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How can I solve this problem in the physics of rotation?

150 kg A happy-go-round as a uniform, solid, horizontal disc 1.50m beam begins by wrapping a rope on the edge of the disc and pull the rope. What force must be constant on the string so that the happy-go-round from rest to speed angle of 0.500 rev / s in 2.00 s? I'm not sure if needed, but the moment of inertia of a solid disk (I) is equal to 1/2MR ^ 2

The moment of inertia (I) of the carousel is: I = 0.5 mR ² = 0.5 (150 kg) (1.50) ² = 169 kg m ∙ The angular acceleration is: ω = Ω ₀ αt + α = (ω - Ω ₀) / t = (0.500rev/min - 0) / = 2.00 0.250rev / s (2πrad / rev) = 1.57rad / s now: Στ RF = F = Iα = Iα / r = [(169 kg ∙ m / s ²) (1.57rad / s ²)] / 1.50 = 177N I hope this helps.

"ON CHRIST THE SOLID ROCK I STAND" REV ISAAC DOUGLAS

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