A thin uniform circular disc of mass M and ra | vedclass

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(A) Given: Mass of the first disc $m_1 = M$, radius $r_1 = R$, and initial angular velocity $\omega_1 = \omega$.The moment of inertia of the first dis

Vedclass · Accepted Answer

$\frac{8}{9} \omega$

Vedclass · Answer

(A) Given: Mass of the first disc $m_1 = M$, radius $r_1 = R$, and initial angular velocity $\omega_1 = \omega$.The moment of inertia of the first disc is $I_1 = \frac{1}{2} M R^2$.The second disc has mass $m_2 = \frac{1}{8} M$ and radius $r_2 = R$.The moment of inertia of the second disc is $I_2 = \frac{1}{2} (\frac{1}{8} M) R^2 = \frac{1}{16} M R^2$.When the second disc is placed on the first, the total moment of inertia of the system becomes $I_{total} = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{16} M R^2 = \frac{8+1}{16} M R^2 = \frac{9}{16} M R^2$.By the principle of conservation of angular momentum, $L_{initial} = L_{final}$.$I_1 \omega = I_{total} \omega_{final}$$\frac{1}{2} M R^2 \omega = \frac{9}{16} M R^2 \omega_{final}$$\frac{1}{2} \omega = \frac{9}{16} \omega_{final}$$\omega_{final} = \frac{16}{18} \omega = \frac{8}{9} \omega$.

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