As shown in the figure, the sphere at P is given a downward velocity $v_0$ and swings in a vertical

Question

As shown in the figure, the sphere at P is given a downward velocity $v_0$ and swings in a vertical plane at the end of a rope of $\ell = 1	ext{ m}$ attached to a support at O. The rope breaks at an angle of $30^\circ$ from the horizontal, knowing that it can withstand a maximum tension equal to three times the weight of the sphere. Then the value of $v_0$ will be: (Take $g = \pi^2 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Let the horizontal be the reference. At $30^\circ$ below horizontal, the sphere has fallen $h = \ell\sin30^\circ$. Work-energy: $\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = mgh \implies v^2 = v_0^2 + 2g\ell(1/2) = v_0^2 + g\ell$. The radial force equation at this point is $T - mg\sin30^\circ = mv^2/\ell$. Given breaking tension $T = 3mg$, we have $3mg - 0.5mg = m(v_0^2 + g\ell)/\ell$. Substituting $\ell = 1	ext{ m}$ yields $2.5mg = m(v_0^2 + g) \implies 2.5g = v_0^2 + g \implies v_0^2 = 1.5g = \frac{3g}{2} \implies v_0 = \sqrt{\frac{3g}{2}}$.