Rotational Motion Question 2
Question 2 - 24 January - Shift 2
A uniform solid cylinder with radius $R$ and length $L$ has moment of inertia $I_1$, about the axis of cylinder. A concentric solid cylinder of radius $R^{\prime}=\frac{R}{2}$ and length $L^{\prime}=\frac{L}{2}$ is caned out of the original cylinder. If $I_2$ is the moment of inertia of the carved out portion ot the cylinder then $\frac{I_1}{I_2}=$
(Both $I_1$ and $I_2$ are about the axis of the cylinder)
Show Answer
Answer: (32)
Solution:
Formula: Moment of inertia
$I_1=\frac{m_1 R^{2}}{2} \quad I_2=\frac{m_2(R / 2)^{2}}{2}$
$\frac{I_1}{I_2}=\frac{4 m_1}{m_2}=\frac{4 \cdot \rho \pi R^{2} \ell}{\rho \cdot \frac{\pi R^{2}}{4} \times \frac{\ell}{2}} \Rightarrow \frac{I_1}{I_2}=32$
