Question

A small body $A$ starts sliding from the height ${}^{\prime }{h}^{\prime }$ down an inclined groove passing into a half-circle of radius $\frac{h}{2}$ (figure shown above). Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory (after breaking off the groove).

Answer 1

To complete a smooth vertical track of radius $R$, the minimum height at which a particle starts, must be equal to $\frac{5}{2}R$ (one can proved it from energy conservation). Thus in our problem body could not reach the upper most point of the vertical track of radius $\frac{R}{2}$. Let the particle $A$ leave the track at some point $O$ with speed $v$ (figure shown below). Now from energy conservation for the body $A$ in the field of gravity
$mg[h-\frac{h}{2}(1+\sin \theta \left)\right]=\frac{1}{2}m{v}^2$
or, $v^2=gh(1-\sin \theta )$ (1)
From Newton's second law for the particle at the point $O$, $F_n=m{w}_n$,
$N+mg\sin \theta =\frac{m{v}^2}{\left(\frac{h}{2}\right)}$
But, at the point $O$ the normal reaction $N=0$
So, $v^2=\frac{gh}{2}\sin \theta$ (2)
From (1) and (2), $\sin \theta =\frac{2}{3}$ and $v=\sqrt{\frac{gh}{3}}$
After leaving the track at $O$, the particle $A$ comes in air and further goes up and at maximum height of it's trajectory in air, it's velocity (say $v^{\prime }$) becomes horizontal (figure shown below). Hence, the sought velocity of $A$ at this point,
$v^{\prime }=v\cos (90-\theta )=v\sin \theta =\frac{2}{3}\sqrt{\frac{gh}{3}}$