Question

The sphere at 'A' is given a downward velocity $v_0$ and swings in a vertical plane at the end of a rope of $l=1$ m attached to a support at O. The rope breaks at angle ${30}^{\circ }$ from horizontal, knowing that it can withstand a maximum tension equal to four times the weight of the sphere. Find the value of $v_0.(g=10m/s^2)$

Answer 1

Solution

$W_{gravity}=mg\left(\frac{l}{2}\right)$

$△KE=w_{gravity}$

[work energy theorem]

$\Rightarrow \frac{m{v}^2}{2}-\frac{mv{o}^2}{2}=mg\frac{l}{2}$ __ (1)

Also,

$T-\frac{mg}{2}=\frac{m{v}^2}{l}$ __ (ii)

From (i) and (ii),

$\frac{1}{2}(T-\frac{mg}{2}{)}^l-\frac{mv{0}^2}{2}=\frac{mgl}{2}$

$\Rightarrow \frac{1}{2}(4mg-\frac{mg}{2})l-\frac{mv{o}^2}{2}=\frac{mgl}{2}$

$\Rightarrow \frac{7}{4}mgl-\frac{mv{o}^2}{2}=\frac{mgl}{2}$

$\Rightarrow \frac{2}{m}[\frac{7mgl}{4}-\frac{mgl}{2}]=v{o}^2$

$\Rightarrow V_o=\sqrt{\frac{5}{2}gl}$