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Torque generated by the motor of ferris wheel?

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    An electric motor can accelerate a Ferris
    wheel of moment of inertia 11300 kg · m2 from
    rest to 9.6 rev/min in 14.6 s. When the mo-
    tor is turned off, friction causes the wheel to
    slow down from 9.6 rev/min to 8.37 rev/min
    in 7.23 s.
    Determine the torque generated by the mo-
    tor to bring the wheel to 9.6 rev/min.
    Answer in units of N · m.


    2. Relevant equations

    [tex]\alpha[/tex] = [tex]\omega[/tex]/t
    [tex]\tau[/tex] = I[tex]\alpha[/tex]


    3. The attempt at a solution

    given:
    I = 11300 kgm2
    [tex]\omega[/tex]o = 0
    [tex]\omega[/tex]f = 9.6 rev/min = 1.005309649 rad/s
    tf = 14.6 s

    The question asks to determine the torque generated by the motor to bring the wheel to 9.6 rev/min so I think that means I don't have to worry about the motor beign turned off or the friction to slow it down... sooooo...

    [tex]\alpha[/tex] = [tex]\omega[/tex]/t
    = (1.005309649 rad/s)/(14.6s)
    = 0.0688568253 rad/s2

    [tex]\tau[/tex] = I[tex]\alpha[/tex]
    = (11300kgm2)(0.0688568253 rad/s2)
    = 778.0821257 Nm

    When I enter this answer, it says it is wrong? What am I doing wrong here?

    Thank you in advance for your help!
     
    Last edited: Oct 25, 2009
  2. jcsd
  3. Oct 26, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    In the first case
    T = I*αo + I*αr
    where I*αr is the torque required to overcome the friction and I*αo is the torque need to accelerate the wheel after overcoming the frictional force.
    In the second case
    Tf = I*αr.
     
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