therefore,
K = 1 Mυ² = 1 MR²ω²
Comparing with Eq. (6.35) we get I = MR² for the ring.
As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body.
Notice from the Table 6.1 that in all cases, we can write I = Mk², where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, k² = L²/12, i.e. k = L/12. Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of gyration. The radius of gyration of a body about an axis may be defined as the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis.
(b) Next, take a rigid rod of negligible mass of length l with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod (Fig. 6.28). Each mass M/2 is at a distance l/2 from the axis. The moment of inertia of the masses is therefore given by
I = (M/2)(l/2)² + (M/2)(l/2)²
Thus, for the pair of masses, rotating about the axis through the centre of mass perpendicular to the rod
I = Ml² / 4
Table 6.1 simply gives the moment of inertia of various familiar regular shaped bodies about specific axes. (The derivations of these expressions are beyond the scope of this textbook and you will study them in higher classes.)
As the mass of a body resists a change in its state of linear motion, it is a measure of its inertia in linear motion. Similarly, as the moment of inertia about a given axis of rotation resists a change in its rotational motion, it can be regarded as a measure of rotational inertia of the body; it is a measure of the way in which different parts of the body are distributed at different distances from the axis. Unlike the mass of a body, the moment of inertia is not a fixed quantity but depends on distribution of mass about the axis of rotation, and the orientation and position of the axis of rotation with respect to the body as a whole.
The property of this extremely important quantity I, as a measure of rotational inertia of the body, has been put to a great practical use. The machines, such as steam engine and the automobile engine, etc., that produce rotational motion have a disc with a large moment of inertia, called a flywheel. Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions, thereby ensuring a smooth ride for the passengers on the vehicle. | 23 | 11 | Physics | 106 |
1392d6df-4a6c-4043-8f70-1e41faec62da | # PHYSICS
# Table 6.1 Moments of inertia of some regular shaped bodies about specific axes
|Z|Body|Axis|Figure|I|
|---|---|---|---|---|
|(1)|Thin circular ring, radius R|Perpendicular to plane, at centre| |M R²|
|(2)|Thin circular ring, radius R|Diameter| |M R²/2|
|(3)|Thin rod, length L|Perpendicular to rod, at mid point| |M L²/12|
|(4)|Circular disc, radius R|Perpendicular to disc at centre| |M R²/2|
|(5)|Circular disc, radius R|Diameter| |M R²/4|
|(6)|Hollow cylinder, radius R|Axis of cylinder| |M R²|
|(7)|Solid cylinder, radius R|Axis of cylinder| |M R²/2|
|(8)|Solid sphere, radius R|Diameter| |2 M R²/5|
# 6.10 KINEMATICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS
We have already indicated the analogy between rotational motion and translational motion. For example, the angular velocity ω plays the same role in rotation as the linear velocity v in translation. We wish to take this analogy further. In doing so we shall restrict the discussion only to rotation about fixed axis. This case of motion involves only one degree of freedom, i.e., needs only one independent variable to describe the motion. | 24 | 11 | Physics | 106 |
e8a9a211-d134-47aa-91b3-1218c0898035 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
This section is limited only to kinematics. We shall turn to dynamics in later sections.
We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.6.29) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 6.29 also shows θ₀, the angular displacement at t = 0.
We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a vector. Further, the angular acceleration, α = dω/dt.
# Example 6.10
Obtain Eq. (6.36) from first principles.
Answer: The angular acceleration is uniform, hence
dω = α = constant
dt (i)
Integrating this equation,
ω = ∫α dt + c = α t + c (as α is constant)
At t = 0, ω = ω₀ (given) ω₀ = 0
From (i) we get at t = 0, c = 0. Thus, ω = αt + ω₀ as required.
Where x₀ = initial displacement and v₀ = initial velocity. The word ‘initial’ refers to values of the quantities at t = 0.
The corresponding kinematic equations for rotational motion with uniform angular acceleration are:
ω = ω₀ + αt (6.36)
θ = θ₀ + ω₀t + 2 αt² (6.37)
and ω² = ω₀² + 2α(θ - θ₀) (6.38)
Where θ₀ = initial angular displacement of the rotating body, and ω₀ = initial angular velocity of the body.
# Example 6.11
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Answer:
(i) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s | 25 | 11 | Physics | 106 |
0cc73676-2220-4964-b763-f4ea8372135a | # PHYSICS
= 2π × angular speed in rev/s It is, however, necessary that these correspondences are established on sound dynamical considerations. This is what we now turn to.
= 2π × angular speed in rev/min = 60 s/min
= 2π × 1200 = 60 rad/s
= 40π rad/s Similarly ω = final angular speed in rad/s
= 2π × 3120 rad/s = 60
= 2π × 52 rad/s = 104 π rad/s
∴ Angular acceleration α = ω − ω⁰ = 4 π rad/s²
The angular acceleration of the engine = 4π rad/s²
(ii) The angular displacement in time t is given by θ = ω t + 1 α t²
0 2 = (40π × 16 + 1 × 4π 2 × 16) rad
= (640π + 512π) rad = 1152π rad
Number of revolutions = 1152π = 576 ⊳ Work done by a torque
# 6.11 DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS
Table 6.2 lists quantities associated with linear motion and their analogues in rotational motion. We have already compared kinematics of the two motions. Also, we know that in rotational motion moment of inertia and torque play the same role as mass and force respectively in linear motion. Given this we should be able to guess what the other analogues indicated in the table are.
