A bob of mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal veloci

Question

A bob of mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$. If the string gets slack exactly at a point $P$ making an angle $	heta$ above the horizontal, the ratio of the initial speed $v_0$ to the speed $v$ at point $P$ ($v_0/v$) is

Professor PC Thomas Classes · Answer

(a) : Let the speed at $P$ be $v$. The height gained by the bob to reach point $P$ is $h = l + l\sin	heta = l(1+\sin	heta)$. By conservation of energy: $\frac{1}{2}mv_0^2 = mgh + \frac{1}{2}mv^2 \implies \frac{1}{2}v_0^2 = gl(1+\sin	heta) + \frac{1}{2}v^2$. At point $P$, the string becomes slack, so tension $T = 0$. The radial component of weight provides the centripetal force: $mg\sin	heta = \frac{mv^2}{l} \implies gl = \frac{v^2}{\sin	heta}$. Substituting this into the energy equation gives: $\frac{1}{2}v_0^2 = \frac{v^2}{\sin	heta}(1+\sin	heta) + \frac{1}{2}v^2$. Solving for $v_0^2$ gives $v_0^2 = v^2 \left( \frac{2 + 2\sin	heta + \sin	heta}{\sin	heta} ight) = v^2 \left( \frac{2 + 3\sin	heta}{\sin	heta} ight)$. Taking the square root yields $\frac{v_0}{v} = \sqrt{\frac{2+3\sin	heta}{\sin	heta}}$.