Question

A smooth semicircular wire track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3R/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 60° with the vertical. Spring constant K = mgl R. The spring force is

• A. mg/3
• B. mg
• C. mg/2
• D. mg/4

Answer 1

Answer: (D)

mg/4

Answer 2

CP = CO = R

∠CPO = ∠POC = 60^o

Thus ∆OCP is an equilateral ∆

∴ OP = R

∴ Extension = R – natural length of spring

= R - $\frac{3R}{4} = \frac{R}{4}$

Thus spring force = kx = $\left( \frac{mg}{R} \right)\left( \frac{R}{4} \right) = \frac{mg}{4}$