For example, we know that in linear motion, work done is given by F dx, in rotational motion about a fixed axis it should be τ dθ, since we already know the correspondence dx → dθ and F → τ.
2024-25 | 26 | 11 | Physics | 106 |
bca32670-f01a-4c97-a3c9-49fafeb7d995 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
# Table 6.2 Comparison of Translational and Rotational Motion
|Linear Motion|Rotational Motion about a Fixed Axis|
|---|---|
|1 Displacement x|Angular displacement θ|
|2 Velocity v = dx/dt|Angular velocity ω = dθ/dt|
|3 Acceleration a = dv/dt|Angular acceleration α = dω/dt|
|4 Mass M|Moment of inertia I|
|5 Force F = Ma|Torque τ = I α|
|6 Work dW = F ds|Work W = τ dθ|
|7 Kinetic energy K = Mv²/2|Kinetic energy K = Iω²/2|
|8 Power P = F v|Power P = τω|
|9 Linear momentum p = Mv|Angular momentum L = Iω|
Figure 6.30 shows a cross-section of a rigid body rotating about a fixed axis, which is taken as the z-axis (perpendicular to the plane of the page; see Fig. 6.29). As said above we need to consider only those forces which lie in planes perpendicular to the axis. Let F₁ be one such typical force acting as shown on a particle of the body at point P₁ with its line of action in a plane perpendicular to the axis. For convenience we call this to be the x′–y′ plane (coincident with the plane of the page). The particle at P₁ describes a circular path of radius r₁ with centre C on the axis; CP₁ = r₁.
In time ∆t, the point moves to the position P₁′. The displacement of the particle ds₁, therefore, has magnitude ds₁ = r₁dθ and direction tangential at P₁ to the circular path as shown. Here dθ is the angular displacement of the particle, dθ = ∠P CP′. The work done by the force on the particle is φ₁ θ α.
dW₁ = F₁. ds₁= F₁ds₁ cos 1= F₁(r₁dθ)sin 1 where φ₁ is the angle between F₁ and the tangent at P₁, and α₁ is the angle between F₁ and the radius vector OP₁; φ₁ + α₁ = 90°.
The torque due to F₁ about the origin is τ = OP₁ × F₁. Now OP₁ = OC + OP₁. [Refer to Fig. 6.17(b).] Since OC is along the axis, the torque resulting from it is excluded from our consideration. The effective torque due to F₁ is τ₁ = CP × F₁; it is directed along the axis of rotation and has a magnitude τ₁ = r₁F₁sinα. Therefore, dW₁ = τ₁dθ.
This expression gives the work done by the total (external) torque τ which acts on the body rotating about a fixed axis. Its similarity with the corresponding expression dW = F ds for linear (translational) motion is obvious.
Dividing both sides of Eq. (6.39) by dt gives P = dW/dt = τ dθ/dt = τω.
This is the instantaneous power. Compare this expression for power in the case of rotational motion about a fixed axis with that of power in the case of linear motion, P = Fv.
In a perfectly rigid body there is no internal motion. The work done by external torques is. | 27 | 11 | Physics | 106 |
5104fea8-2c65-4fc9-85d0-9cb405e7d1d0 | # PHYSICS
therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (6.40). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is
d ⌈ Iω² ⌉ = I (2ω) dω
dt ⌊ 2 ⌋ 2 dt
We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body.
Since α = dω /dt, we get
d ⌈ Iω² ⌉ = I ω α
dt ⌊ 2 ⌋
We use I α = τ
Equating rates of work done and of increase in kinetic energy,
τω = I ω α (6.41)
τ = Iα
Eq. (6.41) is similar to Newton’s second law for linear motion expressed symbolically as
F = ma
Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body. In this respect, Eq.(6.41) can be called Newton’s second law for rotational motion about a fixed axis.
Example 6.12 A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 6.31. The flywheel is mounted on a horizontal axle with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
τ = F R = 25 × 0.20 Nm (as R = 0.20m) = 5.0 Nm
I = Moment of inertia of flywheel about its axis = MR² = 20.0 × (0.2)² = 0.4 kg m²
α = angular acceleration = 5.0 N m/0.4 kg m² = 12.5 s–2
(b) Work done by the pull unwinding 2m of the cord = 25 N × 2m = 50 J
(c) Let ω be the final angular velocity. The kinetic energy gained = 1 Iω2, since the wheel starts from rest. Now, ω2 = ω2 + 2αθ, ω = 0
The angular displacement = length of unwound string / radius of wheel = 2m/0.2 m = 10 rad
ω2 = 2 × 12.5 × 10 = 250 (rad/s)
(d) The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due to friction.
