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I'm trying to solve the following problem: enter image description here

Where the rolling uniform disk of mass $M$ and radius has initial velocity of $v_0$ and angular velocity $\omega_0$ (it is rolling without slipping, but the ground has friction coefficients), and each of the bugs has a mass of $\frac{M}{2}$ and speed $2v_0$ in opposite directions. The bugs collide with the disk as shown in the picture. I'm being asked what is the velocity and angular velocity of the disk a slight moment after the collision. I applied conservation of momentum and got $v=\frac{v_0}{2}$, and I tried conservation of angular momentum and got $\omega = -\omega_0$. This seems to conflict with the rolling without slipping condition as $v \ne \omega R$ anymore. I've applied conservation of momentum on the grounds that a slight moment after the collision the friction's impulse is very very small.

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  • $\begingroup$ How about conservation of energy: $M\frac{(\omega_0R)^2}{2} + I\frac{\omega_0^2}{2} + 2M\frac{(2\omega_0R)^2}{2}= M\frac{(\omega R)^2}{2} + I\frac{\omega^2}{2}$ where $I$ is the moment of inertia of the disc? $\endgroup$
    – denklo
    Apr 18, 2019 at 11:52
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    $\begingroup$ I wouldn't think to use conservation of energy as the bug's collision is plastic $\endgroup$
    – Yizhar Amir
    Apr 18, 2019 at 12:01
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    $\begingroup$ You mean some energy gets lost in "squishing" of the bugs? $\endgroup$
    – denklo
    Apr 18, 2019 at 12:02
  • $\begingroup$ I don't know, but typically when there is a plastic collision conservation of energy does not hold. $\endgroup$
    – Yizhar Amir
    Apr 18, 2019 at 12:05
  • $\begingroup$ Can I assume the bug velocities are with respect to the floor (so they do not have symmetric collisions)? $\endgroup$
    – BowlOfRed
    Apr 19, 2019 at 20:35

2 Answers 2

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I think the solution is quite right since impulse of friction is quite small. Rolling is certainly disturbed, but will be restored by the friction after certain time, which can be calculated by applying the rolling condition and conservation of angular momentum about the point of contact.

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here the way the bugs are colliding,you can directly conserve angular moment and get that the torque of impulse while collision will be opposite to eacg other hence the angular momentum will remain constant and not change. while as talking about conservation of linear momentum you have done something wrong.

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    $\begingroup$ Hello and welcome to Physics Stack Exchange! Would you mind clarifying your answer a bit? It's a little unclear what you mean. $\endgroup$
    – auden
    Oct 29, 2019 at 5:44

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