Exercise 10.31 A playground merry-go-round has radius 2.50 mand moment of inerti

Question

Exercise 10.31

A playground merry-go-round has radius 2.50 mand moment of inertia 2500 kgm2 about a vertical axle through its center, and it turns with negligible friction.

Part A

A child applies an 20.5 N force tangentially to the edge of the merry-go-round for 18.0 s . If the merry-go-round is initially at rest, what is its angular speed after this 18.0 s interval?

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Part B

How much work did the child do on the merry-go-round?

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Part C

What is the average power supplied by the child?

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Exercise 10.31

A playground merry-go-round has radius 2.50 mand moment of inertia 2500 kgm2 about a vertical axle through its center, and it turns with negligible friction.

Part A

A child applies an 20.5 N force tangentially to the edge of the merry-go-round for 18.0 s . If the merry-go-round is initially at rest, what is its angular speed after this 18.0 s interval?

= rad/s

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Part B

How much work did the child do on the merry-go-round?

W = J

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Part C

What is the average power supplied by the child?

P = W

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Explanation / Answer

Here ,

raidius of merry - go -round , r = 2.50 m

moment of inertia , I = 2500 Kg.m^2

part A)

let the angular acceleration is aa

now, using second equation of motion

aa * 2500 = 20.5 * 2.50

aa= 0.0205 rad/s^2

angular velocity = aa * t

angular velocity = 0.0205 * 18

angular velocity = 0.369 rad/s

the angular velocity of wheel is 0.369 rad/s

part B)

work done = change in kinetic energy

work done = 0.5 * I * w^2

work done = 0.5 * 2500 * 0.369^2

work done = 170.2 J

the work done is 170.2 J

part C)

average power = energy/time

average power = 170.2/18

average power = 9.46 W

the average power supplied by the child is 9.46 W

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