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 $\frac{l}{2}$ 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. If highest point to which it rises is $\frac{xl}{y}$. Find (y - x)___

Answer 1

If at point P, tension is zero then $mgcos\theta =\frac{m{v}^2}{r}$
Using conservation of energy,
$v^2=gl(1-cos\theta )$
$\therefore mgcos\theta =\frac{1}{\frac{l}{2}}(1-cos\theta )\Rightarrow \theta =co{s}^{-1}\left(\frac{2}{3}\right)$
$\therefore$ Height of point
$P=\frac{l}{2}+\frac{l}{2}cos\theta =\frac{5l}{6}$, from lowest point
$v_2=gl(1-\frac{2}{3})=\frac{gl}{3}\Rightarrow v=\sqrt{\frac{gl}{3}}$
Now the particle describes parabolic path. The height attained by the particle, from point P.
$h=\frac{(vsin\theta {)}^2}{2g}=\frac{5l}{54}$
$\therefore$ Highest point from lowest point will be
$(\frac{5l}{6}+\frac{5l}{54})=\frac{50l}{54}$
$(\frac{5l}{6}+\frac{5l}{54})=\frac{50l}{54}$