A uniform circular disc of mass 50 kg and ra | vedclass

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(A) The initial moment of inertia of the disc about its central axis is $I_1 = \frac{1}{2} M R^2 = \frac{1}{2} 	imes 50 	imes (0.4)^2 = 4 \ kg \ m^2

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(A) The initial moment of inertia of the disc about its central axis is $I_1 = \frac{1}{2} M R^2 = \frac{1}{2} 	imes 50 	imes (0.4)^2 = 4 \ kg \ m^2$.The initial angular velocity is $\omega_1 = 10 \ rad \ s^{-1}$.When two rings are placed on the disc,their moment of inertia about the axis of rotation must be calculated using the parallel axis theorem. For each ring of mass $m$ and radius $r$,the moment of inertia about its own central axis is $mr^2$. The distance of the center of each ring from the axis of rotation is $r = 0.2 \ m$. Thus,the moment of inertia of one ring about the disc's axis is $I_{ring} = mr^2 + mr^2 = 2mr^2 = 2 	imes 6.25 	imes (0.2)^2 = 0.5 \ kg \ m^2$.The total moment of inertia of the system after placing two rings is $I_2 = I_{disc} + 2 	imes I_{ring} = 4 + 2 	imes 0.5 = 5 \ kg \ m^2$.By the principle of conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.Substituting the values: $4 	imes 10 = 5 	imes \omega_2$.$\omega_2 = \frac{40}{5} = 8 \ rad \ s^{-1}$.

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