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Relative Velocity problem: 2 pulleys riveted together

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    As shown in the picture, two pulleys are riveted together with a known velocity. I need to find the velocity of two other points.

    eq8d5i.jpg

    2. Relevant equations
    V_A = VB + <ω> x <BA>

    V = ωr

    3. The attempt at a solution

    I've honestly been trying to figure this one out for over a week, and I'm sure I'll kick myself when I realize how easy it is, but I'm trying to solve by:

    V_A = (r_OA)*(ω_OA), using V_A = .9 m/s; r_OA = 90mm, which gives me ω_OA = 10 rad/sec (which I assume ω is equal on the entire pulley since it's riveted together).

    Then I used the relative velocity vector formula for A and O:

    V_A = V_O + <-10 rad/sec k> x <-.09 m i>
    which gives me: .9 m/s j = V_O j + .9 m/s j. Solving with this gives me V_O = 0, which is incorrect. I tried solving for B the same way.

    The actual answers are V_O = .6 m/s, and V_B = .849 m/s.

    Thanks in advance!
     
  2. jcsd
  3. Oct 30, 2015 #2

    TSny

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    This would give the speed of A relative to O, not relative to the earth.
     
  4. Oct 30, 2015 #3
    I thought <w>x<OA> was the speed of A relative to O? I thought the equation was sufficient for finding the actual speed of A? How would I solve for thespeed relative to earth?
     
  5. Oct 30, 2015 #4

    TSny

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    Yes, it's the speed of A relative to O. But point O is not at rest relative to the earth.
    Can you find a point of the pulley that is instantaneously at rest relative to the earth? Hint: Think about problems where you have rolling without slipping.
     
  6. Oct 30, 2015 #5
    Ahhhh, I see now. So I just plugged V_O = .18w into the equation and solved for w first. Thanks so much!
     
  7. Oct 30, 2015 #6

    TSny

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    OK. Or, you can get ##\omega## from ##V_A = .27 \omega##, since ##V_A## is given.
     
  8. Oct 30, 2015 #7
    I see that that definitely works mathematically, but I don't understand why it works by looking at the problem. Why is it .27 instead of .18? I don't understand why the distance is .27 when the velocity vector is .18 m away.

    Thanks again!
     
  9. Oct 30, 2015 #8

    TSny

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    Let point D be the point on the outer rim of the pulley where the string on the right meets the pulley.

    Use the fact that ##\vec{V}_A =\vec{V}_D +\vec{V}_{A/D}##.
     

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  10. Oct 31, 2015 #9
    It might help if you consider what happens to a simpler situation where a string raises a single pulley
    with one end of the string attached to the ceiling.
    In the given problem you have:
    V0 = .18 * w where w is the angular velocity
    VA = V0 + .09 w
    and this is easily solved for w, the angular velocity.
     
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