Question

A heavy particle hanging from a fixed point by a light inextensible string of length ' $\ell$ ' is projected horizontally with speed $\sqrt{g \ell}$, the speed of the particle is $\sqrt{\frac{g \ell}{a}}$ and the inclination of the string to the vertical is $\cos ^{-1} \frac{2}{b}$ at the instant of the motion when the tension in the string is equal to the weight of the particle. Determine $\frac{a}{b}$.

Answer 1

Let at the position $C$, the tension becomes equal to weight.
i.e. $T = mg$ ....(1)

$ T - mg \cos \theta =\frac{ mv ^{2}}{\ell} $
$ mg - mg \cos \theta =\frac{ mv ^{2}}{\ell}$
$ \frac{ v ^{2}}{\ell} = g (1-\cos \theta) $
$ v ^{2} =\ell g (1-\cos \theta) $ ...(2)
Where $ v $ is velocity at $C .$
The kinetic energy $(K.E.)$ at $C =\frac{1}{2} mv ^{2}$
Now, $K.E$. at $C=K . E$ at $A-P . E$. gained at $C$
$\frac{1}{2} mv ^{2} =\frac{1}{2} mu ^{2}- mgh ( h = AB )$
$\frac{1}{2} mv ^{2} =\frac{1}{2} mg \ell- mg (\ell-\ell \cos \theta) $
$v ^{2} = g \ell-2 g \ell(1-\cos \theta) \ldots(3)$
From (2) and (3) we get
$g \ell(1-\cos \theta)=g \ell-2 g \ell(1-\cos \theta)$
$1-\cos \theta=1-2+2 \cos \theta$
$\cos \theta=\frac{2}{3}$
From $(2), v ^{2}=\lg \left(1-\frac{2}{3}\right)=\frac{\ell g}{3}$
$v =\sqrt{\frac{\ell g}{3}}$