A smooth semicircular wire track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3 R 4 is attached to the lowest point O of the wire track. A small ring of mass m which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle of 60 ° with the vertical. The spring constant K = m g R . Consider the instant when the ring is released. The normal reaction on the ring by the track is

Question

A smooth semicircular wire track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3 R 4 is attached to the lowest point O of the wire track. A small ring of mass m which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle of 60 ° with the vertical. The spring constant K = m g R . Consider the instant when the ring is released. The normal reaction on the ring by the track is, 3 m g 8, m g, m g 4, 3 m g 4

MARKS App · Accepted Answer

In Δ OCP , OC = CP = R . ∴ The triangle is isosceles ∴ ∠ C O P = ∠ C P O = 60 ° ⇒ ∠ O C P = 60 ° ∴ Δ O C P is an equilateral triangle ⇒ OP = R ∴ Extension of string = R – 3 R 4 = R 4 = x The forces acting are shown in the figure (i) The free-body diagram of the ring is shown in fig (ii) Force in the tangential direction = F cos 30 ° + m g cos 30 ° = [ k x + m g ] cos 30 ° ∴ F t = 5 m g 8 3 ∴ F t = ma t ⇒ 5 m g 3 8 = m a t ⇒ a t = 5 3 8 g Also when the ring is just released N + F sin30º = mg sin 30º = ( m g – m g 4 ) × 1 2 = 3 m g 8

Search any question & find its solution

Looking for more such questions to practice?