A spring with natural length h and spring constant k has one end fixed to the ground. The opposite end is connected to a smooth ring of mass m that can slide along a fixed horizontal rod at a height h , as shown in the figure. The system is released from rest when the spring initially makes an angle of 37^ with the vertical. Determine the speed of the ring at the instant the spring becomes vertical.

Step-by-step solution

Let the initial length of the spring be L . From the given geometry, the vertical height is h and the spring makes an angle of 37^ with the vertical. L 37^ = h L = h 37^ = h 4/5 = 5h 4 The initial extension in the spring is: x = L - h = 5h 4 - h = h 4 The initial mechanical energy of the system consists only of the elastic potential energy of the spring (since the ring is released from rest and gravitational potential energy remains constant): E_i = 1 2 kx^2 = 1 2 k ( h 4 )^2 = kh^2 32 When the spring becomes verti