Question

A solid sphere of mass $100kg$ and radius $10m$ moving in a space becomes a circular disc of radius $20m$ in one hour. Then the rate of change of moment of inertia in the process is

• A. $\frac{40}{9}kg{m}^2{s}^{-1}$
• B. $\frac{10}{9}kg{m}^2{s}^{-1}$
• C. $\frac{50}{9}kg{m}^2{s}^{-1}$
• D. $\frac{25}{9}kg{m}^2{s}^{-1}$

Answer 1

Answer: (A)

Given, mass of solid sphere, $M_s=100kg$
radius of solid sphere, $R_s=10m$
radius of circular disc, $R_c=20m$ and time $=1$ hour $=60$ minute $=60\times 60sec$
Moment of inertia of the solid sphere, $I_s-\frac{2}{5}{M}_s{R}_s^2=\frac{2}{5}\times 100\times (10{)}^2-4000kg/m^2$
Similarly,
moment of inertia of the disc, $I_c=\frac{1}{2}{M}_c{R}^2$
$=\frac{1}{2}\times 100\times (20{)}^2=20,000kg-m^2$
Rate of change of moment of inertia
$=\frac{I_c-I_s}{t}$
$-\frac{20000-4000}{60\times 60}-\frac{16000}{60\times 60}-\frac{160}{36}$
$=\frac{40}{9}kg-m^2{s}^{-1}$