Question

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.

Answer 1

Answer:

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

$E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2$

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:

$E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})}$ (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

$I_c=\frac{1}{2}MR^2\\\\I_r=mR^2$

Finally, by replacing in (1):

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

hope this helps!!