A particle of mass m is whirled in a vertical circle with the help of a string of length L . The ratio of the tension in the string at the lowest point to the tension when the string is horizontal is 3 . The magnitude of the change in velocity of the particle as it moves from the lowest point to the horizontal position is x gL . The value of x is

Step-by-step solution

Let u be the speed at the lowest point and v be the speed at the horizontal position. The tension at the lowest point is T_L = mu^2 L + mg . The tension at the horizontal position is T_H = mv^2 L . By conservation of mechanical energy between the lowest and horizontal points: 1 2 mu^2 = 1 2 mv^2 + mgL v^2 = u^2 - 2gL Substituting v^2 into the expression for T_H : T_H = m(u^2 - 2gL) L = mu^2 L - 2mg We are given that T_L = 3T_H : mu^2 L + mg = 3 ( mu^2 L - 2mg ) mu^2 L + mg = 3mu^2 L - 6mg 7mg = 2mu^2 L u^2 = 3.5gL