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Torque & Angular Speed

  1. Nov 2, 2007 #1
    [SOLVED] Torque & Angular Speed

    1. The problem statement, all variables and given/known data

    The wheel of an engine has a moment of inertia 2.80kgm^2 about its rotation axis.

    What constant torque is required to bring it up to an angular speed of 360 rev/min in a time of 7.60 s, starting from rest?

    What is its final kinetic energy?

    2. Relevant equations

    torque = rF sin(theta)

    3. The attempt at a solution

    First I converted 360 rev/min to seconds


    360 (rev/min) ( 1 min/60s)( 2pi/rev) = 37.699 rad/s

    t = 7.60

    I = 2.80 kgm^2

    so I decided to get angular acceleration, in case I need it is

    alpha = 4.96

    From what I spotted in my book Sum of torque = I(angular acceleration)

    torque = (2.80)(4.96)
    torque = 13.89

    this is right, but is they way I have done it correct or did I just get lucky. I am afraid I just plugged into the "sum of torquez = I (angular accelerationz)"

    without understanding it :O

    part two of the question is

    What is its final kinetic energy?

    for this


    how do I implement this into the problem if I don't have mass?

    omega = omega initial + alpha(time)
  2. jcsd
  3. Nov 2, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's fine. Realize that the equation "Torque = I*alpha" is just the rotational version of Newton's 2nd law; the translational version (which you know and love) is "Force = mass*acceleration".

    You don't need mass. For rotational kinetic energy, use I. What's the rotational version of the kinetic energy formula?

    Read this: Rotational-Linear Parallels
  4. Nov 2, 2007 #3
    Great! Thank you very much Doc Al, you are the greatest. :)
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