Question

A small body starts sliding from the height ${}^{\prime }{h}^{\prime }$ down an inclined groove passing into a half circle of radius $h/2$. Find the speed of the body when it reaches the highest point of its trajectory.

Answer 1

Suppose the particle leaves the groove at $P(x,y)$ relative to the centre of the loop. $x=Rsin\theta$ and $y=Rcos\theta$ where $R=\frac{h}{2}$ and $\theta$ is measured anticlockwise from the vertical.

The top of the loop is level with the initial position of the particle, so the velocity of the particle at P is v where by conservation of energy :

$v^2=2g(R-Rcos\theta )$

$=gh(1-cos\theta ).$

The condition for leaving the groove is that the normal reaction $N$ becomes less than zero. $N$ is given by

$N+mgcos\theta =\frac{m{v}^2}{R}=\frac{2m{v}^2}{h}.$

At point $P$ where $N=0$ we have

$mgcos\theta =\left(\frac{2m}{h}\right)gh(1-cos\theta )$

$cos\theta =2(1-cos\theta )$

$cos\theta =\frac{2}{3}$.

After leaving contact with the groove at P, the horizontal component of velocity remains constant and has value vcosθ where.

$v^2=gh(1-cos\theta )=gh\left(\frac{1}{3}\right)$

$v=\sqrt{\frac{gh}{3}}$

$vcos\theta =\frac{2}{3}\sqrt{\frac{gh}{3}}$

So the speed of the particle at the highest point of its trajectory (where the vertical component is zero) is $\frac{2}{3}\sqrt{\frac{gh}{3}}$