A thin homogeneous rod of mass $M$ and length $l$ is pivotally attached at its upper end. At the end of the rod, a body of mass $M_{1}$ is pivotally connected as shown in the figure. The pivot is exactly at the center of mass of the $M_{1}$ mass body. We fire a projectile of mass $m$ horizontally at the speed $v$ into the $M_{1}$ mass body at the height of the pivot. (The projectile does not penetrate to the axis.) What will be the maximum deflection of the pendulum?

![](https://cdn.mathpix.com/cropped/2024_05_02_4410a200665f1a4c2857g-1.jpg?height=483&width=501&top_left_y=355&top_left_x=801)
Let's assume that the duration of the collision is negligibly small compared to the pendulum's period. Immediately after the collision, the pendulum can still be considered vertical, so the line of action of the external forces acting on the system consisting of the pendulum and the bullet passes through the point $P$ of suspension. Therefore, no external torque acts on this point, meaning that the angular momentum of the system does not change during the collision. The angular momentum before the collision is:

$$
N_{1}=m v l
$$

immediately after the collision:

$$
N_{1}=\Theta \omega
$$

where $\omega$ is the angular velocity of the pendulum, and $\Theta$ is the moment of inertia of the system about the point $P$ after the collision:

$$
\Theta=m l^{2}+M_{1} l^{2}+(1 / 3) M l^{2}
$$

We assume that the bullet stopped near the hinge holding $M_{1}$, so its distance from the point $P$ can be considered as $l$. The angular momentum did not change:

$$
N_{1}=N_{2}
$$

Using (1) and (2), we get the equality:

$$
\omega=\frac{m v l}{\Theta}
$$

Next, we apply the work-energy theorem. The work done by the gravitational force on the system equals the change in the system's kinetic energy. Let $\varphi_{0}$ be the maximum deflection angle - here the pendulum is at rest - and let $E(\varphi)$ be the kinetic energy of the pendulum at an angle $\varphi$!

![](https://cdn.mathpix.com/cropped/2024_05_02_64bdd5e063a38b198c9fg-1.jpg?height=610&width=851&top_left_y=1202&top_left_x=629)

Then the change in kinetic energy is:

$$
E\left(\varphi_{0}\right)-E(0)=-(1 / 2) \Theta \omega^{2}
$$

(We assume that the bullet of mass $m$ stopped near the hinge. The period of the physical pendulum consisting of the masses $M_{1}$ and $m$ is therefore very large. Thus, the rotation of $M_{1}$ at the maximum deflection is still negligible, and it does not have its own rotational energy.)

Considering that the center of mass of the bullet and the mass $M_{1}$ is still at nearly the same height at the maximum deflection, and the work is done against the gravitational force, the work done by the gravitational force (see the first figure):

$$
W=-\left(M_{1}+m\right) g h-M g(h / 2)=-g\left[M_{1}+m+(1 / 2) M\right] l\left(1-\cos \varphi_{0}\right)
$$

According to the work-energy theorem:

$$
W=E\left(\varphi_{0}\right)-E(0)
$$

From (5) and (6), we get the following equality:

$$
-(1 / 2) \Theta \omega^{2}=-g l\left(1-\cos \varphi_{0}\left[M_{1}+m+(M / 2)\right]\right.
$$

From this, using (3) and (4), we get for the maximum deflection:

$$
\cos \varphi_{0}=l-\frac{m^{2} v^{2}}{g l\left[M_{1}+m+(1 / 3) M\right]\left(2 M_{1}+2 m+M\right)}
$$

If this formula yields $\cos \varphi < -1$, we cannot speak of a maximum deflection; in this case, the pendulum rotates completely.

Benkö Zsigmond (Szolnok, Verseghy F. Gymnasium, 2nd grade)

Note. The angular velocity of the pendulum immediately after the collision can also be calculated differently. According to Newton's Second Law, the time integral of the force causing the deceleration of the bullet equals the difference between the final and initial momenta. If $\tau$ is the duration of the collision, $\omega$ is the angular velocity after the collision, and $I(t)$ is the momentum at time $t$, then:

$$
\int_{0}^{\tau} F \mathrm{~d} t=I(\tau)-I(0)=m \omega l-m v
$$

The time integral of the torque acting on the pendulum equals the difference between the final and initial angular momenta. According to Newton's Third Law, the reaction force to the force $F$ decelerating the bullet acts on the pendulum. If $\tau$ is very small, the lever arm of $F$ can be taken as $l$:

$$
\int_{0}^{\tau}(-F l) \mathrm{d} l=N(\tau)-N(0)=\left[M_{1}+(1 / 3) M\right] l^{2} \omega
$$

By comparing equations (7) and (8), we get an expression for $\omega$ that agrees with (4).