Question

A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:

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

Answer 1

Answer: (D)

$\left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}$

Answer 2

Here's the step-by-step explanation:

1. Apply conservation of mechanical energy between the lowest point and point P where the string gets slack.

2. Define the height of point P relative to the lowest point using the angle $\theta$. The height is $h = l(1 + \sin \theta)$.

3. Write the energy conservation equation: $\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgl(1 + \sin \theta)$.

4. At point P, the string gets slack, meaning the tension is zero. Apply Newton's second law in the radial direction. The component of gravity along the string towards the center is $mg \sin \theta$. This provides the centripetal force: $mg \sin \theta = \frac{mv^2}{l}$.

5. From the tension equation, find $v^2 = gl \sin \theta$.

6. Substitute the expression for $v^2$ into the energy conservation equation to find $v_0^2$.

7. Calculate the ratio $v/v_0$ by taking the square root of the ratio of $v^2$ and $v_0^2$.

Therefore, $\frac{v}{v_0} = \left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}$.