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 Kg m², the number of rotations made by the pulley before its direction of motion is reversed is : 

more than 3 but less then 6  

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.