Question

A thin circular disc and a uniform thin circular ring have their masses and radii in the ratio 2 : 1 and 1 : 2 respectively. The ratio of their moments of inertia about their respective diameters is

• A. 4 : 1
• B. 1 : 1
• C. 1 : 4
• D. 2 : 1
• E. 1 : 2

Hint:

Always double check if the axis is "diameter" or "tangent" or "axis perpendicular to plane". For a ring, \(I_{diameter} = \frac{1}{2}I_{axis}\). For a disc, \(I_{diameter} = \frac{1}{2}I_{axis}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The moment of inertia (MI) depends on the mass, radius, and distribution of mass about the axis of rotation. The MI about the diameter is different from the MI about the central axis.

Step 2: Key Formula or Approach:
1. MI of Disc about diameter: \(I_d = \frac{1}{4}mr^2\).
2. MI of Ring about diameter: \(I_r = \frac{1}{2}MR^2\).

Step 3: Detailed Explanation:
Let disc have mass \(m_1, r_1\) and ring have mass \(m_2, r_2\).
Given: \(\frac{m_1}{m_2} = \frac{2}{1}\) and \(\frac{r_1}{r_2} = \frac{1}{2}\).
Ratio of moments of inertia:
\[ \frac{I_{disc}}{I_{ring}} = \frac{\frac{1}{4}m_1r_1^2}{\frac{1}{2}m_2r_2^2} \]
\[ \frac{I_{disc}}{I_{ring}} = \frac{1}{2} \cdot \left(\frac{m_1}{m_2}\right) \cdot \left(\frac{r_1}{r_2}\right)^2 \]
Substitute the given ratios:
\[ \frac{I_{disc}}{I_{ring}} = \frac{1}{2} \cdot \left(\frac{2}{1}\right) \cdot \left(\frac{1}{2}\right)^2 \]
\[ \frac{I_{disc}}{I_{ring}} = 1 \cdot \frac{1}{4} = \frac{1}{4} \]

Step 4: Final Answer:
The ratio of their moments of inertia is 1 : 4.