Answer:
![\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Comega_0%5E2%28%5Cfrac%7BM%2Bm%7D%7BM%7D%29%7D)
Explanation:
The rotational kinetic energy when the cylinder is with the rope is:
![E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7DI_c%5Comega_0%5E2%2B%5Cfrac%7B1%7D%7B2%7DI_r%5Comega_0%5E2)
where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:
(1)
For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:
![I_c=\frac{1}{2}MR^2\\\\I_r=mR^2](https://tex.z-dn.net/?f=I_c%3D%5Cfrac%7B1%7D%7B2%7DMR%5E2%5C%5C%5C%5CI_r%3DmR%5E2)
Finally, by replacing in (1):
![\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Comega_0%5E2%28%5Cfrac%7BM%2Bm%7D%7BM%7D%29%7D)
hope this helps!!
Mass affects a ball's bounce through kinetic energy. The more mass an object has, the more kinetic energy it has when dropped, due to gravity. How much the ball deforms is based on its chemical makeup, or in this case, elasticity. When the ball deforms, the kinetic energy is converted into potential energy. As many kinds of balls have high amounts of elasticity, the potential energy converts back to kinetic energy when the deformation of the ball returns to its normal state. If the force of impact is too great for the ball to absorb, it may collapse and lose its bounce as the energy is dissipated in a different manner.
Option B would be the right one