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A uniform disk with mass 43.9 kgkg and radius 0.280 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.0 NN is applied tangent to the rim of the disk. Part A What is the magnitude vv of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 revolution

Respuesta :

Answer:

1.36 m/s

Explanation:

I = ½mR²

τ = FR

α = τ/I = FR / (½mR²) = 2F/mR

a = Rα = 2F/m

s = θR

v² = u² + 2as

u = 0

v = √2as = √(2(2F/m)(θR)) = 2√(FθR/m)

v = 2√(29.0(0.400)(2π)(0.280) / 43.9) = 1.3636272...

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