Inertia Peak
Please help, I’m so frustrated with this physics problem. moment of inertia, etc.?
So I thought I figured this problem out, but I think my units are off. I just wanted to double check with someone. Please help! I’m sooo frustrated.
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 11.5 cm.
a. 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 91.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Thanks in advance guys!
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 = 91.0 rpm =91.0 rpm * 2Pi rad/rev / 60s/min = 9.53 rad/s
Ke = 14.1E6 J
h = 11.5E-2m
d = 7800 kg/m^3
So
r = (4*14.1E6/(7800*3.14*11.5E-2*9.53^2))^(1/4)
r = 3.85 m
Checking
volume = 11.5E-2*3.14 *3.85^2 = 5.36 m^3
mass = 7800 kg/m^3 * 5.36 m^3 = 41822 kg
I = 41749 * (3.85)^2/2 = 310 501 kg.m^2
Ke = 1/2*310501*(9.53)^2 = 14 100 000 J = 14.1MJ
b) Centripetal acceleration is
Ac = w^2r
Ac = 9.53^2*3.85m = 350 m/s^2
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