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Stuck in this question - 2 spinning disks brought together

  1. Nov 1, 2013 #1
    Stuck in this question -- 2 spinning disks brought together...

    Two disks are spinning freely about axes that run through their respective centres. The larger disk
    (R1 = 1.42 m)
    has a moment of inertia of 1180 kg · m2 and an angular speed of 4.0 rad/s. The smaller disk
    (R2 = 0.60 m)
    has a moment of inertia of 906 kg · m2 and an angular speed of 8.0 rad/s. The smaller disk is rotating in a direction that is opposite to the larger disk. The edges of the two disks are brought into contact with each other while keeping their axes parallel. They initially slip against each other until the friction between the two disks eventually stops the slipping. How much energy is lost to friction? (Assume that the disks continue to spin after the disks stop slipping.)


    What I did was,

    First I found w(angular velocity when the speed becomes same) by:

    w=(w1I1+w2I2)/(I1+I2)

    The the Kf=1/2w^2(I1+I2)
    And Ki=1/2w1^2I1+1/2w2^2I2

    Then I found the energy lost by Kf-Ki but The answer is wrong...
     
  2. jcsd
  3. Nov 1, 2013 #2

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    Hello daveamal,

    Welcome to Physics Forums! :smile:

    Your approach looks correct. (Although your notation is atrocious. It's very difficult to read.)

    If you show more of your work, perhaps we can help find out what went wrong.

    One thing to keep in mind though is when you calculate your
    [tex] \omega = \frac{\omega_1 I_1 + \omega_2 I_2}{I_1 + I_2}, [/tex]
    make sure you pay attention to the signs of [itex] \omega _1 [/itex] and [itex] \omega_2 [/itex]. Are they in the same direction or opposite direction?
     
  4. Nov 1, 2013 #3
    Oh ,they are in opposite direction, it would be:

    ω=ω1I12I2[itex]/[/itex](I1+I2)

    Ryt?
     
  5. Nov 1, 2013 #4

    Doc Al

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    Correct. (Watch those parentheses.)
     
  6. Nov 1, 2013 #5

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    Well, it depends on how you look at it.

    Your original
    ω= (ω1I12I2)/(I1+I2)
    formula is fine the way it is with the "+" sign, with just realizing that either ω1 or ω2 is negative (but not both).

    In other words, you don't need to put a negative in the formula; the negative sign comes from the numerical value of one of the variables.
     
  7. Nov 1, 2013 #6
    The initial angular momentum of disc 1
    is I1 w1
    The initial angular momentum of disc 2
    is I2 w2
    Add those for total angular momentum, which DOES NOT CHANGE in this problem
    because there are no external moments

    Afterwards the no slip condition:
    R1 w1 = R2 W2
    calculate the angular momentum again and find new w s

    Now
    Initial KE = (1/2) I1 w1^2+(1/2)I2 w2^2

    Final Ke = same formula, new w s

    final better be less than initial :)

    find difference
     
  8. Nov 1, 2013 #7
    I did the steps and got the answer as 36900.17 joules, but it says that the answer is still wrong.
     
  9. Nov 1, 2013 #8
    How did you find new w s?
    What did you do with the total angular momentum and R1 w1 = R2 W2 ?
     
  10. Nov 1, 2013 #9

    Doc Al

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    Ah, so they do not rotate about a common center. (I was visualizing the problem incorrectly.)

    Kishlay has the right idea.
     
  11. Nov 1, 2013 #10
    Ok but how to find new ω by using R1ω1=R2ω2 ?
     
  12. Nov 1, 2013 #11

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    For what it's worth that's the same answer I got. Perhaps the program is asking you to enter your answer in terms of kJ? Or maybe it's asking for the answer in fewer significant figures?
     
  13. Nov 1, 2013 #12

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    Oh, now I get it. I originally thought the two disks were on the same axis too (I too was visualizing the problem incorrectly). That does change things. :redface:
     
  14. Nov 1, 2013 #13

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    There isn't a "single" new ω. Instead there are two "new" ω values.

    You already know the initial ω values, ω1i and ω2i .

    Now solve for the final values, ω1f and ω2f .

    (Hint: two equations, two unknowns :wink:).
     
  15. Nov 1, 2013 #14
    I still don't understand what I did wrong? How does the radius come into consideration?
     
  16. Nov 1, 2013 #15

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    The way I'm visualizing the problem now is that the two disks are touching at their edges -- sort of like two gears, except without the teeth.

    Gears_animation.gif
     
    Last edited by a moderator: Apr 15, 2017
  17. Nov 1, 2013 #16
    ok, So will the 2 equations be:

    ω12=newω1/newω2

    and

    Angular momentum= I1newω1+I2newω2

    Then we will find new omega's?
     
  18. Nov 1, 2013 #17
    oh no, That was wrong, even the I's will change den, but how?
     
  19. Nov 1, 2013 #18

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    I don't think that's right.

    Yes, you will need to find "new" omegas.

    One of your equations involves conservation of angular momentum. The total, initial angular momentum is the same as the total, final angular momentum. That should give you one equation (with two unknowns: the final angular velocities, ω1f and ω2f).

    See Kishlay's post for the second equation. The second equation involves the disks' radii. That's where the radii fit in. (Look at the picture of the gears. Notice the small gear has a larger angular velocity than the big gear does). Can you find a relationship between the gears' radii and their respective angular velocities, such that there isn't any slipping?
     
  20. Nov 1, 2013 #19

    haruspex

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    No. The final rates will be in inverse proportion of the radii. The original rates are not (or there would be no skidding).
     
  21. Nov 1, 2013 #20

    haruspex

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    Also need to flip the sign for the 'after' rates, so it would be ω=(ω1I12I2[itex])/[/itex](I1-I2)
    But that's for the coaxial problem anyway, which was a wrong interpretation.
     
    Last edited: Nov 1, 2013
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