Example

Question:

A bob of mass \(m\) is suspended by a light string of length \(L\). It is imparted a horizontal velocity \(v_0\) at the lowest point \(A\) such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, \(C\).
(i) Obtain an expression for \(v_0\).
(ii) Find the speeds at points \(B\) and \(C\).
(iii) Find the ratio of the kinetic energies \(K_B/K_C\) at \(B\) and \(C\).
Comment on the trajectory of the bob after it reaches point \(C\).

Solution:

(i) Expression for \(v_0\):
At point \(A\): Energy is all kinetic, \[ E = \frac{1}{2}mv_0^2 \] At point \(C\) (top), tension is zero and energy is both kinetic and potential: \[ E = \frac{1}{2}mv_C^2 + 2mgL \] Newton’s second law at \(C\): \[ mg = \frac{mv_C^2}{L} \implies v_C = \sqrt{gL} \] Substitute in the energy conservation: \[ \frac{1}{2}mv_0^2 = \frac{1}{2}m(gL) + 2mgL = \frac{5}{2}mgL \implies v_0 = \sqrt{5gL} \] (ii) Speeds at B and C:
At \(B\), using energy conservation (height at \(B\) is \(L\)): \[ E = \frac{1}{2}mv_B^2 + mgL \] \[ \frac{1}{2}mv_B^2 + mgL = \frac{1}{2}mv_0^2 = \frac{5}{2}mgL \] \[ \implies \frac{1}{2}mv_B^2 = \frac{5}{2}mgL - mgL = \frac{3}{2}mgL \implies v_B = \sqrt{3gL} \] From above, at \(C\): \(v_C = \sqrt{gL}\)
(iii) Ratio of kinetic energies: \[ \frac{K_B}{K_C} = \frac{\frac{1}{2}m v_B^2}{\frac{1}{2}m v_C^2} = \frac{v_B^2}{v_C^2} = \frac{3gL}{gL} = 3 \] Comment: At point \(C\), the string becomes slack and the velocity is horizontal and to the left. If the string is cut, the bob will execute projectile motion with this velocity, as in horizontal launch from a cliff.