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 $I = Mk^2 = 120 \ kg 	imes (3 \ m)^2 = 120 	imes 9 = 1080 \ kg \cdot m^2$. When the child jumps o

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) The moment of inertia of the merry-go-round is $I = Mk^2 = 120 \ kg 	imes (3 \ m)^2 = 120 	imes 9 = 1080 \ kg \cdot m^2$. When the child jumps on the rim,the child acts as a point mass at distance $R = 4 \ m$. The moment of inertia of the child is $I_c = mR^2 = 30 \ kg 	imes (4 \ m)^2 = 30 	imes 16 = 480 \ kg \cdot m^2$. The total moment of inertia of the system is $I_{total} = I + I_c = 1080 + 480 = 1560 \ kg \cdot m^2$. By the law of conservation of angular momentum,the initial angular momentum of the child equals the final angular momentum of the system: $L_i = L_f$. $mvr = I_{total} \omega$. $30 \ kg 	imes 5 \ m/s 	imes 4 \ m = 1560 \ kg \cdot m^2 	imes \omega$. $600 = 1560 \omega$. $\omega = \frac{600}{1560} = \frac{60}{156} = \frac{10}{26} \approx 0.3846 \ rad/s$. Rounding to the nearest provided option,the answer is $0.4 \ rad/s$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

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