2024-25 | 28 | 11 | Physics | 106 |
e28800ef-0406-4c5f-b72d-97dd2f74416b | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
# 6.12 ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS
For computing the total angular momentum of the whole rigid body, we add up the contribution of each particle of the body. Thus
We denote by *L⊥ and Lz the components of L* respectively perpendicular to the z-axis and along the z-axis;
*∑ L⊥ = OCi × mivi* (6.42a)
where *m and v are respectively the mass and the velocity of the iᵗʰ particle and Ci* is the centre of the circle described by the particle; and
*L = Iω* (6.42b)
We first consider the angular momentum of a typical particle of the rotating rigid body. We then sum up the contributions of individual particles to get *L* of the whole body.
For a typical particle *l = r × p. As seen in the last section r = OP = OC + CP [Fig. 6.17(b)]. With p = mv*,
*l = (OC × mv) + (CP × mv)*
The magnitude of the linear velocity *v of the particle at P is given by v = ωr where r is the length of CP or the perpendicular distance of P from the axis of rotation. Further, v is tangential at P to the circle which the particle describes. Using the right-hand rule one can check that CP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) is k*. Hence
*kCP × mv = r (mv) k*
Since *υ = ωr*
*l = mr²ω k (since υ = ωr⊥*)
Similarly, we can check that *OC × v is perpendicular to the fixed axis. Let us denote the part of l along the fixed axis (i.e. the z-axis) by lz*, then
*lz = CP × mv = mr²ω k*
and *l = lz + OC × mv*
We note that *lz is parallel to the fixed axis, l is not. In general, for a particle, the angular momentum l is not along the axis of rotation, i.e. for a particle, l and ω are not necessarily parallel. Compare this with the corresponding fact in translation. For a particle, p and v* are always parallel to each other.
Now, Eq. (6.28b) states
*dL/dt = τ*
Referring to Table 6.1, can you tell in which cases *L = Lz* will not apply? | 29 | 11 | Physics | 106 |
e4e3e6d4-fd21-4f1a-bef5-7df01170c41c | # PHYSICS
As we have seen in the last section, only those components of the external torques which are along the axis of rotation need to be taken into account when we discuss rotation about a fixed axis. This means we can take τ = τk.
Since L = Lz + L⊥ and the direction of L (vector k) is fixed, it follows that for rotation about a fixed axis,
L = Iω = constant (6.44)
For symmetric bodies, from Eq. (6.42d), Lz may be replaced by L. (L and Lz are respectively the magnitudes of L and Lz.)
This then is the required form, for fixed axis rotation, of Eq. (6.29a), which expresses the general law of conservation of angular momentum of a system of particles. Eq. (6.44) applies to many situations that we come across in daily life. You may do this experiment with your friend. Sit on a swivel chair (a chair with a seat, free to rotate about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your friend to rotate the chair rapidly. While the chair is rotating with considerable angular speed stretch your arms horizontally. What happens? Your angular speed is reduced. If you bring back your arms closer to your body, the angular speed increases again. This is a situation where the principle of conservation of angular momentum is applicable.
If friction in the rotational
τ = Iα (6.41)
Fig 6.32 (a) A demonstration of conservation of angular momentum. A girl sits on a swivel chair and stretches her arms/ brings her arms closer to the body.
Fig 6.32 (b) An acrobat employing the principle of conservation of angular momentum in her performance.
2024-25 | 30 | 11 | Physics | 106 |
10dce502-7ecb-446a-8d48-e87fdd2e0978 | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
A circus acrobat and a diver take advantage of this principle. Also, skaters and classical, Indian or western, dancers performing a pirouette (a spinning about a tip–top) on the toes of one foot display ‘mastery’ over this principle. Can you explain?
# SUMMARY
1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them.
2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions.
3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time.
4. In pure translation, every particle of the body moves with the same velocity at any instant of time.
5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction.
6. The vector or cross product of two vector a and b is a vector written as a × b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule.
7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin.
8. The centre of mass of a system of n particles is defined as the point whose position vector is
R = ∑ (mi ri) / M
9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant.
10. The angular momentum of a system of n particles about the origin is
L = ∑ (ri × pi)
The torque or moment of force on a system of n particles about the origin is
τ = ∑ (ri × Fi)
The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0. | 31 | 11 | Physics | 106 |
1f03a9c1-f9ab-4cfc-8351-2727cce8adf3 | # PHYSICS
dL = τ
dt τₑₓₜ
# 11.
A rigid body is in mechanical equilibrium if
1. it is in translational equilibrium, i.e., the total external force on it is zero: ∑ Fi = 0, and
2. it is in rotational equilibrium, i.e. the total external torque on it is zero: ∑ τi = ri × Fi = 0.
# 12.
The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero.
# 13.
The moment of inertia of a rigid body about an axis is defined by the formula I = ∑m r2i where ri is the perpendicular distance of the ith point of the body from the axis. The kinetic energy of rotation is K = 1/2 Iω².
# POINTS TO PONDER
1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body.
2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV²/2, K = K′ + MV²/2.
3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles.
4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles.
5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero.
6. The total torque on a system is independent of the origin if the total external force is zero.
7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other.
