A hollow sphere of mass 1 ,kg and radius 10 , | vedclass

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(B) The moment of inertia of a hollow sphere about its diameter is given by $I = \frac{2}{3}MR^2$.Given: Mass $M = 1 \,kg$,Radius $R = 10 \,cm = 0.1 \

Vedclass · Accepted Answer

$450$

Vedclass · Answer

(B) The moment of inertia of a hollow sphere about its diameter is given by $I = \frac{2}{3}MR^2$.Given: Mass $M = 1 \,kg$,Radius $R = 10 \,cm = 0.1 \,m$,Force $F = 30 \,N$.The torque applied is $	au = F 	imes R$.Using the relation $	au = I \alpha$,where $\alpha$ is the angular acceleration:$F 	imes R = \left(\frac{2}{3}MR^2ight) \alpha$$30 	imes 0.1 = \left(\frac{2}{3} 	imes 1 	imes (0.1)^2ight) \alpha$$3 = \left(\frac{2}{3} 	imes 0.01ight) \alpha$$3 = \frac{0.02}{3} \alpha$$\alpha = \frac{3 	imes 3}{0.02} = \frac{9}{0.02} = 450 \,rad/s^2$.

$A$ hollow sphere of mass $1 \,kg$ and radius $10 \,cm$ is free to rotate about its diameter. If a force of $30 \,N$ is applied tangentially to it,its angular acceleration is (in $rad/s^2$):

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