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Moment of Inertia power plant energy

  1. Dec 9, 2014 #1

    462chevelle

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    1. The problem statement, all variables and given/known data
    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?
    2. Relevant 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
    3. 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?
     
  2. jcsd
  3. Dec 9, 2014 #2
    You've taken the volume of a solid cylinder.
     
  4. Dec 9, 2014 #3

    haruspex

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    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##.
     
  5. Dec 10, 2014 #4
    Chevelle probably checked wiki for the moment of inertia of an annulus like I did, haha.
     
  6. Dec 10, 2014 #5

    haruspex

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    Doesn't look that way from what was posted.
     
  7. Dec 10, 2014 #6

    462chevelle

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    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.
    solid cylinder is just 1/2mr^2
    the mass would just be the density*volume
     
  8. Dec 10, 2014 #7
    If we integrate from r2 to r1 we get 1/2m(r1^2 - r2^2) right?
     
  9. Dec 10, 2014 #8

    462chevelle

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    It should be plus, but other than that yes. That's what I'm using
     
  10. Dec 10, 2014 #9

    462chevelle

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    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.
     
  11. Dec 10, 2014 #10

    haruspex

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    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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