Question

The figure shows a thin ring of mass $M=1kg$ and radius $R=0.4m$ spinning about a vertical diameter.(Take $l=\frac{1}{2}M{R}^2)$ A small bead of mass $m=0.2kg$ can slide without friction along the ring. When the bead is at the top of the ring, the angular velocity is $5rad/s$. What is the angular velocity when the bead slips halfway to $\theta ={45}^{\circ }$?

Answer 1

$I_1{\omega }_1=I_2{\omega }_2$
$\therefore {\omega }_2=\left(\frac{I_1}{I_2}\right){\omega }_1$
$=\frac{\left(\frac{1}{2}M{R}^2\right)}{[\frac{1}{2}M{R}^2+m{\left(\frac{R}{\sqrt{2}}\right)}^2]}{\omega }_1$
$=\left(\frac{M}{M+m}\right){\omega }_1$
$=\left(\frac{1}{1+0.2}\right)5$
$=\frac{25}{6}rad/s$