A circular platform of moment of inertia 100 | vedclass

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(B) According to the principle of Conservation of Angular Momentum $(COAM)$,the angular momentum remains constant as no external torque acts on the sy

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$18$

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(B) According to the principle of Conservation of Angular Momentum $(COAM)$,the angular momentum remains constant as no external torque acts on the system.$L_1 = L_2$$I_1 \omega_1 = I_2 \omega_2$Initially,the boy is at the centre,so his moment of inertia is $0$. The total moment of inertia is $I_1 = 100 \ kg\ m^2$.The initial angular velocity is $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{10} \ rad/s$.When the boy moves $2 \ m$ away from the centre,his moment of inertia becomes $I_{boy} = mr^2 = 20 	imes (2)^2 = 80 \ kg\ m^2$.The new total moment of inertia is $I_2 = 100 + 80 = 180 \ kg\ m^2$.Let the new time period be $T_2$,so $\omega_2 = \frac{2\pi}{T_2}$.Substituting these into the $COAM$ equation:$100 	imes \frac{2\pi}{10} = 180 	imes \frac{2\pi}{T_2}$$10 = \frac{180}{T_2}$$T_2 = \frac{180}{10} = 18 \ s$.

$A$ circular platform of moment of inertia $100 \ kg\ m^2$ with a boy of mass $20 \ kg$ standing at the centre completes one rotation in $10 \ s$. If he walks $2 \ m$ away from the centre along the radius,then one rotation (with respect to the ground) will be completed in $.... \ s$.

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