Question

A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is:

 

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

Use the conservation of mechanical energy to find the kinetic energy at different points in the motion.

Answer 1

The Correct Option is A

Using conservation of energy, the total energy at point A is: \[ \frac{1}{2} mv_0^2 = \frac{1}{2} mv_B^2 + mgh \] \[ \frac{1}{2} m(5g\ell) = \frac{1}{2} mv_B^2 + mg\ell/2 \] \[ \Rightarrow KE_B = 2mg\ell \] At point C: \[ \frac{1}{2} mv_C^2 = \frac{1}{2} mv_D^2 + mg\ell/2 \] \[ \Rightarrow KE_C = mg\ell \] Thus, the ratio of the kinetic energies is: \[ \frac{KE_B}{KE_C} = 2 \] Thus, the answer is \( \boxed{2} \).

Answer 2

Step 1 — Apply energy conservation at point A and B:
The total mechanical energy of the system remains constant. Therefore, between the lowest point A and point B:

$$ \frac{1}{2} m v_0^2 = \frac{1}{2} m v_B^2 + m g h_B $$
Substituting $v_0^2 = 5 g \ell$ and $h_B = \dfrac{\ell}{2}$:
$$ \frac{1}{2} m (5 g \ell) = \frac{1}{2} m v_B^2 + m g \frac{\ell}{2} $$
Simplifying, we get:
$$ \frac{1}{2} m v_B^2 = 2 m g \ell $$
Hence, the kinetic energy at B is:
$$ KE_B = 2 m g \ell $$

Step 2 — Apply energy conservation between points C and D:
Similarly, for points C and D:
$$ \frac{1}{2} m v_C^2 = \frac{1}{2} m v_D^2 + m g h_C $$
Since $v_D^2 = g \ell$ and $h_C = \dfrac{\ell}{2}$:
$$ \frac{1}{2} m v_C^2 = \frac{1}{2} m g \ell + m g \frac{\ell}{2} $$
Therefore,
$$ KE_C = m g \ell $$

Step 3 — Compute the ratio of kinetic energies:
$$ \frac{KE_B}{KE_C} = \frac{2 m g \ell}{m g \ell} = 2 $$

Final Answer: The ratio of kinetic energies at points B and C is 2 : 1.
$\boxed{2}$