Question

A rod of mass m and length $2R$ can rotate about an axis passing through $O$ the n vertical plane. A disc of mass $m$ and radius $R$ is hinged to the other end $P$ of the rod and can freely rotate about $P$. When the disc is at the lowest point both rod and disc has angular velocity $w$. If rod rotates by maximum angle ${60}^o$ with downward vertical, then find $w$ in terms of $R$ and $g$. (all hinges are smooth)

• A. $(9g/16R{)}^{1/2}$
• B. $3M{R}^2$
• C. $M{R}^2$
• D. $\frac{199}{196}M{R}^2$

Answer 1

The correct option is A $(9g/16R{)}^{1/2}$

the angular momentum of the rod

$=\frac{1}{12}\times m\times (2R{)}^2$

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

the angular momentum of the disk

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

Now as the rod and disc rotates with same angular momentum omega when the disc is at lowest point and the rod rotates maximum with angle ${60}^{\circ }$ then the angular momentum is,

$\omega =\sqrt{\frac{9g}{16R}}$