A smooth rod PQ is rotated in a horizontal plane about its mid point M which is h = 0.1 m vertically below a fixed point A at a constant angular velocity 14 rad/s. A light elastic string of natural length 0.1 m requiring 1.47 N/cm has one end fixed at A and its other end attached to a ring of mass m = 0.3 kg which is free to slide along the rod. When the ring is stationary relative to rod, then find inclination of string with vertical, tension in string, force exerted by ring on the rod. (g = 9.8 m/s 2 )
Text Solution
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cos θ = 3/5 , T = 9.8 N , N = 
= 2.94 N
Sol. 
T = kx = 147 (0.1 sec θ – 0.1)
T sin θ = m ω 2 r
⇒ 147(0.1 sec θ – 0.1)sin θ = 0.3 × (14) 2 (0.1 tan θ )
1 – cos θ = 0.4
⇒ cos θ = 
⇒ θ = 53º
T = 147 (0.1 sec 53 – 0.1) = 9.8 N
N = T cos θ – mg = 9.8 × 
– 0.3 × 9.8 = 2.94 N
N = 2.94 N Ans.
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