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Twisting Disk - angular speed

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin metal disk of mass m=4.50 g and radius R=2.70 cm is attached at its center to a long fiber. When the disk is turned from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta.

    Suppose the disk is twisted away from equilibrium by an angle of 27 degrees and released from rest. What is the maximum angular speed of the disk as it twists back and forth?


    2. Relevant equations

    w = sqrt(k/I)
    t = -k(theta)

    3. The attempt at a solution

    So i found the first two parts of the problem fine but i am having trouble with this part. I am not sure where i can go with the given information

    I converted the degrees to radians

    27 degrees = .471239 rad

    not sure what to do now
     
  2. jcsd
  3. May 2, 2010 #2

    Doc Al

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    Are you given additional information that would allow you to calculate the spring constant of the fiber?
     
  4. May 2, 2010 #3
    That is all the information i was given.

    I didnt think i could find the spring constant with what is given...
     
  5. May 2, 2010 #4

    Doc Al

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    You can't.
     
  6. May 2, 2010 #5
    so is there any other way i can solve this without finding the spring constant?

    Its a mastering physics question and it asks for an answer in rad/s so i guess there has to be a way to find it :(
     
  7. May 2, 2010 #6
    Not sure if this will help but this was the question it asked before this one.

    The disk, when twisted and released, oscillates with a period T=1.05 s. Find the torsional constant kappa of the fiber.

    i found k to be 5.87 x 10^-5
     
  8. May 2, 2010 #7

    Doc Al

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    Of course it helps! Assuming the question refers to the same disk, you now have the spring constant.
     
  9. May 2, 2010 #8
    How do i have the spring constant now?

    The torsional constant kappa (k) is the same as the spring constant?
     
    Last edited: May 2, 2010
  10. May 2, 2010 #9

    Doc Al

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    Yes, it's the torsional spring constant (measured in Newton-meters per radian).
     
  11. May 2, 2010 #10
    oh it is ok.

    so then can't i just use w = sqrt(k/m)

    so w = 0.1142 rad/s
     
  12. May 2, 2010 #11

    Doc Al

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    No. You have the correct equation in your first post. (One is for linear motion; the other for rotational motion.)
     
  13. May 2, 2010 #12
    Thats what i tried earlier and got

    w = sqrt (k/I)

    w = sqrt (5.87*10^-5 / (.5*.0045kg*.027m^2))

    w = 5.98 rad/s but it says that is wrong?

    do i have to factor the 27 degrees in some how?
     
  14. May 2, 2010 #13

    Doc Al

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    You've found the angular frequency (ω) of the torsional pendulum. How does that relate to the angular speed?

    Of course.
     
  15. May 2, 2010 #14
    It doesnt i was getting my symbols confused :(

    I cant seem to get the angular speed though. i know

    angular speed = V/r

    also V = (angular frequency)(A)

    so i found V

    V = sqrt (k/I) * A

    V = 5.98 rad/s * ( .4712 rad)

    V = 2.819

    then i used angular speed = V/r

    angular speed = 104 rad/s but that cant be right?

    am i using a wrong equation?
     
  16. May 2, 2010 #15

    Doc Al

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    In that equation, V is the tangential speed of something in circular motion. Not what you want here.

    Hint: Write the angle as a function of time like this:

    θ = θmax cos(ωt)
    where θmax is the maximum angle (in radians)

    How would you write the angular speed (dθ/dt) as a function of time? What's the maximum angular speed?
     
  17. May 2, 2010 #16
    would (dθ/dt) = -θmax*ω*sin(ωt)
     
  18. May 2, 2010 #17

    Doc Al

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    Exactly! So what is the maximum value of dθ/dt?
     
  19. May 2, 2010 #18
    I am a little confused on what to plug into where now.

    for ω would i use (2pi)/1.05 ------ 1.05 is the Period given for the earlier part.

    ω = 5.984 rad/s

    and then i am not sure what to use for t?

    is t maybe 1/4*(1.05) = .2625 s

    then for theta max just use 27*pi/180 = .471239 rad right?
     
  20. May 3, 2010 #19

    Doc Al

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    Yes, you've already found ω.

    That's correct, but it's easier than that. What's the maximum value of sinωt?

    Right.
     
  21. May 3, 2010 #20
    the max value of sinωt is always 1. That comes in handy haha :)

    Thank you for all the help Doc Al. I really appreciate it.
     
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