Question

A small ball is suspended from point O by a thread of length $l$. A nail is driven into the wall at a distance of $l/2$ below O, at A. The ball is drawn aside so that the thread takes up a horizontal position at the level of point O and then released. Find
(a) At what point of the ball's trajectory, will the tension in the thread disappear?
(b) What will be the highest point to which it will rise?

Answer 1

(a) If at point P, tension is zero then $mg\cos \theta =\frac{m{v}_p^2}{r}$
Using conservation of energy
$m{v}_p^2=gl(1-\cos \theta )\Rightarrow \theta ={\cos }^{-1}\left(2/3\right)$
Height of point $P=\frac{l}{2}+\frac{l}{2}\cos \theta =\frac{5l}{6}$ from lowest point
(b) Next $v_p^2=gl(1-\frac{2}{3})=\frac{gl}{3}\Rightarrow v=\sqrt{\frac{gl}{3}}$
Now the particle describes parabolic path
The height attained by the particle, from point P
$h=\frac{{\left(v_p\sin \theta \right)}^2}{2g}=\frac{5l}{54}$
Height point from lowest point will be
$(\frac{5l}{6}+\frac{5l}{54})=\frac{50l}{54}$