How to figure out the ANGULAR SPEED of this problem?

A person exerts a tangential force of 38.0 N on the rim of a disk-shaped merry-go-round of radius 2.77 m and mass 156 kg. If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 33.3掳?





Please include any formulas you used and explain how to solve the problem in order to get ';BEST ANSWER';How to figure out the ANGULAR SPEED of this problem?
If this were a problem in uniform linear acceleration you'd use this formula:


v^2 = v0^2 + 2as


where v = final velocity, v0 = initial velocity, a = acceleration and s = distance traveled.





The analogous formula for uniformly accelerated rotational motion is:


w^2 = w0^2 + 2a*theta


where w, w0 = final and initial angular velocities (w0 = 0), a = angular acceleration and theta = total angle moved (33.3 degrees, which you need to convert to radians).





Since w0 = 0, w^2 = 2*a*theta. So all you need to know is the angular acceleration a.





In linear motion, F = ma. The analogous formula is T = I*a, where T = torque, I = moment of inertia, a = angular acceleration. So a = T/I.





The torque is force times radius. You have both those numbers.


The moment of inertia for a disk is (1/2)*m*r^2, and you have m and r.





So you have everything you need to solve the problem. Get the moment of inertia, get the torque, use those to find a, plug into the formula for w^2 to get w.