The bob of mass m of a resting pendulum is given a sharp hit, imparting a horizontal velocity of 10 ,gl , where l represents the pendulum's length. Determine the tension in the string when the string makes an angle of 60^ with the upward vertical.

Step-by-step solution

Let the lowest point of the pendulum be the reference level for potential energy. Initial kinetic energy is given by: K_i = 1 2 m u^2 = 1 2 m (10gl) = 5mgl When the string makes an angle of 60^ with the upward vertical, it makes an angle of = 180^ - 60^ = 120^ with the downward vertical. The height of the bob from the lowest point is: h = l - l = l(1 - 120^ ) = l(1 - (-0.5)) = 1.5l Using conservation of mechanical energy: K_i + U_i = K_f + U_f 5mgl + 0 = 1 2 m v^2 + mg(1.5l) 1 2 m v^2 = 3.5mgl v^2 = 7gl The equatio