A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in the figur…

Question

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in the figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v0 is:

1. A. (sinθ)^1/2
2. B. (1/2+3sinθ)^1/2
3. C. (cosθ/2+3sinθ)^1/2
4. D. (sinθ/2+3sinθ)^1/2 (Correct answer)

Correct Answer

Option D — (sinθ/2+3sinθ)^1/2

Detailed Solution & Explanation

The correct answer is Option D.

Key Points

• At $P$ the string goes slack, so tension $=0$ and the component of gravity along the string supplies the centripetal force: $mg\sin\theta=\dfrac{mv^2}{l}$ ... (i)
• Energy conservation from the lowest point to $P$ (rise $=l+l\sin\theta$): $\dfrac{1}{2}mv_0^2=\dfrac{1}{2}mv^2+mg\,l(1+\sin\theta)$.
• Using (i), $gl=\dfrac{v^2}{\sin\theta}$, this simplifies to $v_0^2=v^2\!\left(3+\dfrac{2}{\sin\theta}\right)$.
• Hence $\dfrac{v}{v_0}=\left(\dfrac{\sin\theta}{3\sin\theta+2}\right)^{1/2}$.