Posted by **Frank** on Thursday, November 12, 2009 at 6:05pm.

Hi I'm kinda new here, and hope someone can please help me out.

A person exerts a tangential force of 38.1 N on the rim of a disk-shaped merry-go-round of radius 2.74 m and mass 207 kg.

If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 32.5degrees?

Any help would be appreciated!

- Physics -
**bobpursley**, Thursday, November 12, 2009 at 6:15pm
Torque= I alpha

38.1*2.74=I alpha

For I,use a solid disk.

solve for angular acceleation, alpha

wf^2 = 2*alpha*displacement, where displacement is 32.5deg in RADIANS

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