# Confusion regarding rolling without slipping

I'm trying to solve the following problem: 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.

• 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? Apr 18, 2019 at 11:52
• I wouldn't think to use conservation of energy as the bug's collision is plastic Apr 18, 2019 at 12:01
• You mean some energy gets lost in "squishing" of the bugs? Apr 18, 2019 at 12:02
• I don't know, but typically when there is a plastic collision conservation of energy does not hold. Apr 18, 2019 at 12:05
• Can I assume the bug velocities are with respect to the floor (so they do not have symmetric collisions)? Apr 19, 2019 at 20:35