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Final angle speed?

  1. Dec 11, 2005 #1
    I am trying to do a pratice test and I am stuck on this problem.
    a disk of radius r=0.5m,mass m=0.25kg has an initial angular speed of w=23rad/s.assume a torque of 0.33Nm is applied against the rotation for an angular distance of 20 rad. what is the final angular speed?
    thanks for any help.:surprised :surprised :bugeye:
     
  2. jcsd
  3. Dec 12, 2005 #2

    Tide

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    What have you tried so far? Do you know how torque relates to angular acceleration?
     
  4. Dec 12, 2005 #3
    distance?

    I am getting thrown off by the 20 rad distance. I thought you just find the two torques and add them ,one being neg and the other pos . I dont know what to do with the distance?:grumpy: I am useing the formula t=mr^2*tnet/I thanks
     
  5. Dec 12, 2005 #4

    Tide

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    Two torques? I only see one torque in your statement of the problem.

    You should recognize that the torque is related to angular acceleration:

    [tex]\tau = - I \alpha[/tex]

    where [itex]\alpha[/itex] is the angular acceleration and I use the negative sign to indicate the torque "is applied against the rotation."

    It follows that [itex]\omega = \omega_0 - (\tau/I) t[/itex] and

    [tex]\theta = \theta_0 + \omega_0 t + \frac {1}{2} \frac {\tau}{I} t^2[/tex]

    and you should be able to take it from there.
     
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