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Question: Answered & Verified by Expert
A solid sphere of \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~m}\) moving in a space becomes a circular disc of radius \(20 \mathrm{~m}\) in one hour. Then the rate of change of moment of inertia in the process is
PhysicsRotational MotionAP EAMCETAP EAMCET 2019 (20 Apr Shift 1)
Options:
  • A \(\frac{40}{9} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\)
  • B \(\frac{10}{9} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\)
  • C \(\frac{50}{9} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\)
  • D \(\frac{25}{9} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\)
Solution:
1410 Upvotes Verified Answer
The correct answer is: \(\frac{40}{9} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\)
Given, mass of solid sphere, \(M_s=100 \mathrm{~kg}\)
radius of solid sphere, \(R_s=10 \mathrm{~m}\)
radius of circular disc, \(R_c=20 \mathrm{~m}\) and time \(=1\) hour \(=60\) minute \(=60 \times 60 \mathrm{sec}\)
Moment of inertia of the solid sphere,
\(I_s=\frac{2}{5} M_s R_s^2=\frac{2}{5} \times 100 \times(10)^2=4000 \mathrm{~kg}-\mathrm{m}^2\)
Similarly,
moment of inertia of the disc, \(I_c=\frac{1}{2} M_c R^2\)
\(=\frac{1}{2} \times 100 \times(20)^2=20,000 \mathrm{~kg}-\mathrm{m}^2\)
Rate of change of moment of inertia \(=\frac{I_c-I_s}{t}\)
\(\begin{aligned}
& =\frac{20000-4000}{60 \times 60}=\frac{16000}{60 \times 60}=\frac{160}{36} \\
& =\frac{40}{9} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}\)

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