2024-25 | 32 | 11 | Physics | 106 |
50e9f606-d159-4322-a229-bcd0ce653a1c | # SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
# 8. The angular momentum L and the angular velocity ω
ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis.
# EXERCISES
1. Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
5. Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.
6. Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components pₓ, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
7. Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
2024-25 | 33 | 11 | Physics | 106 |
685b28a7-59ee-41c7-8748-079cef5ffe09 | # PHYSICS
# 6.10
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
# 6.11
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
# 6.12
(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
# 6.13
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
# 6.14
To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
# 6.15
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
# 6.16
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
# 6.17
The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 × 10-46 kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
2024-25 | 34 | 11 | Physics | 106 |
65af2097-fd26-4d7d-8bfc-f9372587c908 | # CHAPTER TWELVE
# KINETIC THEORY
# 12.1 INTRODUCTION
Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of gases by considering that gases are made up of tiny atomic particles. The actual atomic theory got established more than 150 years later. Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules. This is possible as the inter-atomic forces, which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and others. It has been remarkably successful. It gives a molecular interpretation of pressure and temperature of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory.
# 12.2 MOLECULAR NATURE OF MATTER
Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus. | 0 | 11 | Physics | 205 |
698a396f-2a51-4bcb-b5a9-d1146ff2884a | # KINETIC THEORY
# Atomic Hypothesis in Ancient India and Greece
Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10–10 m.
In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science.
In Greece had suggested that matter may consist of indivisible constituents. The scientific ‘Atomic Theory’ is usually credited to John Dalton. He proposed the atomic theory to explain the laws of definite and multiple proportions obeyed by elements when they combine into compounds. The first law says that any given compound has a fixed proportion by mass of its constituents. The second law says that when two elements form more than one compound, for a fixed mass of one element, the masses of the other elements are in ratio of small integers.
To explain the laws Dalton suggested, about 200 years ago, that the smallest constituents of an element are atoms. Atoms of one element are identical but differ from those of other elements. A small number of atoms of each element combine to form a molecule of the compound. Gay Lussac’s law, also given in early 19ᵗʰ century, states: When gases combine chemically to yield another gas, their volumes are in the ratios of small integers. Avogadro’s law (or hypothesis) says: Equal volumes of all gases at equal temperature and pressure have the same number of molecules. Avogadro’s law, when combined with Dalton’s theory explains Gay Lussac’s law. Since the elements are often in the form of molecules, Dalton’s atomic theory can also be referred to as the molecular theory.
The size of an atom is about an angstrom (10 -10 m). In solids, which are tightly packed, atoms are spaced about a few angstroms (2 Å) apart. In liquids the separation between atoms is also about the same. In liquids the atoms are not as rigidly fixed as in solids, and can move around. This enables a liquid to flow. In gases the interatomic distances are in tens of angstroms. The average distance a molecule can travel without colliding is called the mean free path. The mean free path, in gases, is of the order of thousands of angstroms. The atoms are much freer in gases and can travel long distances without colliding. If they are not enclosed, gases disperse away. In solids and liquids the closeness makes the interatomic force important. The force has a long range attraction and a short range repulsion. The atoms attract when they are at a few angstroms but repel when they come closer. The static appearance of a gas. | 1 | 11 | Physics | 205 |
b284ac71-0b88-42e7-8790-ce8fff9ac637 | # PHYSICS
is misleading. The gas is full of activity and the equilibrium is a dynamic one. In dynamic equilibrium, molecules collide and change their speeds during the collision. Only the average properties are constant.
Atomic theory is not the end of our quest, but the beginning. We now know that atoms are not indivisible or elementary. They consist of a nucleus and electrons. The nucleus itself is made up of protons and neutrons. The protons and neutrons are again made up of quarks. Even quarks may not be the end of the story. There may be string like elementary entities. Nature always has surprises for us, but the search for truth is often enjoyable and the discoveries beautiful.
In this chapter, we shall limit ourselves to understanding the behaviour of gases (and a little bit of solids), as a collection of moving molecules in incessant motion.
# 12.3 BEHAVIOUR OF GASES
Properties of gases are easier to understand than those of solids and liquids. This is mainly because in a gas, molecules are far from each other and their mutual interactions are negligible except when two molecules collide.
Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation between their pressure, temperature and volume given by (see Chapter 10)
PV = KT (12.1)
for a given sample of the gas. Here T is the temperature in kelvin or (absolute) scale. K is a constant for the given sample but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then K is proportional to the number of molecules, (say) N in the sample. We can write K = N k. Observation tells us that this k is same for all gases. It is called Boltzmann constant and is denoted by kB.
As 1/N1T1 = 1/N2T2 = constant = kB (12.2)
if P, V and T are same, then N is also same for all gases. This is Avogadro’s hypothesis, that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure. The number in 22.4 litres of any gas is 6.02 × 1023. This is known as Avogadro number and is denoted by NA.
The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P (standard temperature 273 K and pressure 1 atm). This amount of substance is called a mole (see Chapter 1 for a more precise definition). Avogadro had guessed the equality of numbers in equal volumes of gas at a fixed temperature and pressure from chemical reactions. Kinetic theory justifies this hypothesis.
