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itsme24
Nov15-06, 12:38 AM
Ok the problem is:

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of 1.30 m and a mass of 72.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600 m/s^2.

What I did was solve for the angular velocity through the radial acceleration:

3600m/s^2 = rw^2

w = 52.6 rad/s

Then I solved for the moment of inertia:

I = mr^2 = 72.0kg(1.30m)^2 = 122 kg*m^2

Finally I plugged it all into the rotational kinetic energy equation:

K = (1/2)(122m*m^2)(52.6rad/s)^2 = 1.68*10^5 J

The actual answer is 8.42*10^4, exactly half of what I got. I don't suppose someone could explain where the *(1/2) is coming from?

OlderDan
Nov15-06, 12:48 AM
The moment of inertia of a disk is??????

itsme24
Nov15-06, 12:56 AM
hehe (1/2)mr^2

swifty
Apr7-10, 01:13 AM
Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.16m and a mass of 73.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s^2.

CAN SOMEONE PLEASE SOLVE THIS FFS IVE TRIED SO MANY OPTIONS AND ITS NOT WORKING!@@!_#!@#!@!@#)!@(#*)!@(#)((!@*