Question

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotating in similar way, is \( n \) times higher than the moment of inertia of the given ring. Here, \( n = \):

Hint:

In rotational motion, the moment of inertia depends on the mass distribution relative to the axis of rotation.

Answer 1

Correct Answer: 0.8

Step 1: Moment of Inertia of the Ring

For a ring rotating about its diameter: \[ I_{\text{ring}} = \frac{MR^2}{2} \]

Step 2: Moment of Inertia of the Disc

For a solid disc rotating about its diameter: \[ I_{\text{disc}} = \frac{MR^2}{4} \]

According to the problem, the moment of inertia of the disc is 2.5 times the moment of inertia of the ring: \[ I_{\text{disc}} = 2.5 I_{\text{ring}} \]

Substituting the values: \[ \frac{MR^2}{4} = 2.5 \times \frac{MR^2}{2} \]

Solving this will confirm the relationship.

Step 3: Moment of Inertia of the Solid Sphere

For a solid sphere rotating about its diameter: \[ I_{\text{sphere}} = \frac{2MR^2}{5} \]

Step 4: Finding the Ratio \( n \)

The moment of inertia of the solid sphere is \( n \) times the moment of inertia of the ring. \[ I_{\text{sphere}} = n I_{\text{ring}} \]

Substituting the values: \[ \frac{2MR^2}{5} = n \times \frac{MR^2}{2} \] \[ n = \frac{2}{5} \div \frac{1}{2} = \frac{2}{5} \times \frac{2}{1} = \frac{4}{5} \]

Therefore, \[ \boldsymbol{n = \frac{4}{5}} \]

Answer 2

Step 1: Understand the given information.
We are comparing the moments of inertia (M.I.) of three objects — a ring, a solid disc, and a solid sphere — all rotating about a diameter. It is given that:
- The M.I. of the solid disc about its diameter is 2.5 times the M.I. of the ring about its diameter.
We need to find the ratio \( n \) such that:
\[ I_{\text{sphere}} = n \, I_{\text{ring}} \]

Step 2: Write standard formulas for moments of inertia about diameter.
- For a ring of mass \( M \) and radius \( R \):
\[ I_{\text{ring, diameter}} = \frac{1}{2} M R^2 \]
- For a solid disc about its diameter:
\[ I_{\text{disc, diameter}} = \frac{1}{4} M R^2 \] - For a solid sphere about its diameter:
\[ I_{\text{sphere, diameter}} = \frac{2}{5} M R^2 \]

Step 3: Use the given condition for the disc and the ring.
We are told that: \[ I_{\text{disc, diameter}} = 2.5 \times I_{\text{ring, diameter}} \] Substitute the known expressions: \[ \frac{1}{4} M R^2 = 2.5 \times \frac{1}{2} M R^2 \] Since both have same mass \( M \) and radius \( R \), this confirms that the comparison is based on the rotational axis and proportional relationships hold true.

Step 4: Find the ratio for the sphere.
We need: \[ n = \frac{I_{\text{sphere, diameter}}}{I_{\text{ring, diameter}}} \] Substitute: \[ n = \frac{\frac{2}{5} M R^2}{\frac{1}{2} M R^2} = \frac{\frac{2}{5}}{\frac{1}{2}} = \frac{2}{5} \times \frac{2}{1} = \frac{4}{5} = 0.8 \]

Step 5: Conclusion.
The moment of inertia of the solid sphere is \( 0.8 \) times that of the ring.

Final Answer:
\[ \boxed{n = 0.8} \]