The perfect gas equation can be written as
PV = μRT (12.3)
where μ is the number of moles and R = NAkB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale for absolute temperature, R = 8.314 J mol–1K–1.
Here
µ = M = N (12.4)
where M0 is the mass of the gas containing N molecules, M0 is the molar mass and NA the Avogadro’s number. Using Eqs. (12.4) and (12.3) can also be written as
PV = kBNT or P = kBnT
where n is the number density, i.e. number of molecules per unit volume. kB is the Boltzmann constant introduced above. Its value in SI units is 1.38 × 10–23 J K–1.
Another useful form of Eq. (12.3) is
P = ρRT (12.5)
M0 | 2 | 11 | Physics | 205 |
fa911085-54ef-4751-ae6e-4d0b614caafe | # KINETIC THEORY
where ρ is the mass density of the gas. A gas that satisfies Eq. (12.3) exactly at all pressures and temperatures is defined to be an ideal gas. An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal.
Fig. 12.1 shows departures from ideal gas behaviour for a real gas at three different temperatures. Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures.
At low pressures or high temperatures the gas molecules are far apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal one.
If we fix μ and T in Eq. (12.3), we get
PV = constant (12.6)
i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 12.2 shows comparison between experimental P-V curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures.
Next, if you fix P, Eq. (12.1) shows that V ∝ T i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (Charles’ law). See Fig. 12.3.
Fig. 12.3 Experimental T-V curves (solid lines) for CO₂ at three pressures compared with Charles’ law (dotted lines). T is in units of 300 K and V in units of 0.13 litres.
We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule.
# Example 12.1
The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m–3. The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.
Finally, consider a mixture of non-interacting ideal gases: μ1 moles of gas 1, μ2 moles of gas 2, | 3 | 11 | Physics | 205 |
15f9806f-8e5f-48e1-a929-9ac3beb43b68 | # PHYSICS
Answer For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 × 10-4) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10-4.
# Example 12.2
Estimate the volume of a water molecule using the data in Example 12.1.
Answer In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kg m–3. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. Since 1 mole contains about 6 × 1023 molecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = 3 × 10–26 kg. Therefore, a rough estimate of the volume of a water molecule is as follows:
|Volume of a water molecule|ρ1 = m1 /V = m1 = μ1 × M1|
|---|---|
|= (3 × 10–26 kg)/ (1000 kg m–3)|ρ2 = m2 /V = m2 = μ2 × M2|
|= 3 × 10–29 m3|μ2 = 3 × 20.2 = 0.947|
|= (4/3) π (Radius)3|32.0|
|Hence, Radius ≈ 2 × 10–10 m = 2 Å| |
# Example 12.3
What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples 12.1 and 12.2.
Answer: A given mass of water in vapour state has 1.67×103 times the volume of the same mass of water in liquid state (Ex. 12.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å.
# Example 12.4
A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i)
# 12.4 KINETIC THEORY OF AN IDEAL GAS
Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a collection of a large number of molecules (typically of the order of Avogadro’s number) that are in incessant random motion. At ordinary pressure and temperature, the average distance between molecules is a factor of 10 or more than the typical size of a molecule (2 Å). Thus, interaction between molecules is negligible and we can assume that they move freely in straight lines according to Newton’s first law. However, occasionally, they come close to each other, experience intermolecular forces and their velocities change. These interactions are called collisions. The molecules collide incessantly against each other or with the walls and change. | 4 | 11 | Physics | 205 |
229e7586-1c69-40cb-8ea1-9e5907cf201a | # KINETIC THEORY
their velocities. The collisions are considered to be elastic. We can derive an expression for the pressure of a gas based on the kinetic theory. We begin with the idea that molecules of a gas are in incessant random motion, colliding against one another and with the walls of the container. All collisions between molecules among themselves or between molecules and the walls are elastic. This implies that total kinetic energy is conserved. The total momentum is conserved as usual.
The total momentum transferred to the wall by these molecules in time ∆t is:
Q = (2mvₓ) (½ n A vₓ ∆t) (12.10)
The force on the wall is the rate of momentum transfer Q/∆t and pressure is force per unit area:
P = Q /(A ∆t) = n m vₓ² (12.11)
Actually, all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation, therefore, stands for pressure due to the group of molecules with speed vₓ in the x-direction and n stands for the number density of that group of molecules. The total pressure is obtained by summing over the contribution due to all groups:
P = n m v² (12.12)
Consider a gas enclosed in a cube of side l. Take the axes to be parallel to the sides of the cube, as shown in Fig. 12.4. A molecule with velocity (vₓ, vᵧ, v𝓏) hits the planar wall parallel to the x-plane of area A (= l²). Since the collision is elastic, the molecule rebounds with the same velocity; its y and z components of velocity do not change in the collision but the x-component reverses sign. That is, the velocity after collision is (-vₓ, vᵧ, v𝓏). The change in momentum of the molecule is: –mvₓ – (mvₓ) = –2mvₓ. By the principle of conservation of momentum, the momentum imparted to the wall in the collision = 2mvₓ.
