Question

A solid sphere of mass $100\, kg$ and radius $10\, m$ moving in a space becomes a circular disc of radius $20\, m$ 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, $M_{s}=100 \,kg$
radius of solid sphere, $R_{s}=10\, m$
radius of circular disc, $R_{c}=20 \,m$ and time $=1$ hour $=60$ minute $=60 \times 60 \,sec$
Moment of inertia of the solid sphere, $I_{s}-\frac{2}{5} M_{s} R_{s}^{2}=\frac{2}{5} \times 100 \times(10)^{2}-4000 \,kg /m ^{2}$
Similarly,
moment of inertia of the disc, $I_{c}=\frac{1}{2} M_{c} R^{2}$
$=\frac{1}{2} \times 100 \times(20)^{2}=20,000\, kg - m ^{2}$
Rate of change of moment of inertia
$=\frac{I_{c}-I_{s}}{t}$
$-\frac{20000-4000}{60 \times 60}-\frac{16000}{60 \times 60}-\frac{160}{36}$
$=\frac{40}{9} kg - m ^{2} \,s ^{-1}$