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 given by: \[ I_{\text{disc}} = \frac{1}{4} M R^2 \] where: \( M \) is the mass of the disc, \( R \) is the radius of the disc.
Step 1: Moment of inertia of the solid sphere about PQ
The moment of inertia of a solid sphere about its center is given by: \[ I_{\text{sphere}} = \frac{2}{5} M R^2 \] Since the sphere is rotating about point \( P \), the parallel axis theorem applies. The moment of inertia about the axis passing through \( P \) is: \[ I_{\text{sphere, PQ}} = I_{\text{sphere}} + M d^2 \] where \( d \) is the distance between the center of the sphere and the point \( P \), which is \( R \). Therefore, the moment of inertia for the sphere about PQ is: \[ I_{\text{sphere, PQ}} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 \]
Step 2: Moment of inertia of the spherical shell about PQ
For a spherical shell, the moment of inertia about its center is: \[ I_{\text{shell}} = \frac{2}{3} M R^2 \] Again, using the parallel axis theorem, the moment of inertia about PQ 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 the moments of inertia of the disc, sphere, and spherical shell: \[ I_{\text{total}} = I_{\text{disc, PQ}} + I_{\text{sphere, PQ}} + I_{\text{shell, PQ}} \] Substituting the values: \[ I_{\text{total}} = \frac{1}{4} M R^2 + \frac{7}{5} M R^2 + \frac{5}{3} M R^2 \]
Step 4: Simplifying the total moment of inertia
The common denominator is 60, so we can rewrite each term as: \[ I_{\text{total}} = \frac{15}{60} M R^2 + \frac{84}{60} M R^2 + \frac{100}{60} M R^2 = \frac{199}{60} M R^2 \]
Step 5: Comparing with the given moment of inertia
We are told that the moment of inertia about PQ is \( \frac{x}{15} I \), where \( I = \frac{1}{4} M R^2 \). Therefore: \[ \frac{x}{15} I = \frac{199}{60} M R^2 \] Substituting \( I = \frac{1}{4} M R^2 \) into the equation: \[ \frac{x}{60} = \frac{199}{60} \quad \Rightarrow \quad x = 199 \]
Thus, the value of \( x \) is 199.

Answer 2

Given the moments of inertia for different objects: For a disk: \[ I_A = \frac{mR^2}{4} \quad I = \frac{mR^2}{4} \] For a solid sphere: \[ I_B = \frac{7}{5} mR^2 \] For a spherical shell: \[ I_C = \frac{5}{3} mR^2 \] The combined moment of inertia \( I_{PQ} \) is: \[ I_{PQ} = mR^2 \left[ \frac{1}{4} + \frac{7}{5} + \frac{5}{3} \right] \] \[ I_{PQ} = mR^2 \left( \frac{199}{4} \right) \times \frac{1}{15} \] Thus, solving for \( x \): \[ \frac{x}{15} \times \frac{mR^2}{4} = \frac{mR^2 \times 199}{4 \times 15} \] \[ \boxed{x = 199} \]