Calculating Maximum Kinetic Energy in a Flywheel

[makeAwish]

Homework Statement

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!

 

[rl.bhat]

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

 

[sArGe99]

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

 

[LowlyPion]

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?

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

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.

 

[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.

 

[LowlyPion]

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.

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.

 

[makeAwish]

i got it. thanks!