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Rotational Mechanics

  1. Dec 6, 2007 #1
    Hello,

    Hopefully this is in the correct place.
    I am presented with a the following problem.
    "A hamster running on an exercise wheel, exterts a torque on the wheel. If the wheel has an angular velocity that can be expressed as:
    [tex]\omega[/tex](t)= 3.0 rads/s + (8.0 rad/s[tex]^{}2[/tex])t + (1.5 rad/s[tex]^{}4[/tex])t[tex]^{}3[/tex]. Calculate the torque on the wheel as a function of time. Assume that the moment of inertia is 500 kg*m[tex]^{}2[/tex] and is constant."

    [tex]\tau[/tex]=Fr F=m[tex]\alpha[/tex] and I=mr[tex]^{}2[/tex]

    I then said that [tex]\tau[/tex]=m[tex]\alpha[/tex]r. Next I set I=mr[tex]^{}2[/tex] equal to m and plugged it into [tex]\tau[/tex]=m[tex]\alpha[/tex]r.
    I got [tex]\tau[/tex]=I[tex]\alpha[/tex]/r.

    After that I differentiated the angular velocity and got [tex]\alpha[/tex](t)=8.0 + 3(1.5)t[tex]^{}2[/tex]. I plugged it in [tex]\tau[/tex]=I[tex]\alpha[/tex]/r and solved. My end result is: [tex]\tau[/tex](t)=2250t[tex]^{}2[/tex] + 4000[tex]/[/tex]r.

    Is this correctly done?
     
  2. jcsd
  3. Dec 6, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    One error is mixing up Newton's 2nd law for rotation and translation.
    For translation:
    [tex]F = m a[/tex]
    For rotation:
    [tex]\tau = I \alpha[/tex]
     
  4. Dec 6, 2007 #3
    Ok, thank you very much.
     
  5. Dec 6, 2007 #4
    NO

    L=IA=Id(w)/dt=I(8+9/2t^2)
    A=angular acceleration
    w=angular velocity
    I=momentum of inertia=mr^2
     
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