Question

A solid uniform disk of mass $m$ and radius $R$ is pivoted about a horizontal axis through its centre and a small body of mass $m$ is attached to the rim of the disk. If the disk is released from rest with the small body, initially at the same level as the centre, find the angular velocity when the small body reaches the lowest point of the disk.

Answer 1

By conservation of energy,

$k_{initial}+u_{initial}=k_{final}+u_{final}$

$u_{final}=0;u_{initial}=mgR$

$0+mgR=k_f+0$

$k_f=\frac{1}{2}I{\omega }_f^2=\frac{1}{2}(\frac{1}{2}m{R}^2+m{R}^2){\omega }_f^2$

$=\frac{1}{2}\times \frac{3}{2}m{R}^2{\omega }^2$

$mgR=\frac{1}{2}\times \frac{3}{2}m{R}^2{\omega }^2$

$\omega =\sqrt{\frac{4g}{3R}}$