Question

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed $\sqrt{gl}.$ Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle.

Answer 1

Let T=mg at angle $\theta$ as shown in figure.
$h=l(1-\cos \theta )$ ...(i)
Applying conservation of mechanical energy between points A and B, we get
$\frac{1}{2}m(u^2-v^2)=mgh$
Here, $u^2=gl$ ...(ii)
and v=speed of particle in position B
$\therefore v^2=u^2-2gh$ ...(iii)
Further, $T-mg\cos \theta =\frac{m{v}^2}{l}$
or $mg-mg\cos \theta =\frac{m{v}^2}{l}$ (T=mg)
or $v^2=gl(1-\cos \theta )$ ...(iv)
Substituting values of $v^2,u^2$ and h from Eqs. (iv), (ii) and (i) in Eq. (iii), we get
$gl(1-\cos \theta )=gl-2gl(1-\cos \theta )$
or $\cos \theta =\frac{2}{3}$
or $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$
Substituting $\cos \theta =\frac{2}{3}$ in Eq. (iv), we get
$v=\frac{\sqrt{gl}}{3}$