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Conceptual question about rolling on a moving floor

  1. Jul 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A hoop sits on a bus. The bus begins to accelerate toward the right with acceleration a1, shown below. The bus tires do not slip on the road. As the bus accelerates, the hoop begins to roll without slipping on the rigid floor of the bus and has rightward center of mass acceleration a2.

    1) The bus wheel's radius is r1. Its center of mass velocity and acceleration are v1 and a1; the bus wheel's angular velocity and angular acceleration around its center of mass are ω1 and α1.

    2) The hoop's radius is r2. Its center of mass velocity and acceleration are v2 and a2; the hoop's angular velocity and angular acceleration around its center of mass are ω2 and α2. Point P on the wheel is in contact with the rigid bus floor.

    Which equations are implied by each of the constraining conditions indicated in the figure?

    RollingHoopCar_zpsd64ace5f.jpg
    The sixth equation that was cut out accidentally is v1 = ω1r1
    3. The attempt at a solution

    My initial thoughts:
    Intuitively, I imagine that the hoop will roll and accelerate to the back of the bus. I do not imagine that it must roll at a constant velocity relative to the bus. No where in the problem is it implied that μstatic's are equal, and we could be at the point where the hoop or wheel is about to slip.

    Condition X Implies:
    a2 = α2r2 But... is this incorrect because the floor moves?
    aP = a2 If the hoop doesn't slip, then the center of mass should be vertically aligned with the center of mass at all times.

    Condition X Eliminates:
    vP = v1 AND v2 = ω1r1 (for t ≠ 0s)
    aP = a1

    If the hoop rolls w/o slipping, and vP = v2, then there must be a non-zero relative velocity between the CoM of the hoop and the bus. Since both start at rest, their linear accelerations cannot be equal.

    Condition Y Implies:
    v1 = ω1r1


    Can anyone help?

    Thanks much!
     
    Last edited: Jul 23, 2014
  2. jcsd
  3. Jul 24, 2014 #2
    You have got one equation right , the last one .There are some more correct options.

    What does rolling without slipping condition tell you about the speed of the surfaces in contact ?
     
  4. Jul 24, 2014 #3
    Well, the speed of the center of mass of the hoop and the location of the contact point on the surface must be the same. xP must equal x2. But I still think that the contact point will move relative to the bus floor since the hoop rolls.

    For example, if a can rolls without slipping down a stationary wedge, the velocity of the point of application of the normal force would equal the linear velocity of the can. However, the velocity of the can would not equal the velocity of the wedge for all t.
     
    Last edited: Jul 24, 2014
  5. Jul 24, 2014 #4
    No . This is simply wrong . The motion of the hoop is a combination of translation+rotation . The speed of the CM and the point of contact would be same when the hoop is in pure translational motion , which is not the case .

    The question clearly states that the hoop rolls without slipping , which means there is no slipping.

    Again this is wrong .The two points have different velocities . In fact the point of contact has zero velocity whereas the can has non zero linear velocity .

    I think you are not clear with the concept of rolling without slipping.

    Consider a simple case of a hoop rolling without slipping on the floor with speed 'v'. What do you think is the speed of the bottommost point of the hoop i.e the point in contact with the floor ?
     
    Last edited: Jul 24, 2014
  6. Jul 24, 2014 #5
    Hi Vibhor,

    Thank you very much for helping explain this concept.

    I think my interpretation of the term contact point is different than what was intended in the problem. I was defining the contact point as the axis of rotation of the hoop, and considering the axis of rotation in my analysis. If the hoop were rolling on a stationary horizontal surface, certainly the axis of rotation of the hoop moves, right?

    After reading your response, and watching some excellent videos on youtube (I like this one: ) I think I arrived at the correct answers.

    The instantaneous velocity of a point on the edge of the wheel in contact with the floor, vP, must be the same as the floor. Since the point in contact with the floor has 0 angular acceleration relative to the floor, its instantaneous acceleration must be the same as that of the floor.
     
    Last edited by a moderator: Sep 25, 2014
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