# Rotational KE problem

1. Mar 17, 2009

### makeAwish

The problem statement, all variables and given/known data

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.30m and a mass of 66.0kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600m/s² .
What is the maximum kinetic energy that can be stored in the flywheel?

The attempt at a solution

I took I = 3/2MR² (at rim)
where R = 2.6m
Then K = 1/2 * I * ω², where ω² = a/R

so K = (1/2)(3/2)(66kg)(2.6m)²(3600/2.6) = 463320J
which is wrong..
The ans is 7.72x10^4 J

can someone pls tell me where i gone wrong?
Thanks!!

2. Mar 17, 2009

### rl.bhat

Moment of inertia of flywheel is 1/2*M*R^2 and radius is 1.3 m.

3. Mar 17, 2009

### sArGe99

Radial acceleration = $$m r^2 \omega$$
Find out $$\omega$$
Calc. K.E using the equation $$\frac{1}{2} I \omega^2$$..

4. Mar 17, 2009

### LowlyPion

Not quite. The equation is correct, but ω = v/r = 3600/1.3

As noted by rl.bhat a uniform disk has a moment of inertia of of 1/2mr².

5. Mar 17, 2009

### sArGe99

The Moment of Inertia isn't always 1/2 mr^2, is it?
It does depend upon the axis chosen, with M.I. of the new axis found out using parallel and perpendicular axis theorems.

6. Mar 17, 2009

### LowlyPion

Rotating about its central axis it is.
http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl2

The || and ⊥ axis theorem are useful of course for other axes of rotation.

7. Mar 20, 2009

### makeAwish

i got it. thanks!