A person of 80, kg mass is standing on the ri | vedclass

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(D) Since there is no external torque acting on the system,the angular momentum is conserved: $L_i = L_f$.Initial angular momentum $L_i = I_i \omega_i

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Since there is no external torque acting on the system,the angular momentum is conserved: $L_i = L_f$.Initial angular momentum $L_i = I_i \omega_i = (I_{	ext{person}} + I_{	ext{platform}}) \omega_i$.$I_{	ext{person}} = mR^2 = 80R^2$ and $I_{	ext{platform}} = \frac{1}{2}MR^2 = \frac{1}{2} 	imes 200 	imes R^2 = 100R^2$.So,$L_i = (80R^2 + 100R^2) \omega_i = 180R^2 \omega_i$.When the person reaches the centre,their distance from the axis becomes $0$,so $I_{	ext{person}} = 0$.Final angular momentum $L_f = (0 + 100R^2) \omega_f = 100R^2 \omega_f$.Equating $L_i = L_f$: $180R^2 \omega_i = 100R^2 \omega_f$.$180 	imes 5 = 100 	imes \omega_f$.$\omega_f = \frac{900}{100} = 9\, rpm$.

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