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Velocities at B and C on a rotating rod

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    In the attached image, find angular velocity of the rod and the velocities at points B and C.

    2. Relevant equations

    vB|A=0.3ω

    3. The attempt at a solution

    How do I know what angle the absolute velocity makes with the other vectors when put in a triangle? How would the velocity vectors look like at point C in the attached image? It's hard for me since I only have one angle. I could calculate ω, vB, vB|A, I'm only left with vC
    vB= 5.95 m/s
    vB|A=7.778 m/s

    Now, vC|A should be (0.3+0.15)*ω, but what do I do after that?
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2012 #2
  4. Nov 30, 2012 #3
    vC|A = Vc - Va by definition.
     
  5. Nov 30, 2012 #4
    But that's in the vector form! The answer is VC=9.28 m/s at an anticlockwise angle of 15.61° with the horizontal. WHA?
     
  6. Nov 30, 2012 #5
    Vc|a = Vc - Va

    You know the direction and magnitude of Va and you know the direction and magnitude of Vc|a so you can solve for both magnitude and direction of Vc using this one equation. You have two unknowns and this vector equation represents two equations in two dimensions.

    Graphically you can draw a vector diagram which will end up being a closed triangle with Vc one of the sides.

    I think we're not sure what difficulty you are having... maybe you are unsure of the direction of Vc|a? AC is a rigid body which means the two points A and C cannot get closer together or further apart. So any velocity one part of the rigid body has with respect to another must be a rotation, ie directed perpendicular to the line connecting the two points. So Vc|a must be directed perpendicular to the line joining AC.
     
  7. Nov 30, 2012 #6
    Please check if I have this right:
    letters in bold are vectors.

    rC|A=0.45sin(40°)i-0.15cos(40°)j

    vA=-5j

    vC|A= ω x rC|A

    vC=vA+vC|A

    I get vC=-5j+(2.979i +(edited) 7.943j
    ) which is obviously wrong.
    I don't get why the vC makes an angle of 15.61° with the horizontal, as given in my book!!
     
    Last edited: Nov 30, 2012
  8. Nov 30, 2012 #7
    That should be 0.45 on the cos as well, maybe that is it.

    I've attached a diagram showing a graphical solution.

    Vc will be up from the horizontal because Vb is flat horizontally. Vc|a is longer than Vb|a because r is larger at C so the resultant Vc will rise above the horizontal.
     

    Attached Files:

  9. Nov 30, 2012 #8
    Oh my god, unbelievable. Thank you so much!!

    Can you also explain the geometry that goes into making the angle of vC with the horizontal?
    EDIT: that angle can easily be found from VC, so nvm...
     
    Last edited: Nov 30, 2012
  10. Nov 30, 2012 #9
    Yes, just in case:

    The magnitude of Vc is found from the law of cosines:

    Vc2 = Va2 + Vc|a2 - 2VaVc|a cos 50

    and then the law of sines gets you alpha:

    Vc/sin 50 = Vc|a/sin alpha

    and

    θc = alpha - 90


    I think it's probably equally difficult to do it this way or the cross product way; maybe a bit easier using your method because there are fewer things to think about.

    The geometry does give some more insight though. All points on the rod AC travel at a velocity Va+Vp|a where Va is constant and pointing downward and Vp|a is always perpendicular to the rod. This means the one vector velocity diagram shows all velocities at every point of the rod :: any point's velocity is simply the sum of Va and a vector along Vc|a. The length you travel along Vc|a to find Vp|a depends on the distance r of the point p from a (the length is rw remember). So the tail of the absolute velocity of p is anchored at A and the head sweeps along the vector Vc|a for each point p. Vb, eg, is the horizontal vector I drew from A to the vector Vc|a in that diagram.
     
    Last edited: Nov 30, 2012
  11. Nov 30, 2012 #10
    YEah! And had the channel not been there for the second block, the vb WOULDN'T be horizontal, right? So, it's just luck that we knew it's direction. This is exactly why I was confused at point c because I didn't know what was the angle made by vc, so I wasn't able to figure out its components!!
     
  12. Nov 30, 2012 #11
    These things are machines so the movements are normally constrained like that on purpose so that the machines will perform motions that are predictable and useful.

    Had the constraint not been there, you'd have to invoke Netwon's laws (ie it would be a kinetics problem) to find out exactly how the rod moved. But from that diagram with no forces acting (including friction and gravity) and assuming no existing rotational motion, the entire rod would just move downward at 5 m/s.
     
    Last edited: Nov 30, 2012
  13. Nov 30, 2012 #12

    ehild

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    Homework Helper
    Gold Member

    The two channels were at right angles: it was not stated, but rather obvious from the picture. You can choose the most convenient coordinate system: here you say that A moves vertically, along the y axis and B moves horizontally, along the x axis. The rod connects them, the length of the rod is fixed so xb2+ya2=L2, and you get the relation between velocities by implicit derivation: xava +ybvb=0, vb=(-ya/xb)vb

    C moves along a circle of radius r about B. Its velocity is perpendicular to the radius r. You can find that the shaded angles are all equal, so the velocity encloses angle θ with the horizontal.

    ehild
     

    Attached Files:

  14. Dec 1, 2012 #13
    thanKs a lot ehild and aralbrec, you guys helped a lot!
     
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