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2nd law of rotation, merry go round

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data
    http://ready2goxtr.googlepages.com/problem1052.jpg

    A uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of 2.00 cm and a mass of 20.0 grams and is initially at rest. Starting at time t = 0 two forces are to be applied tangentially to the rim as indicated, so at time t = 1.25s the disk has an angular velocity of 250rad/s coutnerclockwise. Force F1 has a magnitude of .100 N. What is the magnitude of F2.


    2. Relevant equations
    T(net) = I[tex]\alpha[/tex]
    [tex]\omega[/tex] = [tex]\alpha[/tex] t <---- Equation 2



    3. The attempt at a solution
    r = 2.00cm = .2m m = 20.0g = .02kg t = 1.25s [tex]\omega[/tex] = 250 rad/s

    Using equation 2 I find [tex]\alpha[/tex] = 200 rad/ s^2

    I did try one way converting [tex]\alpha[/tex] into a linear acceleration by multiplying it by r Then said that F1 + F2 = ma

    Didnt work =(
     
  2. jcsd
  3. Jul 11, 2008 #2
    I would try to go the other way around, convert everything to rotation terms: you know [tex]\tau_{net} = I \alpha[/tex] so [tex]F_1 r - F_2 r = I \alpha[/tex].
     
  4. Jul 11, 2008 #3

    Doc Al

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    Staff: Mentor

    Good.
    Don't convert anything. What net torque is required to produce such an angular acceleration? (What's the rotational inertia of the disk?)
     
  5. Jul 11, 2008 #4
    How do i find the rotational inertia of the disk, its uniform, so does mr^2 work?


    or is it... 1/4 mr^2
     
  6. Jul 11, 2008 #5

    Doc Al

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    Don't just guess. Look it up!
     
  7. Jul 11, 2008 #6
    It just gives it to me in x y and z components.
     
  8. Jul 11, 2008 #7

    Doc Al

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  9. Jul 11, 2008 #8
    I think its 1/2 mr^2 in which ase I = .0004 so I (alpha) = .08


    F1r - F2r = .08
    F2 = -.3 N
     
  10. Jul 11, 2008 #9

    Doc Al

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    Staff: Mentor

    Correct formula, but recheck your arithmetic.
     
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