Answer to Question #102391 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

moment of inertia of sphere:

$I=\frac{2}{5} mr \raisebox{0.45em}{2}$

moment of inertia of a circular disk:

$I'=\frac{1}{2} mR\raisebox{0.45em}{2}$

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

$\frac{I'-I}{t}=\frac{\frac{1}{2} mR\raisebox{0.45em}{2} - \frac{2}{5} mr\raisebox{0.45em}{2}}{t} = m\frac{\frac{1}{2} R\raisebox{0.45em}{2} - \frac{2}{5} r\raisebox{0.45em}{2}}{t}$

$\frac{I'-I}{t}= 100\frac{\frac{1}{2}* 20\raisebox{0.45em}{2} - \frac{2}{5}*10\raisebox{0.45em}{2}}{3600} = 4.44$

Answer: 4.44