Question

A thin uniform circular disc of mass $M$ and radius $R$ is rotating with an angular velocity $\omega ,$ in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same dimensions but of mass $\frac{M}{4}$ is placed gently on the first disc co-axially. The final angular velocity of the system is

• A. $\frac{2}{3}\omega$
• B. $\frac{4}{5}\omega$
• C. $\frac{3}{4}\omega$
• D. $\frac{1}{3}\omega$

Answer 1

The correct option is B $\frac{4}{5}\omega$

Initial angular momentum of system $L_i=$ Angular momentum of disc of mass $M$
$\text{MI}$ of disc $M$,
$I_1=\frac{1}{2}M{R}^2$
$\text{MI}$ of disc $\frac{M}{4}$:
$I_2=\frac{1}{2}\times \frac{M}{4}\times R^2=\frac{M{R}^2}{8}$
Final $\text{MI}$ of system:
$I=I_1+I_2=\frac{5M{R}^2}{8}$
$\because$when the second disc of mass $\frac{M}{4}$ is placed on $1^{\text{st}}$ disc ${\tau }_{\text{ext}}=0$, friction will operate between discs to prevent slipping between them, and eventually both will move with same angular velocity ${\omega }_2$.
$\Rightarrow$ Applying conservation of angular momentum on the system
$L_f=L_i$
$I{\omega }_2=I_1{\omega }_1$
$\Rightarrow {\omega }_2=\frac{I_1{\omega }_1}{I}$
given ${\omega }_1=\omega$
$\therefore {\omega }_2=\frac{\left(\frac{M{R}^2\omega }{2}\right)}{\frac{5M{R}^2}{8}}=\frac{4}{5}\omega$