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Rotational period for constant mass but different volume

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    The Earth's radius is 6371 km. If the Earth's radius were to increase by 30 m (0.03 km), but no change in mass, by what percentage would the Earth's rotational period increase? (Model the Earth as a uniform sphere)

    2. Relevant equations
    ∑Torque = Iα
    v = rω
    a = rα
    KE = (1/2)Iω^2

    3. The attempt at a solution
    Moment of inertia for a sphere is (2/5)mr^2, so ∑Torque = (2/5)mr^2 * (a/r)
     
  2. jcsd
  3. Apr 23, 2016 #2

    TSny

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    Welcome to PF!

    ##\sum \tau = I \alpha## only applies to a rigid body rotating about a fixed axis. Here, the earth is not acting as a rigid body.

    Can you relate this problem to any other examples you've been studying lately where a rotating system changes its mass distribution while no external torque acts?
     
  4. Apr 23, 2016 #3
    Can we solve it with the rotational KE and its relation to PE?
     
  5. Apr 23, 2016 #4

    TSny

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    Rather than energy, there's a much more appropriate physical quantity that can be used. Have you read or discussed in class what goes on when an ice skater pulls in her arms in order to spin faster?
     
  6. Apr 23, 2016 #5
    Oh yeah! Conservation of angular momentum! Which is L = Iω. And it will be the same at its original size and after it grows a little bit.
    (2/5)mr^2 ω = (2/5)m(r+.03)^2 ω, m's cancel out and I solve for ω?
     
  7. Apr 23, 2016 #6

    TSny

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    Yes. (Distinguish between the two ω's.)
     
  8. Apr 23, 2016 #7
    Yeah, and then we know one of the ω's already because we know the period of the earth right now, which is 23 hours 56 minutes 4.1 seconds, or 86164.1 s. Take 2π divided by that time and that is the initial ω, and using that I can solve for the one I need.
     
  9. Apr 23, 2016 #8

    TSny

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    OK. But since you are asked to find the % change in ω, you could first work out an expression for the % change and simplify the expression before substituting numbers. That way, you might not even need to know the initial rate of spin.
     
  10. Apr 23, 2016 #9
    Right now what I have is:
    (r_f)^2 * ω_f = (r_i)^2 * ω_i where _f indicates the final state and _i indicates the initial state. Can I turn that into some expression for the percent change?
     
  11. Apr 23, 2016 #10

    TSny

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    Yes. How to you find the % change in a quantity?
     
  12. Apr 23, 2016 #11
    You divide it by the original quantity and multiply by 100
     
  13. Apr 23, 2016 #12

    TSny

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  14. Apr 23, 2016 #13
  15. Apr 23, 2016 #14

    TSny

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    Yes. (Sorry, somehow I was thinking the question asked for percent change in ω rather than T.)
     
  16. Apr 23, 2016 #15

    TSny

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    It might be helpful to note that ##\frac{T_f - T_i}{T_i} = \frac{T_f}{T_i}-1##.
     
  17. Apr 23, 2016 #16
    Is this right?
    ω_f = (r_i)^2 * ω_i / (r_f)^2
    then, (ω_f-ω_i)/ω_i = (ω_f/ω_i) - 1
    Using the expression I got for ω_f above, that gives:
    ((r_i)^2 / (r_f)^2) -1
     
  18. Apr 23, 2016 #17

    TSny

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    Yes, this would give the % change in ω. Likewise, you can determine an expression for the % change in T in terms of just ri and rf.

    (Actually, of course, to get % change you need to multiply by 100.)
     
  19. Apr 23, 2016 #18
    Yeah, multiply by one hundred, and to get that expression for the change in T, would I just have to take 2π divided by the expression I have now?
     
  20. Apr 23, 2016 #19

    TSny

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    No. I would go back to the angular momentum relation and express it in terms of T rather than ω. Then you can easily find an expression for Tf/Ti.
     
  21. Apr 23, 2016 #20
    ω in terms of T is (Δθ)/T right? I substituted that in place of ω and when I solved for it, it gave me
    ((r_f)^2 / (r_i)^2) - 1
    Is that the expression for % change in the period?
     
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