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in Laws of motion by (15.2k points)
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A smooth semicircular wire track of radius R is fixed in a vertical plane (Figure). 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 stationery at point P such that the spring makes an angle 60° with the vertical. the spring constant k = mg/R. consider the ring is making an angle 60°  with the vertical. the spring the released.

(a) Draw the free body diagram of the ring 

(b) Determine the tangential acceleration of the ring and the normal reaction

2 Answers

+1 vote
by (39.0k points)
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Best answer

(a) CP = CO = Radius of circle (R)

∴ ∠CPO = ∠POC = 60°

∴ ∠OCP is also 60°

Therefore, DOCP is an equilateral triangle.

Natural length of spring is 3R/4.

∴ Extension in the spring x = R – \(\frac{3R}{4} = \frac R4\)

⇒ Spring force, F = kx = \(\left(\frac {m g}4\right)\) and N = Normal reaction.

(b) Tangential acceleration aT the ring will move forwards the x-axis just after the release. So, net force along x-axis

Therefore, tangential acceleration of the ring,

Normal reaction N Net force along y-axis on the ring just after the release will be zero.

+1 vote
by (15.7k points)

answer is :

(a) B \(\frac{5\sqrt{3}}{8}g\),

(b) \(\frac{3mg}{8}\)

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