Question

A Pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2)N where t is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 KgM2, the number of rotations made by the pulley before its direction of motion is reversed is :

(A) more than 3 but less then 6

(B) more than 6 but less then 9

(C) more than 9

(D) Less then 3

Answer 1

Correct option: (A) more than 3 but less**then**6

Explanation:

For reversing direction work done = 0

F = 20t – 5t2

τ = F ∙ r = (20t – 5t2) × 2 = 40t – 10t2

also α = (τ/I) = [(40t – 10t2) / (10)] = 4t – t2

now ω = t∫0 α ∙ dt = t∫0 (4t – t2)dt

= 2t2 – (t3 / 3) (1)

at ω = 0, 2t2 – (t3 / 3) = 0

∴ 2t2 = (t3 / 3) = 0

∴ t = 6 sec

θ = ∫ω ∙ dt

= θ∫0 [2t2 – (t3/3)]dt --------- from (1)

= [(2t3/ 3) – (t4 / 12)]60

= 216 [(2/3) / (1/2)]

θ = 36 rad

θ = 2πn = 36 rad

hence n = (36 / 2π) = 5.72 < 6

hence number of revolution should be less than 6.