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3 years ago

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A 125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

bout the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

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