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I was doing the following problem:

A solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface. The cube is constrained to rotate about an axis $AB$. A bullet of mass $m$ and speed $v$ is shot at the face opposite $ABCD$ at a height of $4a/3$. The bullet becomes embedded in the cube. Find the minimum value of $v$ required to tip the cube so that it falls on face $ABCD$. Assume $m \ll M$. Diagram of the problem

I wasn't sure how to do this problem, so I checked this website for hints: https://www.physicsforums.com/threads/conservation-of-angular-momentum-help-needed.53833/

Here's the method they suggested (I'm consolidating the solution into one place for simplicity):

First, we need to determine the minimum angular velocity for which the cube will fall on face $ABCD$. For that, it needs enough angular velocity to rotate an angle of 45 degrees, after which gravity will help it down. So, we get the equation (by applying Conservation of Mechanical Energy):

$$Mga\big( \sqrt{2} - 1\big) = \dfrac{1}{2}\left(\dfrac{8Ma^2}{3}\right)\omega_{\text{min}}^2$$

This implies that

$$\omega_{\text{min}} = \sqrt{\dfrac{3g\big( \sqrt{2} - 1\big)}{4a}}$$

Now, we apply Conservation of Angular Momentum to relate the minimum velocity to the minimum angular velocity:

$$mv_{\text{min}}\left(\dfrac{4a}{3}\right) = \left(\dfrac{8Ma^2}{3}\right)\omega_{\text{min}}$$

(we are justified in neglecting the MOI of the bullet because it is given that $m \ll M$.)

After rearranging, we get that $v_{\text{min}}$ is equal to

$$v_{\text{min}} = \left(\dfrac{M}{m}\right)\sqrt{3ga\big(\sqrt{2} - 1\big)}$$

This answer checks out with the one given in the back of the textbook, so I know that the answer (and therefore the method too, most likely) is correct.


What I don't get is, why is angular momentum conserved in this case? Our system is the cube + bullet, and gravity, which acts through the COM of the cube, is exerting an external torque on the system.

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Before the bullet hits, the torque on the block is $0$. The reaction force of the horizontal surface is uniformly spread over the bottom of the block.

As the block rotates, the reaction force shifts to the edge of the block. This creates a torque opposing that of the bullet. The assumption is that the bullet comes to rest very quickly. The change in angular momentum from the reaction force during the collision is $\Delta L_r = \tau_r \Delta t$, where $\tau_r$ is the torque from the reaction force. $\tau_r << \tau_b$, the torque from the bullet. (No wood is crushed by the horizontal surface.) So during the collision, $\Delta L_r << \Delta L_b$, and can be ignored. This allows you to calculate the initial angular momentum of the block + bullet.

After the collision, angular momentum is not conserved. The $\tau_r$ slows the block. If $v = v_{min}$, the block comes to a stop balanced on its edge.

Not that you need it, but this old answer of mine shows pictures of the reaction force and how it relates to torque.

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  • $\begingroup$ It might be useful to note that this is idea is similar to the momentum conservation used in the ballistic pendulum problem. $\endgroup$ Jun 19 '20 at 15:01
  • $\begingroup$ @mmesser314 So basically, angular momentum isn't conserved? Then how am I supposed to relate the (linear) velocity of the bullet to the angular velocity of the block? As for the other thing, yes, you're right. Strictly speaking, the velocity of the bullet would have to be $v = v_{\text{min}} + \epsilon$ where $\epsilon$ is an extremely small positive real number. However, I've never seen that distinction made in Physics problems, which is why I did not make that distinction either. $\endgroup$
    – wermos
    Jun 20 '20 at 14:54
  • $\begingroup$ @BioPhysicist I can certainly see the similarities in the two problems, but in the ballistic pendulum problem, the net external force on the system is zero because the force of gravity on the block is countered by the tension in the rope that the block is hanging from. But then again, gravity was acting on the bullet while it was travelling towards the block... $\endgroup$
    – wermos
    Jun 20 '20 at 15:21
  • $\begingroup$ @TirthankarMazumder In the ballistic pendulum as soon as the bullet hits the block the block will start moving and thus the net force is technically not $0$ during the collision. $\endgroup$ Jun 20 '20 at 17:37
  • $\begingroup$ Ohhh, I finally got the answer. It just clicked :) So basically, the torque due to these other forces is negligible, and that is why we are justified in ignoring it and saying that angular momentum is conserved. $\endgroup$
    – wermos
    Jun 24 '20 at 5:09

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