Physics - Rotation Question with Solution | TestHub

Suppose the masses $m_1$ and $m_2$ are attached to the ends of a massless rod. $C_1$ is the axis of rotation which is at a distance $R$ from each of the masses. $\therefore$ the moment of inertia of the system, $$\mathrm{I}=m_1{R}^2+m_2{R}^2=\left(m_1+m_2\right)R^2$$ Now if the distance between the masses is doubled, then the moment of inertia $$\Gamma =m_1(2R{)}^2+m_2(2R{)}^2=\left(m_1+m_2\right)4R^2$$ Since the same torque is applied in both the cases, $$\begin{array}{rl}& \tau =I\alpha =I^{\prime }{\alpha }^{\prime }\\ & \Rightarrow {\alpha }^{\prime }=\frac{I\alpha }{I^{\prime }}=\frac{\left(m_1+m_2\right)R^2\alpha }{\left(m_1+m_2\right)4R^2}=\frac{\alpha }{4}\end{array}$$