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Angular velocity of a solid disk

  1. Jul 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A solid disk is spinning around it center with an angular velocity of 35.0 rad/s. The disk has a mass of 3.6kg and a radius of .50m. You drop a thin ring with the same radius and a mass on the disk along the same center of rotation. What is the angular velocity of the disk and ring?

    2. Relevant equations
    Id= 1/2mR^2
    Ir=mR^2
    w=vf/R


    3. The attempt at a solution

    Id=1/2(3.6kg)(.50)^2 =.45 kg m/s
    Ir= (3.6)(.50)^2= .90

    W= (I'd+Ir)/ R
    = (.45+.90)/(.50)
    =2.7 rad/s

    Is that the correct answer?
     
  2. jcsd
  3. Jul 16, 2014 #2

    Nathanael

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    w stands for angular velocity right?

    That's the relationship between linear velocity (at a distance R) and angular velocity.



    For this problem you'll want to use conservation of angular momentum.
     
  4. Jul 16, 2014 #3
    Would it be Li=Lf= i(wf)
     
  5. Jul 16, 2014 #4

    Nathanael

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    Yes, it would be [itex]L_i=L_f[/itex] (initial angular momentum = final angular momentum)

    So what is the initial angular momentum? What then changes? And what does this change do to the angular velocity?
     
  6. Jul 16, 2014 #5
    The initial is 35 rad/s aNd nothing changes because the second object is has the same mass and radius. Therefore would the finial be 35 as well?
     
  7. Jul 16, 2014 #6

    Nathanael

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    35 rad/s is the initial angular velocity, but it is not the same as the initial angular momentum.

    Angular momentum is conserved, not angular velocity.
     
  8. Jul 16, 2014 #7
    So it would be L= I times w

    Since I solved for IDisk and Iring, would I add those two up and mutiple by w?
     
  9. Jul 16, 2014 #8

    Nathanael

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    What would be your reasoning for doing that?

    You know [itex]I_i[/itex] and [itex]ω_i[/itex] and [itex]I_f[/itex] so how can you solve for [itex]ω_f[/itex]?


    (Check posts #3 and #4)
     
  10. Jul 16, 2014 #9
    ((Id)(wd)+(Ir)(wr)) / R
    = answer

    Wf=w initial - answer
    And that would give me the final angular vel?
     
  11. Jul 17, 2014 #10

    Nathanael

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    I'm not sure where you're getting these equations.

    If:
    [itex]L_i=L_f[/itex]

    Then:
    [itex]I_iω_i=I_fω_f[/itex]


    So what is the final rotational inertia?
     
  12. Jul 17, 2014 #11
    I'm a little confused because I solved for the iRing and the iDisk but no I don't know where to plug it in
     
  13. Jul 17, 2014 #12

    Nathanael

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    iDisk is the initial rotational inertia right?

    And after the ring is added, it still has that initial rotational inertia, but now it also has an additional rotational inertia from the ring (iRing)

    How would you write that in math?
     
  14. Jul 17, 2014 #13
    So would it be (.45)(2.7rad/s)=(.45+.90)(Wf)?
     
  15. Jul 17, 2014 #14

    Nathanael

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    Very close, but where did you get 2.7 rad/s?
     
  16. Jul 17, 2014 #15
    I got 2.7 from the first post so would it be (.45)(35m/s)=(.45+.90)(Wf)?
     
  17. Jul 17, 2014 #16

    Nathanael

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    Yes, it would be that.

    Now just solve for the final angular velocity and you're done!
     
  18. Jul 17, 2014 #17
    THANK YOU! Very much appreciated Nathanael
     
  19. Jul 17, 2014 #18

    Nathanael

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    No problem, and welcome to the physics forums!
     
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