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posted by  Sammy S. on 11/4/2009 6:05:28 PM  |  status: Closed  |  Earned Karma: 55

8.36 merry-go-round

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N/A N/A N/A N/A 11/4/2009 at 5:40:00 PM
Question Details:
A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 17 rpm in 9.7 s. Assume the merry-go-round is a uniform disk of radius 2.1 m and has a mass of 680 kg, and two children (each with a mass of 30 kg) sit opposite each other on the edge.
What is the torque required to produce the acceleration, neglecting frictional torque.
What force is required at the edge?
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posted by Namero on 11/4/2009 7:28:57 PM  |  status: Live
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Response Details:
wo = 0
w = (17 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1.7802 rad/s
t = 9.7 s

The moments of inertia isI = 1/2mr2
so 
I = 1/2(680 kg)(2.1 m)2 = 2125 kgm2
But this is just the Merry-go-round, we need the total moment of inertia.  Treating the children as point 30 kg masses at the edge (r = 2.1 m) their moment of inertia would be each:
I = mr2 = (30 kg)(2.1 m)2 = 132.3 kgm2

The total moment of inertia is
I = 2125 kgm2 + 132.3 kgm2 + 132.3 kgm2 = 2389.6 kgm2

Now apply:
w = wo + at : wo = 0; w  = 1.7802 rad/s; t = 9.7 s
a = 0.1835 rad/s/s


And finally the angular equivalent of F = ma:
t = Ia
t = (2389 kgm2)(0.2160 rad/s/s) = 439 Nm of torque

Since force is applied at a 90o angle to the radius,  the torque is:
t = rF 

So F = t/r = (439 Nm)/(2.1 m) = 208.8 N = 209 N
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