A solid cylinder of mass M and radius R has a light flexible rope wound around it. The rope carries a mass m at its free end. The mass is rest at a height h above the floor. Find the angular velocity of the cylinder at the instant the mass m, after release, strikes the floor. Assume friction to be absent and the cylinder to rotate about its own axis.
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As the mass m descends, its initial potential energy gets converted into the kinetic energy of the falling mass M itself and kinetic energy of the cylinder set into rotation. We thus have $\text{mgh}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{I}\omega^2$ Now I, the moment of intertia of the cylinder about its own axis is $\frac{1}{2}\text{MR}^2\text{ Also }\omega=\frac{\text{v}}{\text{R}}$ 
We thus have $\text{mgh}=\frac{1}{2}\text{mv}^2+\frac{1}{4}\text{MR}^2\frac{\text{v}^2}{\text{R}^2}$ $\Rightarrow\text{v}^2=\frac{2\text{mgh}}{\text{m}+\Big(\frac{\text{M}}{2}\Big)}=\Bigg[\frac{2\text{gh}}{1+\Big(\frac{\text{M}}{2\text{m}}\Big)}\Bigg]$ Thus the require angular velocity, $\omega=\frac{\text{v}}{\text{R}}=\frac{1}{\text{R}}\sqrt{\frac{2\text{gh}}{1+\Big(\frac{\text{M}}{2\text{m}}\Big)}}$
We thus have $\text{mgh}=\frac{1}{2}\text{mv}^2+\frac{1}{4}\text{MR}^2\frac{\text{v}^2}{\text{R}^2}$ $\Rightarrow\text{v}^2=\frac{2\text{mgh}}{\text{m}+\Big(\frac{\text{M}}{2}\Big)}=\Bigg[\frac{2\text{gh}}{1+\Big(\frac{\text{M}}{2\text{m}}\Big)}\Bigg]$ Thus the require angular velocity, $\omega=\frac{\text{v}}{\text{R}}=\frac{1}{\text{R}}\sqrt{\frac{2\text{gh}}{1+\Big(\frac{\text{M}}{2\text{m}}\Big)}}$
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