Physics Help Forum Torque on merry-go-round

May 12th 2009, 10:36 PM   #1
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Join Date: Feb 2009
Posts: 24
Torque on merry-go-round

Hey guys, I don't seem to be able to get the right answer for this question eventhough I really thought I was going about it in the right way, here it is:
 A uniform disk can rotate around its center like a merry-go-round. The disk has a radius of 2.00 cm and a mass of 20.0 grams and is initially at rest. Starting at time t = 0, two forces are to be applied tangentially to the rim, so that at time t = 1.25 s the disk has an angular velocity of 250 rad/s counterclockwise. Force F1 has a magnitude of 0.100 N. What is the magnitude of F2?
They answer they got was 0.140 N and I got -0.02 N.

 May 13th 2009, 12:27 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 0.140 N doesn't look right. The two torques have to oppose for the ang accln alpha to be 200 as it works out in this case. If you just consider one force of 0.100 N, that gives an alpha = 500 , so obviously the other force has to counteract for this to fall to 200. I worked it out and got F = -0.06 N. Pl check your calculations.
 May 13th 2009, 02:34 AM #3 Senior Member   Join Date: May 2009 Location: Mumbai,India Posts: 102 well, the question really doesnt specify the direction of the torque due to F1 if i assume the torque to be CLOCKWISE instead of counterclockwise, i get magnitude of F2=0.14N parallel to F1 this is plausible as the the counterclockwise torque due to F2 has to be more than the clockwise torque in order to produce a counterclockwise angular acceleration. if we assume torque due to F1 to be COUNTERCLOCKWISE then we get the magnitude of F2=0.06N parallel to F1 the question should be more specific.
 May 14th 2009, 12:56 AM #4 Junior Member   Join Date: Feb 2009 Posts: 24 Oh sorry! there was a diagram that came with the question, F1 was clockwise and F2 was anti-clockwise.

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