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!

1 answer

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