A disc of mass M and radius R is free to rota | vedclass

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(A) Let the large disc have mass $M$ and radius $R$,and the small disc have mass $M$ and radius $r = R/2$. The small disc is at a distance $d = R$ fro

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Let the large disc have mass $M$ and radius $R$,and the small disc have mass $M$ and radius $r = R/2$. The small disc is at a distance $d = R$ from the axis of the large disc.Since there is no external torque on the system about the vertical axis of the large disc,the total angular momentum is conserved.Initially,both discs are at rest,so the initial angular momentum $L_i = 0$.Let the large disc rotate with angular velocity $\omega'$ in the opposite direction to the rotation of the small disc relative to the motor shaft.The angular momentum of the large disc is $L_1 = I_{large} \cdot \omega' = (\frac{1}{2} M R^2) \omega'$.The angular momentum of the small disc about the axis of the large disc consists of its spin angular momentum and its orbital angular momentum.The spin angular momentum of the small disc is $L_{spin} = I_{small} \cdot \omega = (\frac{1}{2} M (R/2)^2) \omega = \frac{1}{8} M R^2 \omega$.The orbital angular momentum of the small disc is $L_{orbit} = M v_{cm} d = M (\omega' R) R = M R^2 \omega'$.Applying conservation of angular momentum: $L_i = L_f = 0$.Taking the direction of $\omega'$ as positive,the spin of the small disc is in the opposite direction:$L_{spin} - (L_1 + L_{orbit}) = 0$$\frac{1}{8} M R^2 \omega - (\frac{1}{2} M R^2 \omega' + M R^2 \omega') = 0$$\frac{1}{8} M R^2 \omega = \frac{3}{2} M R^2 \omega'$$\omega' = \frac{1}{8} \cdot \frac{2}{3} \omega = \frac{\omega}{12}$.Thus,$\omega' = \omega/n$,which gives $n = 12$.

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