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. If an another disc of same dimensions but $\frac{M}{4}$ in mass is placed coaxially on first disc gently, then the angular velocity of the system now becomes

• A. $\frac{2}{\sqrt{5}}\omega$
• B. $\frac{1}{4}\omega$
• C. $\frac{5}{4}\omega$
• D. $\frac{4}{5}\omega$

Answer 1

Answer: (D)

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

We have $\overrightarrow{\tau }=\frac{dL}{dt}$
since $\overrightarrow{\tau }=0$, $\therefore$ $L=$ constant
$\therefore I_1{\omega }_1=I_2{\omega }_2$
Here $I_1=\frac{1}{2}M{R}^2$
${\omega }_1=\omega$
$I_2=\frac{1}{2}M{R}^2+\frac{1}{2}\frac{M}{4}{R}^2=\left(\frac{4+1}{8}\right)M{R}^2=\frac{5}{8}M{R}^2$
$\therefore {\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{8}{2\times 5}\omega =\frac{4}{5}\omega$