Question

The sphere at A is given a downward velocity $v_0$ of magnitude $5m/s$ and swings in a vertical plane at the end of a rope of length $l=2m$ attached to a support at $O$. Determine the angle $\theta$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

• A. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• B. ${\sin }^{-1}\left(\frac{1}{3}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{2}\right)$
• D. ${\sin }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{1}{4}\right)$

Given : $v_o=5m/s$ $l=2$ m $T=2mg$

From figure, $PM=lsin\theta$

Circular motion equation at P : $\frac{m{v}^2}{l}=T-mgsin\theta$

OR $m{v}^2=l(2mg-mgsin\theta )=2(2m\times 10-m(10\left)sin\theta \right)$

$⟹$ $m{v}^2=m(40-20sin\theta )$

Work-energy theorem : $W=\Delta K.E$

$\therefore$ $mg\left(PM\right)=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_o^2$

OR $2mg\left(lsin\theta \right)=m(40-20sin\theta )-m{v}_o^2$

OR $2m\left(10\right)\left(2sin\theta \right)=m(40-20sin\theta )-m\left(25\right)$ $⟹sin\theta =\frac{1}{4}$