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Physics Question on Moment Of Inertia

A solid sphere of mass 100 kg100\, kg and radius 10 m10\, m moving in a space becomes a circular disc of radius 20 m20\, m in one hour. Then the rate of change of moment of inertia in the process is

A

409kg  m2  s−1\frac{40}{9} kg \; m^2 \; s^{-1}

B

109kg  m2  s−1\frac{10}{9} kg \; m^2 \; s^{-1}

C

509kg  m2  s−1\frac{50}{9} kg \; m^2 \; s^{-1}

D

259kg  m2  s−1\frac{25}{9} kg \; m^2 \; s^{-1}

Answer

409kg  m2  s−1\frac{40}{9} kg \; m^2 \; s^{-1}

Explanation

Solution

Given, mass of solid sphere, Ms=100 kgM_{s}=100 \,kg
radius of solid sphere, Rs=10 mR_{s}=10\, m
radius of circular disc, Rc=20 mR_{c}=20 \,m and time =1=1 hour =60=60 minute =60×60 sec=60 \times 60 \,sec
Moment of inertia of the solid sphere, Is−25MsRs2=25×100×(10)2−4000 kg/m2I_{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, Ic=12McR2I_{c}=\frac{1}{2} M_{c} R^{2}
=12×100×(20)2=20,000 kg−m2=\frac{1}{2} \times 100 \times(20)^{2}=20,000\, kg - m ^{2}
Rate of change of moment of inertia
=Ic−Ist=\frac{I_{c}-I_{s}}{t}
−20000−400060×60−1600060×60−16036-\frac{20000-4000}{60 \times 60}-\frac{16000}{60 \times 60}-\frac{160}{36}
=409kg−m2 s−1=\frac{40}{9} kg - m ^{2} \,s ^{-1}