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Rotational inertia and torque

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A 103 g Frisbee is 16 cm in diameter and has about half its mass spread uniformly in a disk and the other half concentrated in the rim. With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 730 rpm.

    a) What is the rotational inertia of the Frisbee?

    b) What is the magnitude of the torque, assumed constant, that the student applies?

    2. Relevant equations

    [tex]I = \frac{1}{2}MR^{2}[/tex]
    [tex]I = MR^{2}[/tex]
    [tex]\tau = I\alpha[/tex]

    3. The attempt at a solution

    I already solved for the rotational inertia from the combination of [tex]I =\frac{1}{2}MR^{2}[/tex] and [tex]I = MR^{2}[/tex]. But I can't seem to get how to get the torque from [tex]\tau= I\alpha[/tex]. I know what I (rotational inertia) but I don't know how to get [tex]\alpha[/tex]. Help me figure it out! Thanks.
     
  2. jcsd
  3. Mar 15, 2009 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi 4t0mic! Welcome to PF! :smile:

    (have an alpha: α and an omega: ω :wink:)

    You know that after a quarter revolution, ω = 730 rpm …

    so you have a "distance" and an initial and final "speed" …

    so use one of the usual constant acceleration equations :wink:
     
  4. Mar 15, 2009 #3
    Hello Tiny Tim.

    Oh, I see. So with the constant acceleration equation:
    ωf² = ωi² + 2αθ
    I get:
    α = ωf² - ωi² / 2θ
    Giving me:
    τ = Iα
    τ = I (ωf² - ωi² / 2θ)

    Given I = 4.9x10^-4 kg*m² (my answer from part A) and
    730 rpm*(1m/60s)=12.1667rps

    τ = (4.9x10^-4 kg*m²)([12.1667rps]² - 0 / 2*0.25 rev) = 0.145 kg*m²*r/s²

    How do I convert kg*m²*r/s to the SI units for torque, which is N*m? I know N is kg*m/s²... overall I'm stuck.
     
    Last edited: Mar 15, 2009
  5. Mar 15, 2009 #4
    Or maybe the question is... how to convert revolutions (r) into SI units?
     
  6. Mar 16, 2009 #5

    tiny-tim

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    Hello 4t0mic! :smile:
    Exactly! :approve:

    The great thing about SI units is that if all your inputs are in them, then your result is also …

    you don't need to both about what a newton is (if you break it down) … any force in SI units will be in newtons automatically!

    So, as you say, the only question is... how to convert revolutions (r) into SI units …

    well, a revolution is just a dimensionless number, so convert it to radians and you're on a home run :smile:
     
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