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Homework Help: Instantaniouse center of 0 velocity

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve using the mothod of instantaneous center of zero velocity
    Point A = (.5,1.5), B = (1.5,1) and IC = (.5,2), w = 4 rad/s

    Determine the velocity of B with respect to A

    2. Relevant equations


    3. The attempt at a solution

    V(a) = wr(a) where w = 4 rad/s and r(a) = .5
    V(b) = wr(b) hwere w = 4, and r(b) = (x^2+y^2)^1/2
    V(b) = 4.4721...

    How do i find the velocity of B with respect to A

    Attached Files:

  2. jcsd
  3. Apr 27, 2009 #2


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    Homework Helper

    The Velocities are vectors. A is directed positive X and B is directed 45° above positive X.

    So ... Vb relative to Va is Vb - Va isn't it?
  4. Apr 29, 2009 #3
    I'm confused. My book does not mention anything about vectors in this section.

    How are u able to say that A is headed possitive x and B is headed 45 above x.

    But i'll take ur suggestion and see hwat happens

    vector A = 2i, vector B = 3.162i + 3.162j

    vector B - vector A = 2i - (3.162i + 3.162j) = -1.162i + 3.162j

    Doesnt seem right.
  5. Apr 29, 2009 #4
    Look at point A: if the object id 'rotating' counterclockwise about the IC, then what direction is A moving in at this instant ? Surely both points are moving to the right..right?

    Now, relative motion equations say that:

    {the absolute velocity of point B}=

    {the absolute velocity of point A}+{the velocity of point B with respect to point A}

    or in vector notation:

    [tex]\vec V_B=\vec V_A+\vec V_{B/A}[/tex]

    Solve this for

    [tex]\vec V_{B/A}[/tex]

    Then using the IC, find [itex]\vec V_B[/itex] and [itex]\vec V_A[/itex]
    Last edited: Apr 29, 2009
  6. Apr 29, 2009 #5


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    Homework Helper

    Looking at the picture you provided, I'd say that the arrow to the Northwest looks to be counterclockwise to the center of zero motion. Drawing the displacement of A from O, then isn't the direction of motion necessarily ⊥ to the displacement vector?

    That would tell me to direct that velocity along positive X if it is to be ⊥ to OA.
    If this is incorrect, then my understanding of the center of zero velocity must be in error.
  7. Apr 29, 2009 #6
    Ok Ok I understand the vector part now and the direction they are headed.

    Absol V of B = 4.4721
    Absol V of A = 2

    So would the Absol V of B with resp to A = 4.4721 - 2 = 2.4721
  8. Apr 29, 2009 #7
    Unless it is a special case, I do not think that this is correct since [itex]\vec V_B[/itex] and [itex]\vec V_A[/itex] are vectors and thus must be added (or subtracted) as such.

    I am not sure is you will get the same answer.
  9. Apr 29, 2009 #8
    ok now im really confused. If i subtract vectors i get

    vector A = 2i, vector B = 3.162i + 3.162j

    vector B - vector A = 2i - (3.162i + 3.162j) = -1.162i + 3.162j

    If i use the absolute velocity

    Absol V of B = 4.4721
    Absol V of A = 2

    So would the Absol V of B with resp to A = 4.4721 - 2 = 2.4721

    Is either of them right or wrong.
  10. Apr 29, 2009 #9
    I could be making an error, here...but I am getting different numbers for [itex]\vec V_B[/itex]

    Using the exact same equations at the IC to find [itex]\vec V_B[/itex] we have

    [itex]\vec V_B=\vec V_{IC}+\vec V_{B/IC}[/itex]

    [itex]\Rightarrow \vec V_B=0+\vec V_{B/IC}[/itex]

    By definition:

    [itex]\vec V_{B/IC}=\vec \omega \times \vec r_{B/IC}[/itex]

    Where [itex]\vec \omega=4 \hat{k}[/itex] and [itex]\vec \r_{B/IC}=1\hat{i}-1/hat{j}[/itex]


    [itex]\vec V_B=4\hat{i}+4\hat{j}[/itex]
  11. Apr 29, 2009 #10
    ahhhhhhhhhhh i gotcha calculator error

    ok so then the Velocity of B with respect to A is..

    2i+4j m/s in vector notation

    3.6568 m/s in normal notation
  12. Apr 29, 2009 #11
    Sounds good.

    Want to run that by me again?


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