Question

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

• A. $\sqrt{\frac{9 g}{16 R}}$
• B. $\sqrt{\frac{3 g}{23 R}}$
• C. $\frac{1}{3} \sqrt{\frac{g}{R}}$
• D. none of these

Answer 1

Answer: (A)

As hinges are smooth the disc continue to rotate at $\omega$ so by work energy theorem we use
$\frac{1}{2}\left(\frac{1}{3} m(2 R)^{2}\right) \omega^{2}+\frac{1}{2} m(2 R \omega)^{2} $
$=(m g(2 R)+m g R) \frac{1}{2}$
$\Rightarrow 2 R^{2} \omega^{2}+\frac{2 R^{2} \omega^{2}}{3}=\frac{3 g R}{2}$
$\Rightarrow \frac{8 R \omega^{2}}{3}=\frac{3 g}{2}$
$\Rightarrow \omega=\sqrt{\frac{9 g}{16 R}}$