A uniform disc of radius $a$ and mass $m$ is rotating freely with an angular speed of $\omega$ in a…
A uniform disc of radius $a$ and mass $m$ is rotating freely with an angular speed of $\omega$ in a horizontal plane about a smooth fixed vertical axis passing through its centre. A particle, also of mass $m$ is suddenly attached to the rim of the disc and starts to rotate with it. The new angular speed of this system is
$\frac{\omega}{3}$
$\frac{\omega}{6}$
$\frac{\omega}{2}$
$\frac{\omega}{5}$
Solution
Given, mass of disc $=m$
Radius of disc $=a$
Angular speed $=\omega$
Moment of inertia of the disc, $I=\frac{1}{2} m a^2$
Initial angular momentum, $L_i=I \times \omega=\frac{1}{2} m a^2 \omega$
Let $\omega^{\prime}$ be the angular momentum of the disc when the particle of mass $m$ is attached at the rim.
Final angular momentum, $L_f=I \omega^{\prime}+m a v^{\prime}$
$
\begin{aligned}
& =\frac{1}{2} m a^2 \omega^{\prime}+m a^2 \omega^{\prime} \\
& {\left[\therefore v^{\prime}=\omega^{\prime} a\right]}
\end{aligned}
$
Applying conservation of angular momentum,
$
\begin{aligned}
L_i & =L_f \\
\frac{1}{2} m a^2 \omega & =\frac{1}{2} m a^2 \omega^{\prime}+m a^2 \omega^{\prime} \\
\Rightarrow \quad \omega^{\prime} & =\frac{\omega}{3}
\end{aligned}
$