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

By applying the principle of conservation of energy, the total energy at point A can be expressed as: \[ \frac{1}{2} mv_0^2 = \frac{1}{2} mv_B^2 + mgh \] Substituting the given values: \[ \frac{1}{2} m(5g\ell) = \frac{1}{2} mv_B^2 + mg\ell/2 \] This yields the kinetic energy at point B: \[ \Rightarrow KE_B = 2mg\ell \] At point C, the conservation of energy gives: \[ \frac{1}{2} mv_C^2 = \frac{1}{2} mv_D^2 + mg\ell/2 \] This results in the kinetic energy at point C: \[ \Rightarrow KE_C = mg\ell \] Therefore, the ratio of the kinetic energies is: \[ \frac{KE_B}{KE_C} = 2 \] The final answer is \( \boxed{2} \).