A horizontal circular platform of radius R = | vedclass

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(D) The system (platform + two balls) is initially at rest,so the initial angular momentum is $L_i = 0$.After the balls are fired,the platform rotates

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$4$

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(D) The system (platform + two balls) is initially at rest,so the initial angular momentum is $L_i = 0$.After the balls are fired,the platform rotates with an angular velocity $\omega$ in the direction opposite to the angular momentum imparted to the balls to conserve the total angular momentum.The angular momentum of the two balls about the axis of rotation is $L_{balls} = 2 	imes (mvr) = 2mvr$.The angular momentum of the platform is $L_{platform} = I\omega = \left(\frac{MR^2}{2}ight)\omega$.By the law of conservation of angular momentum,the total angular momentum remains zero:$L_{balls} + L_{platform} = 0$$2mvr + \left(\frac{MR^2}{2}ight)\omega = 0$Taking the magnitude,we have:$\omega = \frac{2mvr}{I} = \frac{2mvr}{\frac{MR^2}{2}} = \frac{4mvr}{MR^2}$Substituting the given values:$m = 0.05 \ kg = 5 	imes 10^{-2} \ kg$$M = 0.45 \ kg = 45 	imes 10^{-2} \ kg$$v = 9 \ m/s$$r = 0.25 \ m = 1/4 \ m$$R = 0.5 \ m = 1/2 \ m$$\omega = \frac{4 	imes (0.05) 	imes 9 	imes 0.25}{0.45 	imes (0.5)^2}$$\omega = \frac{4 	imes 0.05 	imes 9 	imes 0.25}{0.45 	imes 0.25}$$\omega = \frac{4 	imes 0.05 	imes 9}{0.45} = \frac{1.8}{0.45} = 4 \ rad/s$

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