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## A horizontal 790-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 45 N applied tangentially to the m

Question

A horizontal 790-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

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Physics
3 years
2021-08-26T05:19:21+00:00
2021-08-26T05:19:21+00:00 1 Answers
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## Answers ( )

Answer:

404.3 J

Explanation:

Given that

Weight of the merry go round = 790 N

Radius if the merry go round = 1.6 m

Horizontal force applied = 45 N

Time taken = 4 s

To find the mass of the merry go round, we divide the weight by acceleration due to gravity. Thus,

m = F/g

m = 790 / 9.8

m = 80.6 kg

We know that the moment of inertia is given as

I = ½mr², on substitution, we have

I = ½ * 80.6 * 1.6²

I = 103.17 kgm²

Torque = Force applied * radius, so

τ = 45 * 1.6

τ = 72 Nm

To get the angular acceleration, we have,

α = τ / I

α = 72 / 103.17

α = 0.70 rad/s²

Then, the angular velocity is

ω = α * t

ω = 0.7 * 4

ω = 2.8 rad/s

Finally, to get the Kinetic Energy, we have

K.E = ½ * Iω², on substituting, we get

K.E = ½ * 103.17 * 2.8²

K.E = 404.3 J

Therefore, the kinetic energy is 404.3 J