Rotational Kinetic Energy

1. Oct 25, 2007

lgmavs41

1. The problem statement, all variables and given/known data

A horizontal 813 N merry-go-round of radius 1.28 m is started from rest by a constant horizontal force of 66 N applied tangentially to the merry-go-round. The acceleration of gravity is 9.8 m/s^2. Assume it is a solid cylinder. Find the kinetic energy of the merry-go-round after 2.68 s.

2. Relevant equations
Weight = mg
F=ma
Vf=Vi + at
Kr = 1/2 (moment of inertia*angular speed^2)
moment of inertia = 1/2 Mass*Radius^2 for solid cylinder

3. The attempt at a solution
Well, using the constant force applied and the weight of the merry-go-round, I found the tangential acceleration: a = F * g / weight...Then I solved for the tangential velocity with Vf = a * t since it started from rest...then I solved for the angular speed:
w = tangential velocity / radius. Then plugged all the numbers I found to solve for moment of inertia and rotational kinetic energy. My final answer is 94.31353 Joules. I entered it in the computer and I got it wrong. I'm not sure where my error is. pls advice. Thanx!

2. Oct 25, 2007

lgmavs41

oh. I should have considered torque...