Question

A uniform disc of radius $a$ and mass $m$ is rotating freely with angular speed $\omega$ in a horizontal plane, about a smooth fixed vertical axis through its centre. A particle of mass $m$ is then suddenly attached to the rim of the disc and rotates with it. The new angular speed is

• A. $\omega /\sqrt{3}$
• B. $\omega /3$
• C. $\omega /\sqrt{5}$
• D. $\omega /5$

Answer 1

Answer: (B)

$\omega /3$

Moment of inertia of the disc, $I=\frac{1}{2}m{a}^2$ moving with angular speed $w$

Initial angular momentum, $L_i=Iw=\frac{1}{2}m{a}^{2}w$

Let $w^{\prime }$ be the angular momentum of the disc when the particle of mass m is attached at the rim.

Final angular momentum, $L_f=Iw=\frac{1}{2}m{a}^2{w}^{\prime }+m{a}^2\left(w^{\prime }\right)$

Applying conservation of angular momentum, $L_i=L_f$

$\frac{1}{2}m{a}^{2}w=\frac{1}{2}m{a}^2{w}^{\prime }+m{a}^2{w}^{\prime }$

$⟹w^{\prime }=\frac{w}{3}$