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AP EAMCET201920 Apr 2019Morning ShiftPhysicsRotational MotionActual

A solid sphere of (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

Options

  1. A( 40 9 ~kg ~m ^2 ~s ⁻¹ )
  2. B( 10 9 ~kg ~m ^2 ~s ⁻¹ )
  3. C( 50 9 ~kg ~m ^2 ~s ⁻¹ )
  4. D( 25 9 ~kg ~m ^2 ~s ⁻¹ )

Correct answer

A. ( 40 9 ~kg ~m ^2 ~s ⁻¹ )

Step-by-step solution

Given, mass of solid sphere, (M_s=100 ~kg ) radius of solid sphere, (R_s=10 ~m ) radius of circular disc, (R_c=20 ~m ) and time (=1 ) hour (=60 ) minute (=60 60 sec ) Moment of inertia of the solid sphere, (I_s= 2 5 M_s R_s^2= 2 5 100 (10)^2=4000 ~kg - m ^2 ) Similarly, moment of inertia of the disc, (I_c= 1 2 M_c R^2 ) (= 1 2 100 (20)^2=20,000 ~kg - m ^2 ) Rate of change of moment of inertia (= I_c-I_s t ) ( aligned & = 20000-4000 60 60 = 16000 60 60 = 160 36 & = 40 9 ~kg - m ^2 ~s ⁻¹ aligned )

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