JEE Main 2023 · Physics · Rotational Motion · hard

A horizontal circular platform of mass 100 kg and radius 2 m rotates freely about its vertical axis at 2 rad s^-1 with a 50 kg man standing at its…

Numerical answer

A horizontal circular platform of mass

100\ \mathrm{kg}

and radius

2\ \mathrm{m}

rotates freely about its vertical axis at

2\ \mathrm{rad\,s^{-1}}

with a

50\ \mathrm{kg}

man standing at its rim. He walks slowly to the centre. The work done by the man (in J) during the walk, treating the platform as a uniform disc and the man as a point mass, is

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Answer: 800

Platform:

I_p=\tfrac12MR^2=\tfrac12(100)(4)=200\ \mathrm{kg\,m^2}

. Man at the rim:

mR^2=50\times4=200

. Initially

I_1=400

\omega_1=2

. No external torque, so

L=400\times2=800

is conserved; at the centre

I_2=200

and

\omega_2=4\ \mathrm{rad\,s^{-1}}

. Work done by the man equals the change in kinetic energy:

\tfrac12(200)(16)-\tfrac12(400)(4)=1600-800=800\ \mathrm J

. (Angular momentum is conserved but kinetic energy is not.)

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