A thin uniform circular dise of mass ' M ' and radius ' $R$ ' is rotating with angular velocity ' $\omega$ '…

A thin uniform circular dise of mass ' M ' and radius ' $R$ ' is rotating with angular velocity ' $\omega$ ' in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass $\left(\frac{M}{3}\right)$ is placed gently on the first disc co-axially. The new angular velocity will be
  1. $\frac{2}{3} \omega$
  2. $\frac{3}{4} \omega$
  3. $\frac{4}{3} \omega$
  4. $\frac{5}{4} \omega$

Solution

Angular momentum $=\mathrm{I} \omega$ By conservation of angular momentum, $\begin{aligned} & \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\ & \text { Here, } I_1=\frac{M R^2}{2}, I_2=\frac{(M+M / 3)}{2} R^2=\frac{2 M R^2}{3} \\ & \therefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{2 \mathrm{MR}^2}{3} \omega_2 \\ & \text {...[From(i)] } \\ & \therefore \quad \omega_2=\frac{3}{4} \omega_1=\frac{3}{4} \omega \quad \ldots\left(\because \omega_1=\omega\right) \end{aligned}$ .

Asked in: MHT CET 2024 (04 May Shift 2)

Practice more Rotational Motion questions on Aicharya