Consider a playground merry-go-round with mass 250 kg and radius 1.42 m. Approxi

Question

Consider a playground merry-go-round with mass 250 kg and radius 1.42 m. Approximate the merry-go-round as a solid, flat cylinder rotating about its axis of symmetry; you can look up the moment of inertia for this geometry in any college physics textbook. (E.g., the OpenStax College Physics figure captioned "Some rotational inertias".)

(a) Calculate the rotational kinetic energy in the merry-go-round when it has an angular velocity of 17.0 rpm.

(b) Find the number of revolutions you would have to push the merry-go-round at the edge with force 18.5 N to achieve this angular velocity starting from rest.

(c) Calculate the force you would need to exert to stop the merry-go-round in 2 revolutions.

Explanation / Answer

here,

mass , m = 250 kg

radius , r = 1.42 m

(a)

angular velocity , w = 17 rpm

w = 1.78 rad/s

the rotation kinetic energy , KE = 0.5 * I * w^2

KE = 0.5 * 0.5 * 250 * 1.42^2 * 1.78^2

KE = 339.3 J

the rotational kinetic energy of the merry go ground is 339.3 J

(b)

F = 18.5 N

let the angular accelration be alpha

F*r = I*alpha

18.5 * 1.42 = 0.5 * 250 * 1.42^2 * alpha

alpha = 0.104 rad/s^2

let theta be the angle

using third equation of motion

2*alpha*theta = 1.78^2

theta = 15.23 rad

the number of revolutions are 2.43 revolutions

(c)

when theta2 = 2 * 6.28 rad

theta2 = 12.56 rad

let the accelration be alpha'

using third equation of motion

2*alpha' * 12.56 = 1.78^2

alpha' = 0.126 rad/s^2

the force exerted , F' = I*alpha'/r

F' = 0.5 * 250 * 1.42^2 * 0.126 /1.42

F' = 22.39 N

the force exerted to stop the merry go ground is 22.39 N

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