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Rotational Inertia of Pulley, Integration Help!

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A pulley, with a rotational inertia of 1.5 10-3 kg·m2 about its axle and a radius of 21 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t[tex]^{2}[/tex], where F is in newtons and t in seconds. The pulley is initially at rest.

    (a) At t = 10.0 s what is its angular acceleration?
    rad/s2


    (b) At t = 10.0 s what is its angular velocity?
    rad/s




    2. Relevant equations
    Tnet=I[tex]\alpha[/tex]

    f=.50t + .30t[tex]^{2}[/tex]

    3. The attempt at a solution
    I allready got the acceleration for the pulley. It turned out to be 4900 rad/sec. But now i assume because the amount of force is reliant on the time (in other words, is accelerating at a non-constant rate), some integration will be needed. The problem for me is, i have only the vaguest idea of how to actually integrate. Would anyone mind helping me with this problem if it does in fact require integration, by showing me the exact steps? And if that is not the case, perhaps point me in the right direction?
     

    Attached Files:

  2. jcsd
  3. Apr 13, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    You are correct. You will need to do an integral to determine the angular velocity.

    You have already evaluated the a(t) at 10 sec to determine a|10 from

    |a(t)| = |r|*|F(t)|/I = 140*|.5*t + .3*t2|

    Since a(t) = dω/dt

    then ω|10 = ∫a(t)*dt

    or

    ω|10 = 140*∫(.5t+.3t2)*dt = 140*(.25*t2 +.1*t3 + c) evaluated from 0 to 10.

    Since you are told it was at rest at t=0, then the constant of integration c = 0.
     
  4. Apr 13, 2009 #3
    Awesome, thank you for the walkthrough, that was incredibly helpful
     
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