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Polar coordinates question

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Here is a picture of the situation http://i48.tinypic.com/vnmi5t.jpg

    2. Relevant equations

    polar coordinate system

    3. The attempt at a solution

    ok so first i'm attempting to find velocity as a function of time,
    first I know V=(dR/dt)er +(R)(d∅/dt)e∅ - this is a vector

    so, R = 20+15cos(∅) and d∅/dt = ∏ and dR/dt = -15sin(∅)

    now this is wher ei get stuck.

    so I'm trying to get velocity as a function of time,
    the only thing i can think of to get there is my finding the magnitude of the vector, so
    I get, ((-15sin(∅))^(2)+(20∏ +125cos(∅))^(2))^(1/2)

    simplifying i get 1 + (8/3)∏cos(∅) + (16/9)∏^(2) = V(∅)
    but i want v as a function of time not theta, and i have no idea to go about getting there. Any help would be greatly appreciated thanks!
     
  2. jcsd
  3. Apr 17, 2013 #2
    If ## R = 20 + 15 \cos \theta ## and ## \theta = \pi t ##, then what is ## \frac {dR} {dt} ##?
     
  4. Apr 17, 2013 #3
    -15*∏*sin(∏*t) = dR/dt
     
  5. Apr 17, 2013 #4
    oh i see, so 1+(8/3)*pi*cos(pi*t)+(16/9)*pi^(2) = V(t) ?
     
  6. Apr 17, 2013 #5
    how did you know theta was pi*t though? My intuition with regard to that statement is lacking.
     
  7. Apr 17, 2013 #6
    So what is the entire velocity vector as a function of time?
     
  8. Apr 17, 2013 #7
    did you see post 4?
     
  9. Apr 17, 2013 #8
    I do not see any vectors in post #4. And I don't really understand what that thing in it really is.
     
  10. Apr 17, 2013 #9
    ok so the velocity vector = -15*pi*sin(pi*t)Er + ((20+15cos(pi*t))*pi)Eo
     
  11. Apr 17, 2013 #10
    That's correct. Can you continue from this on?
     
  12. Apr 17, 2013 #11
    I'll give it a go, thank you for the push in the right direction.
     
  13. Apr 17, 2013 #12
    Do you understand why ## \theta = \pi t ##? Note the problem specified a constant angular velocity.
     
  14. Apr 17, 2013 #13
    Ah i see! the the integral of d(theta) with respect to time is pi*t
     
  15. Apr 18, 2013 #14
    when calculating the magnitude of the velocity and acceleration vectors at t=0.7s do i need to change the equation pi*t into degrees or does it work just as it is?
     
  16. Apr 18, 2013 #15
    Why would you?
     
  17. Apr 18, 2013 #16
    cos(a) equals a different value depending on a being measured in rads or degrees?
     
  18. Apr 18, 2013 #17
    This is not a correct statement. When you compute a trigonometric function, you - or, rather, the calculator - must know what units are used for the angle measure. Then the result is independent of that.
     
  19. Apr 18, 2013 #18
    so just to makesure i understand you the velcoity at time t=0.7s then would be,

    (((-15*pi*sin(pi*0.7))^(2)+(20+15cos(pi*0.7)*pi)^(2))^(1/2) which equals 109.92 m/s?
     
  20. Apr 18, 2013 #19
    I get a different result.
    $$
    \sqrt {(15 \pi \sin \pi t)^2 + (20 + 15 \cos \pi t)^2 \pi^2}

    = 5\pi \sqrt { (3 \sin \pi t)^2 + 4^2 + 24 \cos \pi t + (3 \cos \pi t)^2}

    \\

    = 5\pi \sqrt { 3^2 + 4^2 + 24 \cos \pi t } = 5 \pi \sqrt { 25 + 24 \cos \pi t }
    $$
     
  21. Apr 18, 2013 #20
    when putting in all the values in you're equation i get the same result as when using my equation which is 109.93 m/s :S
     
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