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Torque graph; finding angular velocity

  1. Mar 12, 2007 #1
    An object whose moment of inertia is 4.10 kg/m^2 experiences the torque shown in the graph attached.

    What is the object's angular velocity at 2.70 s? Assume it starts from rest.

    My main problem i think is how to read the graph. How should i interpret the graph?

    What i would do first is find the angular acceleration by using...

    aa = angular acceleration
    t = torque
    I = moment of inertia

    aa= t/I

    I would find the aa for three parts, the part where it is accelerating positively, when it is not increasing, and then when it decreases.

    I would then use the aa for the equation while putting in the velocity...

    w = angular velocity
    T = time

    w(f) = w(i) + aaT

    Attached Files:

  2. jcsd
  3. Apr 3, 2008 #2
    i wish they would reply, i would like help with this also.
  4. Apr 3, 2008 #3
    Well.. you have the Torque as a function of time. Break it up into 3 different functions, one for each of the time interval. At t=2, you can take it to be a part of either of the time interval, just make sure you include it in ONE and ONLY ONE of them.

    Once you have [itex]\tau (t)[/itex], divide it by [itex]I[/itex] to get [tex]\alpha (t) = \frac{1}{I} \tau (t)[/tex].

    Then integrate it w.r.t to get the angular velocity in each of these intervals using proper limits. This might help you:

    \Delta \omega = \int_{t_1}^{t_2} \alpha (t) dt

    Here, [itex]\Delta[/itex] is used to denote that i'm computing the change in angular velocity since the given acceleration was applied. If the initial angular velocity was 0, you can take [itex]\Delta \omega = \omega[/itex]. Once you have done that, you'll need to addup the angular velocities [or their differences rather] attained in each time interval. Do note that, no net torque doesn't mean zero angular velocity, it just means a zero change in angular velocity. Reply if you have any problems with this.
    Last edited: Apr 3, 2008
  5. Apr 3, 2008 #4


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    Welcome to PF!

    Hi animanga008! Welcome to PF! :smile:

    (btw, your original post was at 4am here in London, and just before midnight in New York … if you want a quick answer, you probably need to start a bit earlier!)

    The torque is only in two parts.

    (1) The torque increases steadily from t = 0 to 1, according to the formula tau = 2t (so, for example, at t = 1/4, tau = 1/2)

    (2) The torque is then steady for one second.

    After that, it disappears suddenly (in no time at all)! :smile:
  6. Feb 23, 2010 #5
    Can anyone tell me why does the car engine have maximum power of e.g. 120 horsepower at 4200 revs per minute, and maximum torque of 250 Nm from 1400-2600 revs per minute?

    The car seems to have the best acceleration in the range of the greatest torque (1400-2600 revs per minute), but not at maximum power (4200 revs per minute). What is the catch?
  7. Feb 23, 2010 #6
    Given the date of the post, I would suggest you start a new thread with your question. Please do not bump topics so old when the post content is not explicitly related to that of the op. Thank you.
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