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Velocity and Acceleration Vectors

  1. Jan 23, 2016 #1

    CGI

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    1. The problem statement, all variables and given/known data
    upload_2016-1-23_15-23-7.png

    2. Relevant equations
    Rotational Kinematic Equations
    Kinematic Equations

    3. The attempt at a solution
    I honestly have no clue how to get a vector out of this.
    I thought about an equation:

    Θ = Θ(initial) + ω(initial)*t + .5αt^2

    and how maybe that v = wr could play into this, but there is so much I don't know where to start.

    Any help would be really appreciated!
     
  2. jcsd
  3. Jan 23, 2016 #2

    TSny

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  4. Jan 23, 2016 #3

    CGI

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    Oh okay, so I should be thinking of something along those lines.

    So if my theta = 0 and my R = .8ft, would my position vector just be r = .8êr?
    what would my r dot and theta dot be? Sorry, this is still relatively new to me
    and I've been watching videos on it as well. I just want to make sure I understand
    this.
     
  5. Jan 23, 2016 #4

    haruspex

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    Assuming the answers are wanted in terms of the x, y coordinates:
    Let the diagram position represent time 0. At time t, what will r and theta be? So what will x and y be?
     
  6. Jan 23, 2016 #5

    CGI

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    Would the x just be rcosΘ and y be rsinΘ?
     
  7. Jan 23, 2016 #6

    haruspex

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    Yes, but you are trying to find velocities and accelerations, so you need to express them as functions of time.
     
  8. Jan 23, 2016 #7

    CGI

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    Oh okay. Right. Could that just be rcostheta(t) and rsintheta(t)?
     
  9. Jan 23, 2016 #8

    haruspex

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    According to the problem statement, both r and theta vary with time. Write each as a function of t.
     
  10. Jan 23, 2016 #9

    CGI

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    Hmmm...okay. Could I say that r = r_initial + vt and that theta = theta_inital + wt where w = 45 rev/min?
     
  11. Jan 23, 2016 #10

    haruspex

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    Yes. Now express x and y that way and differentiate as necessary.
     
  12. Jan 23, 2016 #11

    CGI

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    When you say express x and y in that way, do you mean that I can say

    x = (r_initial + vt)cos(Θ_initial + ωt)

    And the same for y, only with a "sin?"
     
  13. Jan 24, 2016 #12

    haruspex

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    Yes. Now differentiate.
     
  14. Jan 24, 2016 #13

    CGI

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    Okay I was just double checking. So when I take the derivative with respect to t I get,

    x = vcos(Θ_inital + ωt) - (r_initial + vt)sin(Θ_initial + ωt)*ω

    y = vsin(Θ_inital + ωt) + (r_inital + vt)cos(Θ_initial + ωt)*ω

    Does this look about right?
     
  15. Jan 24, 2016 #14

    haruspex

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    Yes.
     
  16. Jan 24, 2016 #15

    CGI

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    Okay great. So would these two x and y be the vector for velocity?
     
  17. Jan 24, 2016 #16

    haruspex

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    Small correction: in your post #13 I presume you meant ##\dot x=## etc., not x=. Similarly y.
    If so, yes they would be the x and y components of velocity.
     
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