Is angular momentum conserved if centripetal force is increased?
Let a mass $m$ be attached to the end of a light rigid rod of length $r$. The other end of the rod has a loop of non stretchable thread attached.
I put my finger into the thread loop and rotate the rod (around my finger) with angular velocity w and centripetal force $F$.
Now if I increase the the centripetal force to $4F$, then the angular velocity should increase to $2\omega$, because $F=mr\omega^2$. $m$ and $r$ are constant.
Since there is no torque the angular momentum L should be conserved.
\begin{align} \text{(initially)}\qquad L_1&=mr\omega \\ \text{(later)}\qquad L_2&=mr\, 2\omega \end{align}
$L_1$ is not equal to $L_2$. That means angular momentum is not conserved. Thus there should be a torque applied somewhere but where? I cant't figure this out. Is my finger providing the torque? But the pivot point can't provide torque?
newtonian-mechanics kinematics rotational-dynamics rotational-kinematics
L..L..
$\begingroup$ How would the "increase the centripetal force to $4F$" step be done? It is very difficult to create a constant force in an experiment. $\endgroup$ – BowlOfRed Sep 5 '17 at 20:07
If I understand your description, then when you are trying to "increase the centripetal force", you first have to move your finger around in a circle. When you do this, the direction of the force (from the finger to the loop to the rod to the mass) is no longer perpendicular to the instantaneous velocity of the mass, but it has a component along it.
That component is accelerating the mass (generates torque), and increasing the angular momentum. See this diagram:
FlorisFloris
$\begingroup$ +1 You could also emphasize that when the finger is off the center of the trajectory, the torque (with respect to that point) is not zero. $\endgroup$ – Diracology Sep 5 '17 at 21:12
In your situation, the tangential velocity $v=r\omega$ increases by a factor of two over the process, which means that some force acted on the particle along its direction of travel; it was that force that provided the relevant torque.
Your description of the process is at present too vague to say anything more, though, but it's important to remark that just because
I increase the the centripetal force to $4F$
it does not follow that
the angular velocity should increase to $2\omega$.
The angular velocity does need to increase if the particle is to remain on its circular trajectory, but that increase is not going to happen spontaneously - and it's not going to happen just from the centripetal pull, either. (If all you do is increase the force, then the particle will move into some form of elliptical trajectory, after which its movement will depend on how the force changes if the radius changes.) So, there's something pushing the particle transversely, and that same force is providing the appropriate torque.
Emilio PisantyEmilio Pisanty
So what you have proposed, if I'm reading you correctly, is that you have a mass on a rod rotating with angular velocity $\omega$ and you want to take the force of the rod on the mass $F$ and provide an additional $3F$ from your finger onto the mass, "squeezing it into the rod."
It is your claim that this increases the centripetal force to $4F.$ I claim, it does not. That the rod is rigid is what we call a "constraint force", it forces the mass to remain on the circle of radius $r$. When the mass is orbiting normally, before you press your finger into it, this requires a constant force of $F_r = -m \omega^2 r$ in the radial direction (that is, the minus sign means that this points towards the center).
This "force" does not come from nowhere: it comes from calculus. The circular trajectory is the position vector $(x, y) = r~(\cos\theta, \sin \theta),$ and for constant $r$ its second derivative is by the product rule $$(\ddot x, \ddot y) = r \ddot\theta~(-\sin\theta, \cos\theta) - r \dot\theta^2~(\cos\theta, \sin\theta),$$ where dots are time derivatives. This shows two components to the acceleration: the "tangential" acceleration points along the "tangent line" to the curve, with magnitude $r\ddot\theta$, and the "centripetal" acceleration points "towards the center" (that's what the term means), with magnitude $r \dot\theta^2.$
You cannot change $\omega = \dot\theta$ without a tangential force; your center-pointing force of $4F$ will not do this if it is properly applied. So what happens instead?
(Take a moment to think of what a constraint force does. What do we mean if we velcro your boots to the ground and say you are constrained to move such that your feet must stay in place? Sure, normally it means that the ground pushes up on you so that you don't fall through the ground -- but what if you try to jump? Clearly then it must pull down on you. Etc.)
It's all about net forces. The centripetal acceleration derived from calculus must be the net force applied to the mass (divided by $m,$ of course). Calling it a "force" is kind of a misnomer because it makes you think of it as a special, named force, like a "spring force" or a "gravitational" forces is. It's not like those. It's a logical consequence of the form of the motion, therefore it is the shape of the net force once everything is put together. In this case, the net force is the sum of your finger's force, and the constraint force from the rod.
A constraint force supplies whatever force is needed to enforce the constraint, usually in a direction perpendicular to whatever motion the constraint allows. In this case, the requirement that the mass stay on the circle when you are pushing inward with force $F_r = -3|F|$ and the force needed to stay on the circle is only $-|F|$, it must supply a force of $+2|F|$ to keep the mass on the circle. The rod is no longer under tension (being pulled apart) -- it is now under compression (being pushed in).
You can see this most clearly if you imagine a "rod" which can only handle tension loads, not compression loads: a very good example of this is a metal chain; when you try to pull it taut it resists you mightily; when you try to push the ends back together they just come together with no resistance as the links of the chain run slack. So let's imagine that one plays a rather strange game with one's child: instead of pushing that child from behind as they swing on a swingset, imagine laying underneath them as they swing above you, trying to push so hard that they "jump" up in the air and have to hold onto the swing for dear life. That is the clearest way to visualize how, if the constraint force does not oppose us (as the chain cannot), the mass (our child) will no longer obey the circular trajectory.
Of course, you can push the mass tangentially and increase its angular velocity to $2\omega$ and then it will require four times the centripetal acceleration, and if you don't push on it then that will have to come from the rod or chain. (Similarly when the child is swinging on the swingset, the chain must supply a force $+mg~\cos\theta + m \dot\theta^2 r,$ as otherwise at $\theta=\dot\theta=0$ the child would fall to the ground under the force of gravity.) This much you seem to already understand. But of course, the cause of the change in angular momentum is the fact that you pushed it tangentially, which torques it.
CR DrostCR Drost
protected by Qmechanic♦ Sep 5 '17 at 21:22
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