Consider a playground merry-go-round with mass 242 kg and radius 1.54 m. Approxi

Question

Consider a playground merry-go-round with mass 242 kg and radius 1.54 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. (a) Calculate the rotational kinetic energy in the merry-go-round when it has an angular velocity of 21.5 rpm.
(b) Find the number of revolutions you would have to push the merry-go-round at the edge with force 23.5 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 3 revolutions.

Explanation / Answer

Part A :

rotational KE = 0.5 I w^2

here I = moment of inertia = mr^2

I = 242 * 1.54* 1.54 = 574 kgm^2

so Rotational KE = 0.5 * 574 * 21.5* 21.5* 2pi* 2pi/(60*60)

rot KE = 1454.84 J

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use Work = eenrgy = F s

so theta = 1454.84/23.5

theta = 62 radians

as 1 rev = 2pi radians

62 rad s = 62/2pi

= 9.85 rev or 10 rev

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F = 1454.84/(3* 2* 3.14)

F = 77.2 N

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