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