Therefore, by symmetry, v² = v² = v² where v² is the average of vₓ². Now the gas is isotropic, i.e. there is no preferred direction of velocity of the molecules in the vessel.
Thus, P = (1/3) n m v² (12.14)
Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and ∆t do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in the derivation. Though this assumption is difficult to justify rigorously, we can qualitatively see that it will not lead to erroneous results. The number of molecules hitting the wall in time ∆t was found to be ½ n A vₓ ∆t. Now the collisions are random and the gas is in a steady state. Thus, if a molecule with velocity (vₓ, vᵧ, v𝓏) acquires a different velocity due to collision with some molecule, there will always be some other.
2024-25 | 5 | 11 | Physics | 205 |
1264a78e-dad6-4486-9b90-e701df31e572 | # PHYSICS
molecule with a different initial velocity which P = (1/3) [n₁m₁v² + n₂m₂v² +… ] (12.20) after a collision acquires the velocity (v₁, v₂, v₃).
In equilibrium, the average kinetic energy of the molecules of different gases will be equal. That is,
½ m₁v₁² = ½ m₂v₂² = (3/2) kT
so that P = (n₁ + n₂ +…) kT (12.21) which is Dalton’s law of partial pressures.
# 12.4.2 Kinetic Interpretation of Temperature
Equation (13.14) can be written as
PV = (1/3) nV m v² (12.15a)
PV = (2/3) N x ½ m v² (12.15b)
where N (= nV) is the number of molecules in the sample. The quantity in the bracket is the average translational kinetic energy of the molecules in the gas. Since the internal energy E of an ideal gas is purely kinetic*,
E = N × (1/2) m v² (12.16)
Equation (12.15) then gives:
PV = (2/3) E (12.17)
We are now ready for a kinetic interpretation of temperature. Combining Eq. (12.17) with the ideal gas Eq. (12.3), we get
E = (3/2) kᴮ NT (12.18)
E/N = ½ m v² = (3/2) kBT (12.19)
i.e., the average kinetic energy of a molecule is proportional to the absolute temperature of the gas; it is independent of pressure, volume or the nature of the ideal gas. This is a fundamental result relating temperature, a macroscopic measurable parameter of a gas (a thermodynamic variable as it is called) to a molecular quantity, namely the average kinetic energy of a molecule. The two domains are connected by the Boltzmann constant. We note in passing that Eq. (12.18) tells us that internal energy of an ideal gas depends only on temperature, not on pressure or volume. With this interpretation of temperature, kinetic theory of an ideal gas is completely consistent with the ideal gas equation and the various gas laws based on it.
For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture. Equation (12.14) becomes
* E denotes the translational part of the internal energy U that may include energies due to other degrees of freedom also. See section 12.5.
# Example 12.5
A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27 °C. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed vᵣₘₛ of the molecules of the two gases. Atomic mass of argon = 39.9 u; Molecular mass of chlorine = 70.9 u.
# Answer
The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) kBT. It depends only on temperature, and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1.
(ii) Now ½ m vᵣₘₛ² = average kinetic energy per molecule = (3/2) kBT where m is the mass. | 6 | 11 | Physics | 205 |
265e2bae-45a0-4f16-9134-cf54110856e7 | # KINETIC THEORY
of a molecule of the gas. Therefore,
(v₂ ) (m) (M) ⊳
( rms )ᴬʳ = ( )Cl = ( )Cl = 70.9 =1.77
v² m M 39.9
rms Cl Ar Ar
where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.)
Taking square root of both sides,
(vᵣₘₛ )Ar
(vᵣₘₛ )Cl = 1.33
You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.
Fig. 12.5 Molecules going through a porous wall.
# Example 12.6
Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed? If the atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature.
# Answer
At a fixed temperature the average energy = ½ m <v² > is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So
v₃₄₉ / v₃₅₂ = (352/349)1/2 = 1.0044.
Hence difference ∆V = 0.44 %.
[²³⁵U is the isotope needed for nuclear fission. To separate it from the more abundant isotope 238U, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder (Fig. 12.5). The method is not very efficient and has to be repeated several times for sufficient enrichment.]
# Example 12.7
(a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (Ch.5 will refresh your memory on elastic collisions.)
(b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above.
(c) What happens when a compressed gas pushes a piston out and expands. What would you observe?
(d) Sachin Tendulkar used a heavy cricket bat while playing. Did it help him in any way?
# Answer
(a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature. | 7 | 11 | Physics | 205 |
d034deae-428c-4a0d-b0da-3bed70814adc | # PHYSICS
You should be able to answer (b) (c) and (d) based on the answer to (a).
(Hint: Note the correspondence, piston‡ bat, cylinder ‡ wicket, molecule ‡ ball.)
The molecule thus has two rotational degrees of freedom, each of which contributes a term to the total energy consisting of translational energy εₜ and rotational energy εᵣ.
