moment of inertia of sphere:
I=25mr2I=\frac{2}{5} mr \raisebox{0.45em}{2}I=52mr2
moment of inertia of a circular disk:
I′=12mR2I'=\frac{1}{2} mR\raisebox{0.45em}{2}I′=21mR2
The rate of change of moment of inertia in the process is:
I′−It=12mR2−25mr2t=m12R2−25r2t\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}tI′−I=t21mR2−52mr2=mt21R2−52r2
I′−It=10012∗202−25∗1023600=4.44\frac{I'-I}{t}= 100\frac{\frac{1}{2}* 20\raisebox{0.45em}{2} - \frac{2}{5}*10\raisebox{0.45em}{2}}{3600} = 4.44tI′−I=100360021∗202−52∗102=4.44
Answer: 4.44
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