Question 557: Centre of Mass and Rotational Motion

Question

A thin uniform circular disc of mass '$$\mathrm{M}$$' 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 $$\left(\frac{M}{2}\right)$$ is placed gently on the first disc co-axially. The new angular velocity will be

Accepted Answer

Angular momentum $$=\mathrm{I} \omega$$
By conservation of angular momentum, $$\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2$$
Here, $$I_1=\frac{M R^2}{2}, I_2=\frac{(M+M / 2)}{2} R^2=\frac{3 M R^2}{4}$$
$$\begin{aligned}
& 	herefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{3 \mathrm{MR}^2}{4} \omega_2 \
& 	herefore \quad \omega_2=\frac{2}{3} \omega_1
\end{aligned}$$

Answer

$$\frac{2}{3} \omega$$

Answer

$$\frac{4}{5} \omega$$

Answer

$$\frac{5}{4} \omega$$

Answer

$$\frac{3}{2} \omega$$

Centre of Mass and Rotational Motion

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