# 12.5 LAW OF EQUIPARTITION OF ENERGY
The kinetic energy of a single molecule is
εₜ = 1 mv² + 1 mv² + 1 mv² (12.22)
2 x 2 y 2 z
For a gas in thermal equilibrium at temperature T the average value of energy denoted by < εₜ > is
εₜ = 1 mv₂ + 1 mv₂ + 1 mv₂ = 3k T (12.23)
2 x 2 y 2 z 2 B
Since there is no preferred direction, Eq. (12.23) implies
1 mv₂ = 1 k T, 1 mv₂ = 1 k T,
2 x 2 B 2 y 2 B
1 mv₂ = 1 k T (12.24)
2 z 2 B
A molecule free to move in space needs three coordinates to specify its location. If it is constrained to move in a plane it needs two; and if constrained to move along a line, it needs just one coordinate to locate it. This can also be expressed in another way. We say that it has one degree of freedom for motion in a line, two for motion in a plane and three for motion in space. Motion of a body as a whole from one point to another is called translation. Thus, a molecule free to move in space has three translational degrees of freedom. Each translational degree of freedom contributes a term that contains square of some variable of motion.
In, Eq. (12.24) we see that in thermal equilibrium, the average of each such term is ½ k T.
Molecules of a monatomic gas like argon have only translational degrees of freedom. But what about a diatomic gas such as O₂ or N₂? A molecule of O₂ has three translational degrees of freedom. But in addition it can also rotate about its centre of mass. Figure 12.6 shows the two independent axes of rotation 1 and 2, normal.
* Rotation along the line joining the atoms has very small moment of inertia and does not come into play for quantum mechanical reasons. See end of section 12.6. | 8 | 11 | Physics | 205 |
d294e03d-0768-4203-aff0-8faf9649dce5 | # KINETIC THEORY
At this point, notice an important feature in Eq.(12.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq.(12.26), one vibrational mode contributes two ‘squared terms’: kinetic and potential energies.
The ratio of specific heats γ = Cᵖ/Cᵛ = 5/3 (12.31)
Each quadratic term occurring in the expression for energy is a mode of absorption of energy by the molecule. We have seen that in thermal equilibrium at absolute temperature T, for each translational mode of motion, the average energy is ½ kBT. The most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes ½ kBT to the energy, while each vibrational frequency contributes 2 × ½ kBT = kBT, since a vibrational mode has both kinetic and potential energy modes.
The proof of the law of equipartition of energy is beyond the scope of this book. Here, we shall apply the law to predict the specific heats of gases theoretically. Later, we shall also discuss briefly, the application to specific heat of solids.
# 12.6.2 Diatomic Gases
As explained earlier, a diatomic molecule treated as a rigid rotator, like a dumbbell, has 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is
U = 5 kBT × N = 5 RTA (12.32)
The molar specific heats are then given by
Cv (rigid diatomic) = 5 R, Cp = 7 R (12.33)
γ (rigid diatomic) = 5/7 (12.34)
If the diatomic molecule is not rigid but has in addition a vibrational mode
U = (5/2) kBT + kBT = RTA (12.35)
Cv = 7 R, Cp = 9 R, γ = 9/7 (12.35)
# 12.6.3 Polyatomic Gases
In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has
U = (3/2) kBT × N + (3/2) kBT × N + f kBT × N (12.36)
i.e., Cv = (3 + f) R, Cp = (4 + f) R, γ = (4 + f)/(3 + f) (12.36)
Note that Cp – Cv = R is true for any ideal gas, whether mono, di or polyatomic.
Table 12.1 summarises the theoretical predictions for specific heats of gases ignoring any vibrational modes of motion. The values are | 9 | 11 | Physics | 205 |
42ec4741-7750-4655-93dd-b8e58e72bb59 | # PHYSICS
in good agreement with experimental values of specific heats of several gases given in Table 12.2. Of course, there are discrepancies between predicted and actual values of specific heats of several other gases (not shown in the table), such as Cl₂, C₂H₆ and many other polyatomic gases. Usually, the experimental values for specific heats of these gases are greater than the predicted values as given in Table 12.1 suggesting that the agreement can be improved by including vibrational modes of motion in the calculation. The law of equipartition of energy is, thus, well verified experimentally at ordinary temperatures.
Answer Using the gas law PV = μRT, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 × 10⁵ Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, Cv = (3/2) R, and molar specific heat at constant pressure, Cp = (3/2) R + R = (5/2) R. Since the volume of the cylinder is fixed, the heat required is determined by Cv. Therefore,
Heat required = no. of moles × molar specific heat × rise in temperature = 2 × 1.5 R × 15.0 = 45 R = 45 × 8.31 = 374 J.
# 12.6.4 Specific Heat Capacity of Solids
We can use the law of equipartition of energy to determine specific heats of solids. Consider a solid of N atoms, each vibrating about its mean position. An oscillation in one dimension has average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT. For a mole of solid, N = NA, and the total energy is U = 3 kBT NA = 3 RT.
