Moment of Inertia power plant energy

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Homework Help Overview

The discussion revolves around the calculation of the moment of inertia for a flywheel designed to store energy in a power plant context. The flywheel is specified as a hollow cylinder with given inner and outer radii, and the problem involves determining its length based on energy storage requirements.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the moment of inertia formula and its implications for energy storage. There are attempts to clarify the correct volume calculation for the hollow cylinder, with some questioning the initial approach taken by the original poster.

Discussion Status

Several participants have provided feedback on the original poster's calculations, suggesting that the volume of the hollow cylinder may have been miscalculated. There is an ongoing exploration of the correct formulas and methods to approach the problem, with some participants offering insights into the properties of solid and hollow cylinders.

Contextual Notes

Participants note the importance of correctly identifying the shape of the flywheel as an annulus and the implications this has for calculating mass and moment of inertia. There is also mention of using density and volume relationships in the context of the problem.

462chevelle
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Homework Statement


It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large flywheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius 0.420m and outer radius 1.45m , using concrete of density 2150kg/m3

If, for stability, such a heavy flywheel is limited to 1.35 second for each revolution and has negligible friction at its axle, what must be its length to store 2.20MJ of energy in its rotational motion?

Homework Equations


KE=1/2Iw^2
I=1/2(m)(R^2+R^2) the 2 different R's here
d=m/v
V=pir^2*L

The Attempt at a Solution


So I plugged the moment of inertia formula into KE.
to get: KE=1/2(1/2*m(R^2+R^2))w^2
to get w i took 2pi/1.35
so i end up with 2.2e6=1/2(1/2)m(.42^2+1.45^2)*4.65^2
solving for m i end up with
(4*2.2e6)/(2.2789*4.65^2)=m
m=178588kg
since density is 2150 i took
2150=178588/v
solved for v and got 83m^3
took V=pi*1.45^2*L
I got 12.5 for my length, seems to be the incorrect answer
ive tried a few different things, but this method is the most logical for me
Does anyone see anything obvious wrong?
 
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You've taken the volume of a solid cylinder.
 
Mickey Tee said:
You've taken the volume of a solid cylinder.
Not exactly. chevelle has taken two solid cylinders and added them.
chevelle, the shape is an annulus, a solid cylinder with a concentric cylinder removed. The easiest way to deal with this, if you don't know the specific formula, is to treat it as the difference of two solid cylinders of the same density. What are the mass and moment of inertia of a solid cylinder radius r, length l, density ##\rho##.
 
Chevelle probably checked wiki for the moment of inertia of an annulus like I did, haha.
 
Mickey Tee said:
Chevelle probably checked wiki for the moment of inertia of an annulus like I did, haha.
Doesn't look that way from what was posted.
 
I used the formula in the book that is given for a hollow cylinder, I=1/2m(1R^2+2R^2). The 1 and 2 aren't coefficients, I'm just showing r1 and r2.
haruspex said:
What are the mass and moment of inertia of a solid cylinder radius r, length l, density ρ\rho.
solid cylinder is just 1/2mr^2
the mass would just be the density*volume
 
If we integrate from r2 to r1 we get 1/2m(r1^2 - r2^2) right?
 
It should be plus, but other than that yes. That's what I'm using
 
Alright, I figured it out. Instead of using the volume formula like I was I should have been using.
V=V(outside)-V(inside)

thanks for the help.
 
  • #10
Apologies, your formula for the MoI of the annulus is correct. I should have checked.Mickey Tee was right in the first place - it's your calculation of the volume that's wrong.
 

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