Question

Find the ratio of moment of inertia of ring to disc, about an axis passing through its centre?

• A. 1 : 2
• B. 2 : 1
• C. 3 : 2
• D. 2 : 3
• E. 1 : 1

Hint:

A ring has all its mass concentrated at the maximum distance $R$ from the center, so it has a higher moment of inertia ($MR^2$) compared to a solid disc ($0.5MR^2$) where the mass is spread continuously from the center out to $R$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The moment of inertia (\(I\)) is a measure of an object's resistance to changes in its rotation rate. It depends on the object's mass and how that mass is distributed relative to the axis of rotation. The standard "axis passing through its centre" usually implies an axis perpendicular to the plane of the ring or disc.
Step 2: Key Formula or Approach:
Recall the standard formulas for moment of inertia for objects of mass \(M\) and radius \(R\):
- For a thin circular ring about central perpendicular axis: \(I_{ring} = M R^2\)
- For a uniform circular disc about central perpendicular axis: \(I_{disc} = \frac{1}{2} M R^2\)
We must calculate the ratio \(I_{ring} / I_{disc}\).
Step 3: Detailed Explanation:
Assuming both the ring and the disc have the same total mass (\(M\)) and the same radius (\(R\)) for a fair comparison:
Moment of inertia of the ring, \(I_{ring} = M R^2\)
Moment of inertia of the disc, \(I_{disc} = \frac{1}{2} M R^2\)
Calculate the ratio:
\[ \text{Ratio} = \frac{I_{ring}}{I_{disc}} \]
\[ \text{Ratio} = \frac{M R^2}{\frac{1}{2} M R^2} \]
Cancel out the common \(M R^2\) terms:
\[ \text{Ratio} = \frac{1}{\frac{1}{2}} \]
\[ \text{Ratio} = 2 \]
Expressed as a ratio, this is 2 : 1.
Step 4: Final Answer:
The ratio of the moment of inertia of a ring to a disc is 2 : 1.