Question

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

Answer 1

In this case, the angular momentum of the rod is $\frac{1}{12}m{\left(2R\right)}^2=\frac{1}{3}m{R}^2$ and angular momentum of disk $=\frac{1}{2}m{R}^2.$

Now as the rod and and the disk rotates with same angular momentum, when the disc is at lowest point and the rod rotates maximum with angle $60°$ then the angular momentum is $\frac{\sqrt{9}g}{16R}.$

Hence, the answer is $\frac{\sqrt{9}g}{16R}.$