Now at constant pressure ∆Q = ∆U + P∆V = ∆U, since for a solid ∆V is negligible. Hence,
Cp = ∆Q / ∆T = ∆U / ∆T = 3R (12.37)
# Table 12.1 Predicted values of specific heat capacities of gases (ignoring vibrational modes)
|Nature of Gas|Cv (J mol−1 K−1)|Cp (J mol−1 K−1)|Cp - Cv (J mol−1 K−1)|g|
|---|---|---|---|---|
|Monatomic|12.5|20.8|8.31|1.67|
|Diatomic|20.8|29.1|8.31|1.40|
|Triatomic|24.93|33.24|8.31|1.33|
# Table 12.2 Measured values of specific heat capacities of some gases
As Table 12.3 shows the prediction generally agrees with experimental values at ordinary temperature (Carbon is an exception).
# Table 12.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure
# Example 12.8
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 °C? (R = 8.31 J mol−1 K−1).
# 12.7 MEAN FREE PATH
Molecules in a gas have rather large speeds of the order of the speed of sound. Yet a gas leaking | 10 | 11 | Physics | 205 |
98e2e8c0-daaf-448a-a24a-70811f00a5da | # KINETIC THEORY
from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot move straight unhindered; their paths keep getting incessantly deflected.
Thus we need to replace <v> by <vᵣ> in Eq. (12.38). A more exact treatment gives
l = 1/ ( 2 nπd² ) (12.40)
Let us estimate l and τ for air molecules with average speeds <v> = ( 485m/s). At STP
(0.02 ×₁₀₂₃)
n = (22.4 ×10–3) = 2.7 × 1025 m-3.
Taking, d = 2× 10–10 m, τ = 6.1 ×10–10 s
vt and l = 2.9 × 10–7 m ≈ 1500 d (12.41)
As expected, the mean free path given by Eq. (12.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube n is rather small and the mean free path can be as large as the length of the tube.
# Example 12.9
Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 12.1 and Eq. (12.41) above.
Answer: The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.
So n = 2.7 × 1025 × 273 = 2 × 1025 m–3 373
Hence, mean free path l = 4 × 10–7 m
Note that the mean free path is 100 times the interatomic distance ~ 40 Å = 4 × 10–9 m calculated earlier. It is this large value of mean free path that leads to the typical gaseous behaviour. Gases cannot be confined without a container.
Using the kinetic theory of gases, the bulk measurable properties like viscosity, heat conductivity and diffusion can be related to the microscopic parameters like molecular size. It is through such relations that the molecular sizes were first estimated. | 11 | 11 | Physics | 205 |
824e76fc-c573-44fd-a628-b1aaf298ee2e | # PHYSICS
# SUMMARY
1. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is
<math>PV = \mu RT = k_B NT</math>
where <math>\mu</math> is the number of moles and <math>N</math> is the number of molecules. <math>R</math> and <math>k_B</math> are universal constants.
<math>R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}, \quad k_B = N_A = 1.38 \times 10^{-23} \, \text{J K}^{-1}</math>
Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures.
2. Kinetic theory of an ideal gas gives the relation
<math>P = \frac{1}{3} n m \langle v^2 \rangle</math>
where <math>n</math> is number density of molecules, <math>m</math> the mass of the molecule and <math>\langle v^2 \rangle</math> is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature.
<math>\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T, \quad v_{rms} = \sqrt{\frac{3k_B T}{m}}</math>
This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed.
3. The translational kinetic energy
<math>E = \frac{3}{2} k_B NT.</math>
This leads to a relation
<math>PV = \frac{2}{3} E</math>
4. The law of equipartition of energy states that if a system is in equilibrium at absolute temperature <math>T</math>, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to <math>\frac{1}{2} k_B T</math>. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy <math>\frac{1}{2} k_B T</math>. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to
<math>2 \times \frac{1}{2} k_B T = k_B T.</math>
5. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion.
6. The mean free path <math>l</math> is the average distance covered by a molecule between two successive collisions:
<math>l = \frac{1}{\pi 2} \frac{1}{2 n d}</math>
where <math>n</math> is the number density and <math>d</math> the diameter of the molecule. | 12 | 11 | Physics | 205 |
099bbdb1-5db3-4865-a783-709959a77d5a | # KINETIC THEORY
# POINTS TO PONDER
1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.
2. We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule.
3. The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is ½ kB T. Each quadratic term in the total energy expression of a molecule is to be counted as a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy 2 × ½ kB T = kB T.
4. Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy ½ mv² of the molecules.
5. < v² > is not always equal to (< v >)². The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement.
# EXERCISES
1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
3. Figure 12.8 shows plot of PV/T versus P for 1.00 × 10–3 kg of oxygen gas at two different temperatures.
- (a) What does the dotted plot signify?
- (b) Which is true: T1 > T2 or T1 < T2?
- (c) What is the value of PV/T where the curves meet on the axis? | 13 | 11 | Physics | 205 |
0e9f0424-2eba-42a9-bdfa-ca7e9fa66e09 | # PHYSICS
# 12.4
If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mol–1 K–1.)
# 12.5
An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u).
# 12.6
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
# 12.7
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
# 12.8
Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
# 12.9
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
# 12.10
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at –20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
# 12.11
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u). | 14 | 11 | Physics | 205 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.