Angular velocity of stick after bullet hits it

Question

A weightless rigid stick with length $d$ lying on friction-less surface. Mass $m_1$ is attached to one end of the stick and mass $m_2$ is attached to its another end. Bullet with mass $m_0$ and velocity $v_0$ flies perpendicular to the stick, hits mass $m_2$ and gets stuck inside it. As result stick starts rotating around its centre of mass ($CM$) (it gets translation as well, but it's not relevant to my questions).

After the bullet hits the stick, $CM$ of the resulting system locates at distance $l$ from $m_2$ (and $m_0$ inside it): $$l = \frac {m_1 d}{m_1+m_2+m_0}$$ Moment of inertia of the resulting system is: $$I = \frac {m_1 (m_2+m_0) d^2}{m_1+m_2+m_0}$$

Before the bullet hits, its angular momentum relatively to $CM$ is: $$L_{before} = m_0 v_0 l = m_0 v_0 \frac {m_1 d}{m_1+m_2+m_0}$$ After the bullet hits, the system angular momentum relatively to $CM$ is: $$L_{after} = I \omega = \frac {m_1 (m_2+m_0) d^2}{m_1+m_2+m_0} \omega$$ Due to angular momentum conservation $L_{before} = L_{after}$ and thus: $$\omega = \frac {\frac {m_0 v_0 m_1 d}{m_1+m_2+m_0}} {\frac {m_1 (m_2+m_0) d^2}{m_1+m_2+m_0}}$$ and from here I got: $$\omega = \frac {m_0 v_0} {d (m_2+m_0)}$$ For me it seems strange that resulting $\omega$ doesn't depend on $m_1$.

What do I miss here?

Answer 1

You are correct.

The angular momentum acquired after the impact is a result of the impulse exchange $J$ at a distance $c = d \frac{m_1}{m_1+m_2}$ from the center of mass (distance between point of impact and COM). So the equation for the change in rotation is

$$ \Delta \omega = {I}_C^{-1} c J $$ where ${I}_C = \left( \frac{m_1 m_2}{m_1 + m_2} \right) d^2$ is the mass moment of inertia of the system of two masses. This simplifies to $$ \Delta \omega = \frac{1}{m_2 d} J $$

Now you need to find the impulse $J$ (linear momentum exchange). One easy way of doing this is finding the reduced mass $\mu$ of the exchange and stating the law of impact $ J = (1+\epsilon) \mu\, v_0$ with a coefficient of restitution $\epsilon =0$.

The reduced mass between a particle with mass $m_0$ and a rigid body with properties stated above is

$$ \mu = \frac{1}{ \frac{1}{m_0} + \frac{1}{m_1+m_2} + \frac{c^2}{\mathrm{I}_C} } = \frac{m_0 m_2}{m_0+m_2} $$

Putting together the impulse is $$ J = \frac{m_0 m_2}{m_0 + m_2} v_0 $$ and thus the change in rotational speed

$$ \boxed{ \Delta \omega = \frac{m_0}{m_0 + m_2 } \frac{v_0}{d} } $$

And yes, the result does not depend on $m_1$. This is a direct result of $m_1$ simplifying out of the reduced mass calculation. This is probably because the resulting center of rotation is at $m_1$ which means it has zero motion and zero contribution to the kinetic energy.

The relationship between the impact axis (movement line of $m_0$) and the instant pivot point is called the pole-polar relationship. The location of the pole is $$ \ell = c + \frac{I_C}{(m_1+m_2) c} = \frac{m_1}{m_1+m_2} d + \frac{ \frac{m_1 m_2}{m_1+m_2} d^2 }{ (m_1+m_2) \frac{m_1}{m_1+m_2} d } = d$$

This confirms that the impact point is on the percussion axis when pivoting about $m_1$.

_{Link to another similar post with the same method.}