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2 years ago

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A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent

ially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.34 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J.

1 answer:

nexus9112 [7]2 years ago

4 0

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F = 61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of merry-go-round, m = W/g

= 649/9.8 = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

= ¹/₂ x 66.225 x (1.05)²

= 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

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