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

Answer: 1.00

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

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