Question

A small particle of mass $m$ is projected horizontally from the top of a smooth and fixed hemisphere of radius $r$ and speed $u$ as shown in figure. For values of $u\ge u_0(u_0=\sqrt{gr})$ the particle does not slide on the hemisphere (i.e; leaves the surface at the top itself). For $u=\frac{u_0}{3}$, the height from the ground at which the particle leaves the surface is $\frac{Pr}{27}$. Then the value of P is .

Answer 1

Let the particle leave the contact at point $P$ making an angle $\theta$ with the vertical.
Let the speed of particle at point $P$ be $v$.


Applying work-energy theorem between points $O$ and $P$:
$W_{net}=K_f-K_i$

$W_N+W_{mg}=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

$mgr(1-\cos \theta )=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$ $[\because W_N=0,W_{mg}=mgy\text{and}y=r(1-\cos \theta \left)\right]$

$v^2=u^2+2g(r-h)[\because \cos \theta =\frac{h}{r}]$

$v^2=\frac{u_0^2}{9}+2g(r-h)[\because u=\frac{u_0}{3}]$

$v^2=\frac{19}{9}gr-2gh[\because u_0=\sqrt{gr}]$

Now, using Newton's second law at point P:
$mg\cos \theta -N=\frac{m{v}^2}{r}$

Using condition of just leaving the surface, $N=0$

$mg\cos \theta =\frac{m{v}^2}{r}\Rightarrow g\left(\frac{h}{r}\right)=\frac{v^2}{r}\Rightarrow gh=v^2$

$\Rightarrow gh=\frac{19}{9}gr-2gh$

$\Rightarrow h=\frac{19r}{27}=\frac{Pr}{27}$

$\therefore P=19$