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

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Question

(D) The angular momentum of the system about the axis of rotation is conserved because no external torque acts on the system.Initial angular momentum

Vedclass · Accepted Answer

$\frac{M\omega}{M - m}$

Vedclass · Answer

(D) The angular momentum of the system about the axis of rotation is conserved because no external torque acts on the system.Initial angular momentum $L_i = I_i \omega = (\frac{1}{2}MR^2)\omega$.When a piece of mass $m$ breaks off,the mass of the remaining disc is $(M - m)$.The moment of inertia of the remaining disc is $I_f = \frac{1}{2}(M - m)R^2$.Since the piece $m$ breaks off and flies vertically upwards,it does not exert any torque on the disc about the axis of rotation.By the law of conservation of angular momentum,$L_i = L_f$.$(\frac{1}{2}MR^2)\omega = (\frac{1}{2}(M - m)R^2)\omega'$.Solving for $\omega'$,we get $\omega' = \frac{M\omega}{M - m}$.

$A$ uniform disc of mass $M$ and radius $R$ is rotating about a horizontal axis passing through its centre with angular velocity $\omega$. $A$ piece of mass $m$ breaks from the disc and flies off vertically upwards. The angular speed of the disc will be

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