- #1

Treefolk

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## Homework Statement

It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 10.6 cm.

What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.4 MJ when spinning at an angular velocity of 90.0 rpm about an axis perpendicular to the disk at its center?

## Homework Equations

V = h[itex]\pi[/itex]r^2 <--Volume of a cylinder

I = .5mr^2 <-- Moment of an inertia of a solid cylinder

KE = .5I[itex]\omega[/itex]^2 <-- Kinetic Energy of a Cylinder

## The Attempt at a Solution

[itex]\delta[/itex] = 7800

h = .106m

m = [itex]\delta[/itex]*V = 7800*.106*[itex]\pi[/itex]*r^2 = 2597.5r^2

I = .5(2597.5r^2)r^2 = 1298.7r^4

KE = .5(1298.7r^4)[itex]\omega[/itex]^2 = 649.4r^4[itex]\omega[/itex]^2

[itex]\omega[/itex] = 90rpm = 1.5rps = 3rad/s

KE = 649.4r^4(3)^2

KE = 14.4x10^6

14.4x10^6 = 649.4r^4*(3)^2

r^4 = 2463.9

r=7.05

diameter = 14.1

That sorts through my logic, but the online homework spits back the answer as wrong, so I'm rather at a loss for what I did wrong. Any advice would be quite welcome.