Answer:
[tex]\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}[/tex]
Explanation:
The rotational kinetic energy when the cylinder is with the rope is:
[tex]E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2[/tex]
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:
[tex]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})}[/tex] (1)
For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:
[tex]I_c=\frac{1}{2}MR^2\\\\I_r=mR^2[/tex]
Finally, by replacing in (1):
[tex]\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}[/tex]
hope this helps!!
