Physics - Rotation Question with Solution | TestHub

The moment of inertia of the solid cylinder about its central axis is $$I_{cyl}=\frac{1}{2}M{R}^2$$. When the point mass $m$ is attached to the rim, its moment of inertia about the same axis is $$I_m=m{R}^2$$. By conservation of angular momentum, $I_1{\omega }_1=I_2{\omega }_2$. So, $$\left(\frac{1}{2}M{R}^2\right){\omega }_0=\left(\frac{1}{2}M{R}^2+m{R}^2\right){\omega }_{new}$$. Dividing by $R^2$, we get $$\frac{1}{2}M{\omega }_0=\left(\frac{1}{2}M+m\right){\omega }_{new}$$. Solving for ${\omega }_{new}$ gives $${\omega }_{new}=\frac{\frac{1}{2}M{\omega }_0}{\frac{1}{2}M+m}=\frac{M{\omega }_0}{M+2m}$$.