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Final translational and spin velocities of spinning disc?

  1. May 6, 2015 #1
    1. Problem: Determine the final translational speed of a spinning disc that is dropped on an infinite horizontal plane. Think of spinning up an automobile wheel suspended over a road (wheel spinning normal to the road), then dropping the wheel on the road. When the wheel rim touches the road, friction will create a torque on the wheel that will cause it to begin rolling down the road. It will accelerate to some translational velocity. How fast will it go?

    Everything is ideal: uniform solid disc, no air friction, no loss of energy through frictional heating at the disc/plane contact, no wear, etc.


    2. Relevant equations
    The disc has a mass (m), a radius (r) and an initial spin rate Wi (Omega initial).
    The disc has an initial kinetic energy KEi = IWi^2/2, where I is the disc moment of inertia, mr^2.
    At equilibrium, the disc has a translational velocity (v), and a final spin rate Wf (Omega final). Its translational kinetic energy is mv^2/2. Its final spin KE is IWf^2/2.


    3. The attempt at a solution
    I'm modeling this under these assumptions:
    a. The system's initial kinetic energy of spin is the only energy that is available to accelerate the wheel along the plane.
    b. As frictional torque accelerates the disc in translation, the disc spin rate decreases and the disc's spin kinetic energy decreases, as the disc's translational kinetic energy increases.
    c. When the disc's translational kinetic energy = the disc's remaining spin kinetic energy, translational acceleration stops and the system is in its final equilibrium state.
    d. The sum of the final translational KE and spin KE = the initial spin KE.

    Under these assumptions, the final translational KE (mv^2/2) = the final spin KE (IWf^2/2) and each of these is 1/2 of the initial spin KE (IWi^2/2).

    Substituting mr^2 for I, the final KE equivalence is mv^2/2 = mr^2Wf^2/2. From this, t
    he translational velocity v should = rWf, or the disc radius times its final angular rate.

    My uncertainty is in assumption (c.). I assume that the initial disc spin KE decreases to as the disc transnational KE increases, and that when they are equal, no further KE transfer occurs. Is this correct? If not, what determines the value of either spin KE or translational KE at which no further change in KE occurs? Thank you.
     
  2. jcsd
  3. May 6, 2015 #2

    ehild

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    The energy is not conserved. When the spinning wheel gets into contact with the road, its surface skids with respect to the road. The force of friction decelerates rotation and accelerates translational motion. When the translational speed v is equal to the linear speed of the rim, ωR, the wheel begins to roll without slipping, the friction becomes static.

    skiddingwheel.JPG
     
  4. May 7, 2015 #3
    Thanks for replying. I understand that when the disc peripheral speed equals (-) the disc translational speed, the disc spin and translation speeds are constant (in our perfect lossless system).

    What I don't understand is how to calculate that *actual* speed.

    My thought is that spin KE converts to translation KE through friction and resultant torque produced at the contact zone between disc and road. As the disc translates faster, the disc spin KE must decrease when the initial disc spin KE is partitioned into residual spin KE and translation KE components as the disc accelerates down the road. When KEspin = KE translation, no further spin KE can be transferred to translation KE (assumption). This is the point at which disc peripheral speed = (-) translation speed of the disc center of mass along the road. So, how does one calculate that speed, at which there is no slip between the disc periphery and the road?

    If my assumption that at equilibrium, KEspin = KEtranslation,is correct, then given the disc mass, radius, and initial spin rate, the final speeds should be calculable. Maybe I should analyze this from the perspective of the torque produced at contact and see how that affects the two KE components?
     
  5. May 7, 2015 #4

    rcgldr

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    This is only possible if the interaction is similar to an 100% elastic collision. The surface of the disc and or plane would deform at "impact" and then recover with zero loss of energy. Since the interaction is 100% elastic, the disc would not stabilize, but instead oscillate through a combination of linear and angular acceleration / deceleration cycles. I don't think there's a solution here. The total energy is constant if the potential energy of the deformed surfaces are included as part of the total energy.

    The more realistic and more common form of this problem assumes that energy is lost to heat while the disc surface slides on the plane. For this case, you assume that kinetic friction is constant, generating a linear force in the direction of motion, and a torque reducing the angular velocity of the disc until it transitions from slipping into rolling.
     
    Last edited: May 7, 2015
  6. May 7, 2015 #5

    ehild

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    As you imagine the process it is correct, with one exception: The kinetic friction consumes energy. Some of the rotational energy is converted to translational energy and heat. The mechanical energy is not conserved.

    The motion of the wheel can be considered as translation of its centre of mass (CM) and rotation about the CM. The CM is accelerated by the net force F according to
    F=ma
    and the rotation is accelerated by the net torque τ according to
    τ=Iα
    where a is the acceleration of the CM, α is the angular acceleration, m is the mass and I is the moment of inertia with respect to the CM.

    The force is the kinetic friction now. It is constant so the CM performs uniformly accelerating motion, How does the velocity V change with time?
    The torque of the friction accelerates the rotation uniformly. How does the angular velocity ω change in time?
    Compare the linear velocity of the CM to the angular velocity of the rotation. When V=rω the wheel starts to roll without slipping.
     
    Last edited: May 7, 2015
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