Question

A hollow cylinder of mass m= 1kg and radius $R=\sqrt{3}m$ is placed on a smooth horizontal surface. Two small blocks of equal mass m = 1kg are sliding on the inner surface of the cylinder without friction. At the initial moment both the blocks are moving in the vertically upward direction with speed $u=\sqrt{56}m/s$ . A vertical force F is applied on the cylinder to keep it in contact with ground.What should be the minimum value of F (in newton) at the instant when $\theta ={60}^o$

Answer 1

$\frac{m{V}^2}{R}=\frac{\sqrt{3}}{2}mg+N$

$\frac{1}{2}m{V}^2=\frac{1}{2}m{u}^2=-\frac{\sqrt{3}}{2}mgR$

$\Rightarrow \frac{m{V}^2}{R}=\frac{m{u}^2}{R}-\sqrt{3}mg$

$\frac{m{u}^2}{R}-\sqrt{3}mg=\frac{\sqrt{3}}{2}mg+N$

$\Rightarrow N=\frac{m{u}^2}{R}-\frac{3\sqrt{3}}{2}mg$

$F_{Min}=2Ncos{30}^o-mg$

$=\frac{\sqrt{3}m{u}^2}{R}-\frac{9}{2}mg-mg\Rightarrow$

$F_{Min}=1newton$