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Determining Torque

  1. Aug 11, 2004 #1
    I am attempting to the following question:
    How large is the torque that act on an armature?

    Particulars: (can be approximated as a solid cylinder)
    radius of .081 m
    length of .124 m
    mass of 13.13 kg
    Accelerated from REST to operating speed of 3530 rpm in 5.57 second

    I know T = F*r
    I'm not sure where to start.
     
    Last edited: Aug 11, 2004
  2. jcsd
  3. Aug 11, 2004 #2
    You are given the mass, the radius, and a change in velocity over a given time interval.

    Think of the problem another way:
    T = F x r
    from Newton's 2nd law: F = m*a
    a = dv/dt
    so you have: T = m dv/dt x r
    start by converting the given velocity into meters per second, then calculating the acceleration. from there it is plug and chug
     
  4. Aug 11, 2004 #3

    Doc Al

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    Newton's 2nd law - for rotational motion

    You'll need to apply Newton's 2nd law for rotational motion:
    [tex]\tau = I \alpha[/tex]

    You are given all the information needed to calculate I (the rotational inertia) and [itex]\alpha[/itex] (the rotational acceleration).
     
  5. Aug 11, 2004 #4
    Just to make sure you know that [itex]\tau = r \times F[/itex] not [itex]\tau = F \times r[/itex], it makes a difference. And [itex]||\tau||= rF\sin \phi[/itex]
     
  6. Aug 12, 2004 #5
    Now ya'll have me confused again, I thought I was all set with the first formula,
    T = f * r. Now where does Newtons Second Law fit into it to get the rotational movement.

    Take me step by step if you would.
    Thanks for the help so far
     
  7. Aug 12, 2004 #6

    Doc Al

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    Good point, Corneo. (But it won't matter for this particular problem.)
     
  8. Aug 12, 2004 #7

    Doc Al

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    T = r X F is true, but not helpful in this problem. Are you given the force? No.

    This is just the rotational analog to an "F = ma" problem. Instead of m, you calculate I; instead of a, you calculate α. Then apply T = I α.

    Give it a shot.
     
  9. Aug 12, 2004 #8
    Ok check to see if I am OK on this;
    we know T = I α
    I = 1/2 mr^2 and α = a / r
    so plug it in
    I = (1/2) 13.13 * .081^2 = .043072965
    α = (29.9425 / 5.57)/.081 = 66.36633641
    Multiply the two together for T and you get 2.8585 Nm

    How'd I do??
     
  10. Aug 12, 2004 #9

    Doc Al

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    Looks good. Note that you can calculate [itex]\omega[/itex] directly from the rpm, then use it to calculate [itex]\alpha = \omega/t[/itex]. Just convert rpm to radians/sec.
     
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