Question

A uniform solid cylinder with radius \(R\) and length \(L\) has a moment of inertia \(I_1\) about the axis of the cylinder. A concentric solid cylinder of radius \(R' = \frac{R}{2}\) and length \(L' = \frac{L}{2}\) is carved out of the original cylinder. If \(I_2\) is the moment of inertia of the carved-out portion, then \(\frac{I_1}{I_2} =\)
(Both \(I_1\) and \(I_2\) are about the axis of the cylinder.)

Answer 1

Correct Answer: 32

The moment of inertia for a uniform solid cylinder about its axis is:

\[ I = \frac{1}{2} m R^2 \]

Step 1: Moment of inertia of the original cylinder (\(I_1\)):

\[ I_1 = \frac{1}{2} m_1 R^2 \]

Step 2: Moment of inertia of the carved-out cylinder (\(I_2\)):

The mass of the carved-out portion (\(m_2\)) is proportional to its volume:

\[ m_2 = \rho \cdot \text{Volume} = \rho \cdot \pi \left(\frac{R}{2}\right)^2 \cdot \frac{L}{2} = \frac{\rho \pi R^2 L}{8} \]

Moment of inertia:

\[ I_2 = \frac{1}{2} m_2 \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{\rho \pi R^2 L}{8} \cdot \frac{R^2}{4} = \frac{\rho \pi R^4 L}{64} \]

Step 3: Ratio of moments of inertia (\(I_1/I_2\)):

The mass of the original cylinder (\(m_1\)) is:

\[ m_1 = \rho \cdot \pi R^2 L \]

Moment of inertia of the original cylinder:

\[ I_1 = \frac{1}{2} m_1 R^2 = \frac{1}{2} \cdot \rho \pi R^2 L \cdot R^2 = \frac{\rho \pi R^4 L}{2} \]

The ratio is:

\[ \frac{I_1}{I_2} = \frac{\frac{\rho \pi R^4 L}{2}}{\frac{\rho \pi R^4 L}{64}} = \frac{64}{2} = 32 \]

Answer 2

The correct answer is 32.

I1​=2m1​R2​I2​=2m2​(R/2)2​ 
I2​I1​​=m2​4m1​​=ρ⋅4πR2​×2ℓ​4⋅ρπR2ℓ​⇒I2​I1​​=32