Question 532: Centre of Mass and Rotational Motion

Question

A thin uniform circular disc 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 $\left(\frac{\mathrm{M}}{3}\right)$ 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,
$$\begin{array}{llc} 
& \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \
& 	ext { Here, } \mathrm{I}_1=\frac{\mathrm{MR}^2}{2}, \mathrm{I}_2=\frac{(\mathrm{M}+\mathrm{M} / 3)}{2} \mathrm{R}^2=\frac{2 \mathrm{MR}^2}{3} \
	herefore \quad & \frac{\mathrm{MR}^2}{2} \omega_1=\frac{2 \mathrm{MR}^2}{3} \omega_2 \quad \ldots[	ext { From }(\mathrm{i})] \
	herefore \quad & \omega_2=\frac{3}{4} \omega_1=\frac{3}{4} \omega \quad \ldots\left(\because \omega_1=\omegaight)
\end{array}$$

Answer

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

Answer

$\frac{3}{4} \omega$

Answer

$\frac{4}{3} \omega$

Answer

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

Centre of Mass and Rotational Motion

Choose the correct answer: