Circular MotionHard
Question
A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60o with the vertical. Then (g = 9.8 m/s2)
Options
A. its period of revolution is 4π/7 sec.
B.the tension in the string is double the weight of the particle
C.the speed of the particle = 2.8√3m/s
D.the centripetal acceleration of the particle is 9.8√3 m/s2.
Solution
= 
.............. (1)
= mg .............. (2)Hence T = 2 mg , So (B) holds
From (1) & (2) V2 = 3 gl/2
∴ V =
∴ V = 2.8√3 m/s . So (C) hold
ac = V2/r =
= √3 × g = 9.8 √3 m/s3∴ (D) holds
t =
= 
t = 4π/7 ∴ (A) holds.
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