Moment of Inertia power plant energy

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SUMMARY

The discussion focuses on calculating the length of a hollow cylindrical flywheel designed to store 2.20 MJ of energy for a power plant. The flywheel has an inner radius of 0.420 m and an outer radius of 1.45 m, constructed from concrete with a density of 2150 kg/m³. The moment of inertia (I) is calculated using the formula I = 1/2 m (R₁² + R₂²), and the kinetic energy (KE) is derived from KE = 1/2 I ω². The initial calculations led to an incorrect volume determination, which was later corrected by using the volume difference between the outer and inner cylinders.

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  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with the concepts of kinetic energy and angular velocity
  • Knowledge of volume calculations for hollow cylinders
  • Basic proficiency in algebra and physics equations
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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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