Answer to Question #93456 in Mechanics | Relativity for Sridhar

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

Answer 1

$$I=\frac{2}{5}mr^2$$

$$I'=\frac{1}{2}mR^2$$

The rate of change of moment of inertia in the process is

$$\frac{I'-I}{t}=\frac{\frac{1}{2}mR^2-\frac{2}{5}mr^2}{t}=m\frac{\frac{1}{2}R^2-\frac{2}{5}r^2}{t}$$

$$\frac{I'-I}{t}=100\frac{\frac{1}{2}20^2-\frac{2}{5}10^2}{3600}=4.44\frac{kg\cdot m^2}{s}$$