A ring of mass M and radius R is rotating wit | vedclass

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(D) The initial moment of inertia of the system is $I_i = I_{ring} + I_{masses} = MR^2 + 0 = MR^2$.Initial angular momentum $L_i = I_i \omega = MR^2 \

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$\frac{4}{5} R$

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(D) The initial moment of inertia of the system is $I_i = I_{ring} + I_{masses} = MR^2 + 0 = MR^2$.Initial angular momentum $L_i = I_i \omega = MR^2 \omega$.At the given instant,the angular speed is $\omega' = \frac{8}{9} \omega$.The moment of inertia of the system at this instant is $I_f = I_{ring} + I_{masses} = MR^2 + \frac{M}{8} r_1^2 + \frac{M}{8} r_2^2$,where $r_1 = \frac{3}{5} R$ and $r_2$ is the distance of the other mass.Using the conservation of angular momentum,$L_i = L_f \Rightarrow I_i \omega = I_f \omega'$.$MR^2 \omega = (MR^2 + \frac{M}{8} (\frac{3}{5} R)^2 + \frac{M}{8} r_2^2) 	imes \frac{8}{9} \omega$.$MR^2 = (MR^2 + \frac{M}{8} 	imes \frac{9}{25} R^2 + \frac{M}{8} r_2^2) 	imes \frac{8}{9}$.$\frac{9}{8} R^2 = R^2 + \frac{9}{200} R^2 + \frac{1}{8} r_2^2$.$\frac{9}{8} R^2 - R^2 - \frac{9}{200} R^2 = \frac{1}{8} r_2^2$.$\frac{225 - 200 - 9}{200} R^2 = \frac{1}{8} r_2^2$.$\frac{16}{200} R^2 = \frac{1}{8} r_2^2$.$r_2^2 = \frac{16 	imes 8}{200} R^2 = \frac{128}{200} R^2 = \frac{16}{25} R^2$.$r_2 = \frac{4}{5} R$.

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