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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 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 at 12:01
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    $\begingroup$ You mean some energy gets lost in "squishing" of the bugs? $\endgroup$ – denklo Apr 18 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 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 at 20:35
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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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