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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?

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closed as off-topic by John Rennie, Jon Custer, user197851, Kyle Kanos, ZeroTheHero Dec 21 '18 at 16:05

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ – John Rennie Dec 20 '18 at 16:37
  • $\begingroup$ Hi, while I've got the mentioned result working on some homework problem, I thought I "narrowed it down to the specific concept giving a trouble". Sorry, if it still looks like "do my homework" question. $\endgroup$ – vov Dec 20 '18 at 16:52
  • $\begingroup$ Your solution is right. As a suggestion, what's the meaning of $\omega\,d$? Could this be used to find solution in a much simpler way? $\endgroup$ – Elio Fabri Dec 20 '18 at 16:53
  • $\begingroup$ $\omega d$ looks like linear velocity of $m_2+m_0$ relatively to $m_1$? Don't get how it help here $\endgroup$ – vov Dec 20 '18 at 17:11
  • $\begingroup$ Probably I see now what you mean. Resulting $\omega$ of stick rotating around $CM$ is the same as $\omega$ of rotation around $m_1$. So, it doesn't depend on $m_1$ ($m_1$ could be even fixed). Something like this? $\endgroup$ – vov Dec 20 '18 at 17:17
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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.

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  • $\begingroup$ Thanks! Your answer and provided links made it clear. $\endgroup$ – vov Dec 21 '18 at 17:46

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