A heavy particle, suspended by a 1.5 m long string, is imparted a horizontal velocity of 57 m/s . Calculate the speed of the particle at the instant the string becomes slack. Take g = 10 m/s ^2 .

Step-by-step solution

Let the string become slack when it makes an angle with the upward vertical. At this instant, the tension in the string is zero. The component of gravity along the string provides the necessary centripetal acceleration: mg = mv^2 l v^2 = gl Applying conservation of mechanical energy between the lowest point and the point where the string becomes slack: 1 2 mu^2 = 1 2 mv^2 + mgh The height h reached by the particle from the lowest point is h = l + l = l(1 + ) . Substituting h and v^2 into the energy equation: u^2 =