Question

A solid sphere of 80 kg and radius 15 m moving in a space becomes a circular disc of radius 20 m in 1 hour. The rate of change of moment of inertia in this process is:

• A. \( \frac{30}{9} \, {kg} \, {m}^2 \, {s}^{-1} \)
• B. \( \frac{25}{9} \, {kg} \, {m}^2 \, {s}^{-1} \)
• C. \( \frac{10}{9} \, {kg} \, {m}^2 \, {s}^{-1} \)
• D. \( \frac{22}{9} \, {kg} \, {m}^2 \, {s}^{-1} \)

Hint:

When calculating the rate of change of moment of inertia, use the formula for the moment of inertia for the corresponding object (sphere or disc) and divide the change by the time taken to get the rate.

Answer 1

The Correct Option is D

Given, mass of solid sphere \( = 80 \) kg
Radius of solid sphere, \( R_s = 15 \) m
Radius of circular disc, \( R_c = 20 \) m
Time \( = 1 \) hour \( = 60 \) minutes \( = 60 \times 60 \) sec
Step 1: Moment of Inertia of Solid Sphere
\[ I_s = \frac{2}{5} M R^2 \] \[ I_s = \frac{2}{5} \times 80 \times (15)^2 \] \[ I_s = 7200 { kg} \cdot {m}^2 \] Step 2: Moment of Inertia of Circular Disc
\[ I_c = \frac{1}{2} M R_c^2 \] \[ I_c = \frac{1}{2} \times 80 \times (20)^2 \] \[ I_c = 16000 { kg} \cdot {m}^2 \] Step 3: Rate of Change of Moment of Inertia
\[ \frac{dI}{dt} = \frac{I_c - I_s}{t} \] \[ \frac{dI}{dt} = \frac{16000 - 7200}{60 \times 60} \] \[ \frac{dI}{dt} = \frac{22}{9} { kg} \cdot {m}^2 {s}^{-1} \]