Question

A, B and C are disc, solid sphere and spherical shell respectively with the same radii and masses. These masses are placed as shown in the figure.

The moment of inertia of the given system about PQ is $ \frac{x}{15} I $, where $ I $ is the moment of inertia of the disc about its diameter. The value of $ x $ is:

Hint:

When dealing with systems involving different objects, use the parallel axis theorem to calculate the moment of inertia about any point other than the center of mass.

Answer 1

Correct Answer: 199

The moment of inertia of a disc about its diameter is \( I_{\text{disc}} = \frac{1}{4} M R^2 \), where \( M \) is the disc's mass and \( R \) is its radius.
Step 1: Moment of inertia of the solid sphere about PQ
The moment of inertia of a solid sphere about its center is \( I_{\text{sphere}} = \frac{2}{5} M R^2 \). Applying the parallel axis theorem for rotation about point \( P \), with distance \( d=R \), the moment of inertia is \( I_{\text{sphere, PQ}} = I_{\text{sphere}} + M d^2 = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 \).
Step 2: Moment of inertia of the spherical shell about PQ
The moment of inertia of a spherical shell about its center is \( I_{\text{shell}} = \frac{2}{3} M R^2 \). Using the parallel axis theorem for rotation about PQ, the moment of inertia is \( I_{\text{shell, PQ}} = \frac{2}{3} M R^2 + M R^2 = \frac{5}{3} M R^2 \).
Step 3: Moment of inertia of the system about PQ
The total moment of inertia of the system is the sum of individual moments of inertia: \( I_{\text{total}} = I_{\text{disc, PQ}} + I_{\text{sphere, PQ}} + I_{\text{shell, PQ}} = \frac{1}{4} M R^2 + \frac{7}{5} M R^2 + \frac{5}{3} M R^2 \).
Step 4: Simplifying the total moment of inertia
With a common denominator of 60, the total moment of inertia simplifies to \( I_{\text{total}} = \frac{15}{60} M R^2 + \frac{84}{60} M R^2 + \frac{100}{60} M R^2 = \frac{199}{60} M R^2 \).
Step 5: Comparing with the given moment of inertia
Given that the moment of inertia about PQ is \( \frac{x}{15} I \), where \( I = \frac{1}{4} M R^2 \), we have \( \frac{x}{15} I = \frac{199}{60} M R^2 \). Substituting \( I = \frac{1}{4} M R^2 \) yields \( \frac{x}{60} = \frac{199}{60} \), thus \( x = 199 \).
The value of \( x \) is 199.