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Flywheel moment of inertia(torque)

  1. Sep 23, 2011 #1
    An electric motor turns a flywheel through a drive belt that joins a pulley on the motor
    and a pulley that is rigidly attached to the flywheel, as shown in Figure P10.39. The flywheel is a solid disk with a mass of 80.0 kg and a diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. If the tension in the upper (taut) segment of the belt is 135 N and the flywheel has a clockwise angular acceleration of 1.67 rad/s2, find the tension in the lower (slack) segment of the belt.

    Hi guys, I've managed to work this question via the method below.

    torque about pulley = torque about flywheel
    Tensile force x radius of pulley = moment of inertia of flywheel x angular acceleration
    (T1 -T2) x radius= 0.5 x mass of flywheel x square of radius of flywheel x angular acceleration
    (T1-135) 0.23 = 0.5 x 80 x 0.625^2 x 1.67
    T1 =21.5N

    However, can anyone explain to me why the torque about the pulley and flywheel would be the same?

    Shouldn't the torque about the flywheel be greater due to its greater moment of inertia and similar angular acceleration?
     
  2. jcsd
  3. Sep 23, 2011 #2

    gneill

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    Staff: Mentor

    Torque is taken to be applied around some center of rotation, and the center of rotation of the pulley and flywheel are one and the same.

    Torque is the circular motion analog of force. Moment of inertia is the circular motion analog of mass. You don't expect an applied force to increase because it's applied to a more massive object; a force is whatever values it's specified to be. Similarly for torque

    Now it's certainly true that it takes more force to produce a given acceleration if the mass is larger, but changing the mass doesn't automatically change the force; the force is applied by some outside agency that is separately specified. The same holds for circular motion and torque.

    In the present problem the pulley is taken to be bonded to the flywheel and has negligible mass. So the two together are taken to have the same moment of inertia as the flywheel alone.
     
  4. Sep 26, 2011 #3
    What if it is in the case where the pulley's mass isn't negligible?

    How would things change if that is the case?
     
  5. Sep 26, 2011 #4

    gneill

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    Staff: Mentor

    If the pulley's mass is not negligible then you would add its moment of inertia to that of the flywheel and use this sum in the equations where previously you used just the moment of inertia of the flywheel.

    To use a "linear" analogy, suppose you were told that a certain mass M is being pulled by a force applied via massless rope. The rope is attached via a metal ring that is bolted to M. If the mass of the ring is negligible then you just use M as the mass being accelerated by the force. If, on the other hand, you're told that the metal ring has mass m and cannot be ignored, then you'd take M+m as the mass being accelerated.
     
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