Question

A thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of same thickness and radius but of mass $\frac{1}{8}M$ is placed gently on the first disc co-axially. The angular velocity of the system is now

• A. $\frac{8}{9}\omega$
• B. $\frac{5}{9}\omega$
• C. $\frac{1}{3}\omega$
• D. $\frac{2}{9}\omega$

Answer 1

The correct option is A $\frac{8}{9}\omega$

Moment of inertia of the disc of mass $M$ about the axis passing through its centre and perpendicular to its plane.

$I_1=\frac{1}{2}M{R}^2$

Its angular velocity, $w_1=w$

Moment of inertia of the disc of mass $M/8$ about the axis passing through its centre and perpendicular to its plane.

$I_2=\frac{1}{2}\frac{M}{8}{R}^2=\frac{1}{16}M{R}^2$

Total final moment of inertia of the system, $I^{\prime }=I_1+I_2=\frac{9}{16}M{R}^2$

Let the final angular velocity of the system be $w^{\prime }$.

Using conservation of angular momentum: $L_i=L_f$

$\therefore$ $I_1{w}_1+I_2{w}_2=I^{\prime }{w}^{\prime }$

where $w_2=0$

$\Rightarrow \frac{1}{2}M{R}^{2}w+0=\frac{9}{16}M{R}^2{w}^{\prime }$

$⟹w^{\prime }=\frac{8}{9}w$