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Instantaneous center problem

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data

    2e401042868e57db85fa7d9b4768c7af.png

    2. Relevant equations

    v = rw

    3. The attempt at a solution

    I am not sure where to begin, honestly. For part (a), the angular velocity w will be v/r for the lower tire and upper tire both so all I need is the right values of v and r. Now for r, what is the reference point supposed to be? If I measured r from point D, rc would be r and rA would be 3r. But these are obviously wrong according to the solutions. And I know the velocity at point P will be 0 since there is no slipping, but I have no idea how to get the velocity of the other points (and what about other arbitrary points on the tires..?)

    Another thing that bothers me is that according to the solution, the upper tire is going CCW. How is that possible when the velocity is going towards the right? Shouldn't the upper tire be going CW and the lower tire CCW?

    Also how do we find the instantaneous center of points A, B and C? When they all are parallel and we don't know the magnitudes.

    I've been struggling to solve this problem and haven't gotten anywhere in 2 hours so help would greatly be appreciated
     
  2. jcsd
  3. Nov 25, 2014 #2

    haruspex

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    Consider just the lower tire first. The presence of the upper tire does not affect the motion of the lower tire, agreed?
    What is the forward velocity of the bit of tire at point P? What is the forward velocity of point A?
    You can concentrate on PAB as though it is just a rod, briefly. What do the above answers tell you about the forward velocity of point B?
    Which way is the tire rotating, and how fast?
    When you've got all those right, we can move onto the upper tire.
     
  4. Nov 25, 2014 #3

    NascentOxygen

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    The mechanic is walking forwards, so the lower wheel must rotate CW because there is no slipping at point P. The upper wheel progresses forwards along with the mechanic but without any slipping of contact surfaces at point B. To see which way these must turn, try this using a pair of coins or jam jar lids if you find it difficult to run as a thought experiment.

    r is the distance from the periphery to the wheel's centre of rotation, meaning it's the wheel radius
     
  5. Nov 26, 2014 #4
    How do you know or why do you have to consider the lower tire first? If I do that I got Va, Vb and Vc right, but I am unsure why Vd is zero.

    Also what exactly does "no slipping" mean? I don't think I am understanding that clearly as well.
     
  6. Nov 26, 2014 #5

    haruspex

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    The movement of the lower tire is determined by its contact with the ground and the given overall forward motion, so you can consider it independently of the upper tire. But the movement of the upper tire depends on what the lower tire is doing.
    In the lower tire, P is instantaneously stationary (it is the instantaneous centre of rotation), and A is moving forward at speed v. B is twice as far from P as A is, so it moves forward at speed 2v. Likewise, C is only moving forward at speed v, so to keep BCD straight D must be moving backwards at speed v relative to C, so is instantaneously stationary. Indeed, you could place a ceiling at height 4r and the upper tire would happily roll along that.
    It means that the two surfaces in contact have no relative motion parallel to the contact. In consequence, the frictional force between them does no work.
     
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