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Find watts from rotational kinetic energy

  1. Apr 12, 2013 #1
    a)I used the equation KE=1/2Iw^2

    I=mR^2 ---> 1.5x10^30*20000^2
    w=2.1*2pi
    KE=1/2(1.5x10^30*20000^2)(2.1*2pi)^2=5.2x10^40 J
    t=2.1 rev/s / 1.4x10^-15 rev/s^2=1.5x10^15 seconds

    5.2x10^40 J/1.5*10^15 seconds =3.4x10^25 watts

    B)1.15*10^15 s=1.1*10^18 years

    I feel like this method is correct.
     
  2. jcsd
  3. Apr 12, 2013 #2
    edit

    a)I used the equation KE=1/2Iw^2

    I=mR^2 ---> 1.5x10^30*20000^2
    w=2.1*2pi
    KE=1/2(1.5x10^30*20000^2)(2.1*2pi)^2=5.2x10^40 J
    t=2.1 rev/s / 1.4x10^-15 rev/s^2=1.5x10^15 seconds

    5.2x10^40 J/1.5*10^15 seconds =3.4x10^25 watts

    B)1.15*10^15 s=1.1*10^18 years

    I feel like this method is correct.
     

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  4. Apr 12, 2013 #3

    Dick

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    There some problems there. For one thing the moment of inertia of a sphere is (2/5)MR^2. For another you are supposed to assume the rate of energy loss is constant, not that the rate of angular deceleration is constant. Finally saying "1.15*10^15 s=1.1*10^18 years" is just plain silly. A year is much longer than a second.
     
  5. Apr 12, 2013 #4
    using the inertia equation I have 2.4x10^38.
    making KE=2*10^40
    new answer =1.4*10^25 watts
     
  6. Apr 12, 2013 #5

    Dick

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    I don't think you are rounding correctly for the first answer if you want two significant figures. And you didn't pay any attention to my second comment. The rate of deceleration isn't a constant. The rate of energy loss is. You can't divide the energy by the expected lifetime given constant deceleration. Differentiate the KE expression to get dKE/dt in terms of dw/dt.
     
  7. Apr 12, 2013 #6
    the derivative would be Iw
     
  8. Apr 12, 2013 #7

    Dick

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    No, that's the derivative with respect to w. You want the derivative with respect to t. Use the chain rule.
     
  9. Apr 12, 2013 #8
    I got (w^2)/2
     
  10. Apr 12, 2013 #9

    Dick

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    How? KE=(1/2)*I*w^2. Take d/dt of both sides. I is constant. w isn't. It's a function of t.
     
  11. Apr 12, 2013 #10
    would it be just w?
     
  12. Apr 12, 2013 #11

    Dick

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    No!, why would you think that? You're just guessing. Use the chain rule!
    dKE/dt=d((1/2)*I*w^2)/dw*dw/dt.
     
  13. Apr 12, 2013 #12
    derivative of the outside 1/2Iw^2=Iw
    derivative of the inside Iw=I

    so final derivative = I^2W
     
  14. Apr 12, 2013 #13

    Dick

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    The derivative with respect to w is Iw, that's the outside. You've got that. Now the 'inside' is just w. You want to differentiate that with respect to t. What is it?

    You should really review the chain rule.
     
  15. Apr 12, 2013 #14
    would this involve a derivative being differentiating something with a y leaving y' on one side?
    if so it would be 2Iw
     
  16. Apr 12, 2013 #15

    Dick

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    I'll give you a closely related example. If F=Ky^3, then dF/dt=K*3*y^2*dy/dt. The K*3*y^2 is the derivative of the 'outside', the dy/dt is the derivative of the 'inside'.

    Like I said you are really thrashing around on this. Review it. For now can you apply that pattern to this problem?
     
  17. Apr 12, 2013 #16
    dKE/dt=Iw*dw/dt
     
  18. Apr 12, 2013 #17

    Dick

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    Right. Once more review this. Now use that to find watts.
     
  19. Apr 12, 2013 #18
    what am I suppose to plug in for dw/dt?
     
  20. Apr 12, 2013 #19

    Dick

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    What they gave you in the problem statement.
     
  21. Apr 12, 2013 #20
    can dw/dt also be power=F*v?
    I dont know it it is given or if I should solve for it.
     
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