In a playground,there is a merry-go-round of | vedclass

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(C) The moment of inertia of the merry-go-round is $I = Mk^2 = 120 \ kg 	imes (3 \ m)^2 = 120 	imes 9 = 1080 \ kg \cdot m^2$. Since the child jumps

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) The moment of inertia of the merry-go-round is $I = Mk^2 = 120 \ kg 	imes (3 \ m)^2 = 120 	imes 9 = 1080 \ kg \cdot m^2$. Since the child jumps tangentially,the initial angular momentum of the system is provided by the child: $L = mvr = 30 \ kg 	imes 5 \ m/s 	imes 4 \ m = 600 \ kg \cdot m^2/s$. The final moment of inertia of the system (merry-go-round + child) is $I_{total} = I + mr^2 = 1080 + 30 	imes (4)^2 = 1080 + 30 	imes 16 = 1080 + 480 = 1560 \ kg \cdot m^2$. By the law of conservation of angular momentum,$L_{initial} = L_{final} \implies 600 = I_{total} \omega$. Therefore,$\omega = \frac{600}{1560} = \frac{60}{156} = \frac{10}{26} \approx 0.3846 \ rad/s$. Rounding to the nearest provided option,the value is $0.4 \ rad/s$.

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