It has been argued that power plants should make use of off-peak hours (such as

Question

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/m3 , 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 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center? What would be the centripetal acceleration of a point on its rim when spinning at this rate?

Explanation / Answer

a) The kinetic energy of a rotating disk is

Ke = 1/2 I w^2 where

Ke is energy in joules

I is moment of inertia in kg.m^2

w is angular velocity in rad/s

The moment of inertia of a rotating disk is

I = mr^2/2

And the mass of the disk is going to be

m = d * Pi * r^2 * h where

d is density

r is radius

h is thickness

So

Ke = 1/4 * d * Pi * r^2 * h * r^2 * w^2

Ke = 1/4 * d * Pi * h * w^2 * r^4

r = (4Ke/(d * Pi * h * w^2)^(1/4)

with

w = 93.0 rpm =93.0 rpm * 2Pi rad/rev / 60s/min = 9.74 rad/s

Ke = 14.1E6 J

h = 10.6e-2 m

density = 7800 kg/m^3

So

r = (4*14.1E6/(7800*3.14*10.6E-2*9.53^2))^(1/4)

r = 3.892 m

diameter=7.784 m

b) Centripetal acceleration is

Ac = w^2r

Ac = 9.73^2*3.894m = 368.65 m/s^2

ac= 369m/s2

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