# Rotational Kinetic energy

1. Aug 4, 2010

### mizzy

1. The problem statement, all variables and given/known data
A horizontal 800N merry go round of radius 1.50m is started from rest by a constant horizontal force of 50.0N applied tangetially to the merry go round. Find the kinetic energy of the merry go round after 3.00s. (assume it is a solid cylinder).

2. Relevant equations
I = MR^2

KE (rotational) = I (omega)^2

3. The attempt at a solution
I know this is a straight forward question. I don't know where to start. I know there are a few unknowns: omega, angular acceleration, velocity.

can someone guide me please? thanks

2. Aug 4, 2010

### Staff: Mentor

Hint: Use the torque to find the angular acceleration. Then use some kinematics. (That's just one way to go.)

3. Aug 4, 2010

### mizzy

K. thanks. Here's what I did:

I found moment of inertia (I = mr^2). For m, i found that using the given weight, 800N.

I used the torque equation to find angular acceleration. (torque = I * angular acceleration) Where torque is equal to the Force * r. Once i got the angular acceleration, i solved for tangential acceleration (a = r * angular acceleration).

Then I found v using the equation, v = a*t.

Once I got v, i found angular velocity from the equation, v = r * omega.

THen finally I can solve for Kinetic energy! KE = 1/2 * I (omega)^2

my answer came up to 2.76 x 10^4J, but in the book it's 276J!!!

4. Aug 4, 2010

### Staff: Mentor

That should be: I = 1/2 mr^2.

That's OK, but there's no need to convert from angular quantities to linear then back to angular! The kinematic formulas work just fine for angular quantities:
Use ω = alpha*t instead of v = a*t.

(The fewer 'conversions' the fewer chances for arithmetic errors.)

Give it one more shot.

5. Aug 4, 2010

6. Aug 4, 2010

### mizzy

ooops...i accidentally clicked on post reply.

In calculating the angular acceleration, i wrote down the wrong Force creating the torque. Instead of 50.0N, I used 500N!! silly mistake!

Thanks Doc Al =D