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Determine the absolute Velocity and Acceleration of Particle A

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The disk rolls without slipping on the horizontal surface, and at the instant represented, the center O has the velocity vO = 2.2 m/s and acceleration aO = 5.5 m/s2 with directions shown in the figure. For this instant, the particle A has the indicated speed u = 3.2 m/s and time rate of change of speed = 6.6 m/s2, both relative to the disk with directions shown in the figure. Determine the absolute velocity vA and acceleration aA of particle A.

    I have attached an image of the question

    2. Relevant equations

    VA = VO + wXr + Vrel

    3. The attempt at a solution

    I'm honestly confused as to how to start this question. I'm not given the angular velocity so initially I tried to solve for it.

    v = wXr

    I know that the velocity at point O is -2.2i but how should I determine r? Or should I use the velocity at A instead?

    Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 9, 2013 #2

    haruspex

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    Forget the particle for the moment. You have a disc of known radius rolling at known linear speed....
     
  4. Apr 9, 2013 #3
    So, you're saying that the angular velocity, w, is;

    v = wXr

    w = v/r

    w = 2.2/0.39

    w = 5.641 CCW

    Right?
     
  5. Apr 9, 2013 #4
    After this part I'm fairly certain that I need to use:

    VA = VO + wXr + vrel

    Which I think should give me:

    VA = -2.2i + (5.641k)X(0.3j) + 1i

    vA = -2.8923 i

    But it says this is not correct. I'm thinking that my mistake is with the vrel, how should I interpret this term, because I thought it was 1 because part A is traveling faster in the x direction by 1i. Am I mistaken?
     
  6. Apr 9, 2013 #5

    haruspex

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    vrel is given as 3.2 m/s to the right.
     
  7. Apr 9, 2013 #6
    I've managed to get VA = -0.6923i + 0j

    But I'm stuck on aA

    I think I need to use the equation:

    aA = aO + αXr + wXwXr + 2wXvrel + arel

    I found α with

    α = a/r

    α = 5.5/0.3

    α = -14.1k

    aA = 5.5i + (-14.1k)X(0.3j) + (5.641k)X[(5.641k)X(0.3j)] + 2(5.641k)X(3.2i) - 6.6i

    aA = 3.13i + 23.69j

    I know the j part is incorrect but I don't know about the i part. I don't understand where I'm going wrong. Am I misinterpreting the arel?
     
  8. Apr 9, 2013 #7
    My i component is correct at 3.13i and I finally managed to get the j component but it doesn't make sense.

    the values for my j component are:

    <5.641k>X[<5.641k>X<0.3k>] + 2<5.641k>X<3.2i> - 3.22/0.3

    = -7.57719 j which is correct

    I can see where the first two terms come from in the equation I gave above, but where does the 3.22/0.3 come from?
     
  9. Apr 9, 2013 #8

    haruspex

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    That's the centripetal acceleration from the fact that it is moving at 3.2m/s relative to the disc but constrained to move in an arc radius .3m relative to the disc centre.
     
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