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Math Help - Torque question

  1. #1
    s3a
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    Torque question

    I just started torque so excuse me for asking a stupid question but I'm stuck trying to do the following question:

    "A wheel starts from rest and rotates through 150 rad in 5 s. The net torque due to the motor and friction is constant at 48Nm. When the motor is switched off, the wheel stops in 12s. Find the torque due to (a) friction and (b) the motor."

    Answers:
    (a) -20Nm
    (b)68Nm

    My problem is I do not know how to get (a). As for (b), I'm assuming it's just sigmaTorque = Torque_(motor) + Torque_(friction) = 48Nm and we know that Torque_(friction) is -20Nm so 48+20 = 68.

    What I did was:
    For motor on:
    deltaTheta = (Wi + Wf)/2 * t = Wf(5)/(2) = 150 ==> Wf = 60rad/s

    For motor off:
    Wi = 60rad/s
    Wf = 0rad/s
    t = 12s

    and I can find whatever else I want but the problem, again, is that I have no idea what to do from here in order to get the torque of the friction.

    Any help would be GREATLY appreciated!
    Thanks in advance!
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  2. #2
    MHF Contributor
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    Quote Originally Posted by s3a View Post
    "A wheel starts from rest and rotates through 150 rad in 5 s. The net torque due to the motor and friction is constant at 48Nm. When the motor is switched off, the wheel stops in 12s. Find the torque due to (a) friction and (b) the motor."

    Answers:
    (a) -20Nm
    (b)68Nm
    on the spin up ...

    \alpha = \frac{2\Delta \theta}{t^2} = 6 \, rad/s^2

    \tau_{net} = I\alpha

    \tau_m - \tau_f = 6I = 48

    I = 8 \, kg\,m^2<br />

    \omega_f = \alpha t = 30 \, rad/s<br />

    on the spin down ...

    \omega_0 = 30 \, rad/s

    \alpha = \frac{\Delta \omega}{\Delta t} = -\frac{30}{12} = -2.5 \, rad/s^2

    \tau_{net} = I\alpha<br />

    \tau_f = -2.5I = -2.5(8) = -20 \, Nm

    finally ...

    \tau_m = 8.5I = 8.5(8) = 68 \, Nm
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