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A question about work/rotational Ke

  1. Apr 4, 2004 #1
    Here is the question thats been giving me trouble:

    A electric motor can accelerate a ferris wheel of moment of inertia I = 25300 kgm^2 from rest to 11.9 rev/min in 11.5 s. when the motor is turned off, friction causes the wheel to slow down from 11.9 rev/min to 6.33 rev/min in 7.53 s.

    Determine the torque generated by the motor to bring the wheel to 11.9 rev/min.

    Determine the power needed to maintain a rotational speed of 11.9 rev/min.


    The 2nd part seems fairly straight forward. When i find the torque i use
    Power = (torque)(rotational speed).

    The rotational speed = 11.9 rev/min = .198333 rev/sec = 1.24617 rad/sec.

    So the power should be easy to find.

    I assume i can find the change in kinetic energy (and therefore the work) by using the following equation:

    The final rotating speed = 6.33 rev/min = .662876 rad/sec

    change in K = W = .5(25300)(.662876^2) - .5(25300)(1.24617^2)
    = -14086.2 J

    Im not sure, however, how to find the torque after this?

    Any help would be greatly appreciated!
  2. jcsd
  3. Apr 5, 2004 #2
    I was thinking about using Work = (torque)(rotational speed) to find the torque required for the first part of the problem.

    I guess i didnt calculate the work/change in k correctly above, as i tried using this approach but got the problem incorrect.'

    Still puzzled as to what im doing wrong...
  4. Apr 5, 2004 #3


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    The problem of using an energy argument in this problem, is that the work done by the frictional torque during the slow-down phase is not the same as the work done by the frictional torque during the acceleration phase. While you readily can find the net work done by the friction during the slow-down, I don't think this would help you at all..
    What you first of all need to solve the problem, is the instantenaeous torque provided by the frictional forces, not the time-integrated power of that torque (i.e the work done).
    The frictional torque F typically depends on the rotational speed, i.e. F=F(w), where w is the rot. speed at a given time t. For low speeds, it is reasonable to assume F=-Aw, where A is constant.
    The moment-of-momentum equation in the slow-down phase then reads:
    -Aw=I(dw/dt), where I is the moment of inertia.
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