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Homework Help: Kinematics motion - Given position, find velocity and acceleration

  1. Dec 17, 2013 #1
    Kinematics motion -- Given position, find velocity and acceleration

    1. The problem statement, all variables and given/known data

    A particle's position as a function of time is given by xf = xi . sin wt, where xi and w are constants.

    a) Find expressions for velocity and acceleration.
    b) What are the maximum values of velocity and acceleration.

    2. Relevant equations
    xf = xi. sin(wt)

    3. The attempt at a solution

    a) xf = xi. sin(wt)

    velocity v = xf' = d/dt xi sin(wt) + xi d/dt sin(w.t) . d/dt w.t

    xf' = xi cos(wt).w = xi.w.cos(wt)

    acceleration = v' or xf " = [d/dtxi.w + xi d/dt w]cos(w.t) + xi.w d/dt (cos(w.t) . d/dt (w.t)

    xf " = x dw/dt cos (wt) - xw^2 sin(w.t)

    Can someone verify the above steps for me before I proceed to part (b)?
     
    Last edited: Dec 17, 2013
  2. jcsd
  3. Dec 17, 2013 #2

    Doc Al

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    Looks OK, but use the fact that ω is constant to simplify that expression. (Your notation is difficult to read. Consider using Latex in the future.)
     
  4. Dec 17, 2013 #3

    Is it possible to simplify further?

    On a tangent, how do I utilize Latex? Matlab?
     
    Last edited: Dec 17, 2013
  5. Dec 17, 2013 #4

    adjacent

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    On the advanced editor of physicsforums,you can use latex by pressing the top right button.
     
  6. Dec 17, 2013 #5
    I just realized. But the nitty gritty details would have to wait.
     
  7. Dec 17, 2013 #6

    Doc Al

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    Yes.
     
  8. Dec 17, 2013 #7
  9. Dec 17, 2013 #8

    Doc Al

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    dω/dt = ?
     
  10. Dec 17, 2013 #9
    Capture.JPG

    Why do I have to express the equation as dω/dt?
     
    Last edited: Dec 17, 2013
  11. Dec 17, 2013 #10

    Doc Al

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    Huh? Your expression for the acceleration contains dω/dt.

     
  12. Dec 17, 2013 #11
    Capture.JPG
     
    Last edited: Dec 17, 2013
  13. Dec 17, 2013 #12

    Doc Al

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  14. Dec 17, 2013 #13
    I believe I'm stuck. Even if ω were a constant, hasn't the equation been differentiated to it's second derivative? I could factor out some variables in the second derivative but I don't think that's what you were going at.
     
  15. Dec 17, 2013 #14

    Doc Al

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    What do you mean "even if"? You are told that ω is a constant.

    Answer this. What's the derivative of a constant?
     
  16. Dec 17, 2013 #15

    Doc Al

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    Another hint: The point I am making is so trivial you will slap yourself once you realize it.
     
  17. Dec 17, 2013 #16

    berkeman

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  18. Dec 17, 2013 #17
    it's zero

    y" = [d/dt(xw) + (x)d/dt(w)]cos(wt) + (xw) d/dt cos(wt) d/dt(wt)
    =
    dw/dt cos(wt) - (xw)sin(wt)w
    =
    dw/dt cos(wt) - xw^2 sin(wt)

    Sent from iPhone
     
    Last edited: Dec 17, 2013
  19. Dec 18, 2013 #18
    dw/dt cos (wt) - xw sin(wt)w
    dw/dt =0
    So,
    y" = -xw^2 sin(wt)
    Is the workings above correctly reduced?
     
    Last edited: Dec 18, 2013
  20. Dec 18, 2013 #19
    I would suggest you that while calculating derivative of an function if you see any constant in that equation ,you can substitute 0 directly which makes your equation simple.
    In your equations
    substitute 0 in place of dw/dt of your acceleration equation .
     
    Last edited by a moderator: Dec 18, 2013
  21. Dec 18, 2013 #20

    Doc Al

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    Of course!

    Keep going.

    Exactly.
     
  22. Dec 18, 2013 #21
    Part (b)

    In find maximum velocity:

    y' = xw cos (wt)
    xw cos (wt) = 0
    Cos(wt) = 0
    Cos(pi/2) = 0 A cos(3pi/2) = 0
    wt = pi/2
    t = pi/2w

    wt = 3pi/2
    t = 3pi/2w

    y"(pi/2w)= -xw^2 sin(wt)
    = -xw^2 sin(pi/2)
    = xw^2 < 0 local max

    y"(3pi/2w) = -xw^2
    = xw^2 > 0 local min

    Max velocity occurs at t = pi/2w


    In finding maximum acceleration:

    To determine maximum acceleration, we first determine the rate of change of jerk.

    Jerk is defined as the 3rd derivative of displacement with respect to time, t.

    y'' = -xw^2 sin(wt)

    y"' = [d/dt(-xw^2) + (-x)d/dt(w^2)] sin(wt) + (-xw^2) d/dt sin(wt) d/dt(wt)

    = [0 + 0 ]sin(wt) - (xw^2) cos(wt)w

    = -(xw^3)cos(wt)

    y" = 0 = -xw^2 sin(wt)

    -xw^2 sin(wt)= 0

    sin(wt) = 0

    sin(0) = 0 and sin(pi)= 0

    wt = 0 and wt = pi

    t = 0 and t = pi/w

    y"'(0) = -xw^3 cos(w.0)
    = -xw^3(1)
    = -xw^3


    y"'(pi/w) = -xw^3 cos(w.pi/w)
    = -xw^3 cos(pi)
    = -xw^3(-1)
    = xw^3

    y"'(0) = -xw^3 < 0
    local max

    therefore, maximum acceleration occurs at t = 0
     
    Last edited: Dec 18, 2013
  23. Dec 18, 2013 #22

    Doc Al

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    Why are you setting that to zero?

    No.

    OK.

    OK.

    Why are you setting the acceleration = zero? You calculated the jerk, so use it!

    No.

    Try once more.
     
  24. Dec 18, 2013 #23
    What was I thinking. I was rushing to type these.

    Will retry
     
  25. Dec 18, 2013 #24
    But in finding maximum velocity, shouldn't I find the derivative of velocity, which gives me acceleration, then set it to zero to determine the roots(max velocity implies y"(root)<0 )?
    or are you implying that because I have derived the function for jerk, I should work on jerk instead?
     
  26. Dec 18, 2013 #25
    In finding maximum velocity:

    y''= 0 = -xw^2 sin (wt)
    sin(wt) = 0 if sin(0) or sin(pi)

    wt = 0
    t = 0
    and
    wt = pi
    t = pi/w

    roots are t = 0 and t = pi/w

    y'(0) = xw cos(wt) = xw cos(0) = xw
    y''(pi/w) = xw cos(w. pi/w) = xw cos(pi) = -xw

    max velocity occurs at t = 0

    In finding max acceleration:

    y''' = -xw^3 cos(wt) = 0
    cos(wt) = 0 if cos(pi/2) and cos(3pi/2)

    wt = pi/2
    t = pi/2w
    and
    wt = 3pi/2
    t = 3pi/2w

    y''(pi/2w) = -xw^2 sin (w . pi/2w)
    = -xw^2 sin (pi/2)
    = -xw^2

    y''(3pi/2w) = -xw^2 sin(w. 3pi/2w)
    = -xw^2 sin(3pi/2)
    xw^2

    max acceleration occurs at t = 3pi/2w
     
    Last edited: Dec 18, 2013
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