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Stuck on calculating jerk in UCM

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    The question asks to calculate the jerk of an object (both its direction and magnitude) in uniform circular motion.

    2. Relevant equations
    j=d(a)/dt

    3. The attempt at a solution
    I know that the direction is opposite the velocity vector (I did this by drawing a circle and taking a limit of average acceleration as t approaches 0). I also know that the parallel component of jerk is 0 because the parallel component of acceleration in UCM is 0. I am stuck on the magnitude of the perpendicular component. I know that
    [tex] \vec{a}_{\perp} = \frac{v^2}{2}(-r) [/tex] (the r is a unit vector)... can I just take the derivative of this? wouldn't that be 0? because v^2/r is a constant and there is no t. i am suck.
     
    Last edited: Oct 4, 2015
  2. jcsd
  3. Oct 4, 2015 #2

    gneill

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    Staff: Mentor

    Perhaps start with a parameterized version of the motion?

    ##r(t) = R (cos(\omega t) \vec{i} + sin(\omega t) \vec{j})##

    Should be easy enough to differentiate repeatedly...
     
  4. Oct 4, 2015 #3
    where did you get that parameterized version?
     
  5. Oct 4, 2015 #4

    gneill

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    Staff: Mentor

    It's just a conversion from polar form to rectangular form of a circle. x = R cos(θ), y = R sin(θ), where θ = ωt to make it time dependent.
     
  6. Oct 4, 2015 #5
    ok so, j_x = Rw^3sin(wt) and j_y = -Rw^3cos(wt) right?

    the problem is however that my prof wanted us to express it interms of v and r. i dont really know how to convert from w to r.... we havent really talked about w yet.
     
  7. Oct 4, 2015 #6

    gneill

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    Staff: Mentor

    It seems strange that you'd be learning about a concept like jerk without having covered the basics of rotational motion.

    ##v = \omega r~~ ; ~~a = \alpha r## are the basic relationships between angular and linear velocities and accelerations. You would profit from taking the magnitudes of each of the vectors along the differentiation path: position → velocity → acceleration → jerk. For example, the magnitude of the velocity vector is v = ω R.
     
  8. Oct 4, 2015 #7
    yeah, it was just one problem out of a list of 10 or so. our teacher kinda talked about w in the last 5 minutes of class on friday, but we haven't touched rotations yet... we havent even begun f=ma (school just started)
     
  9. Oct 4, 2015 #8

    Mister T

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    Science Advisor

    Did your teacher mention that v2/r = rw2?

    Most classes have reading assignments in addition to classroom lectures, so maybe it's discussed in more depth there.
     
  10. Oct 4, 2015 #9
    he wrote on the board that v=ds/dt = r d(theta)/dt ... and d(theta)/dt is the same as w. that is all he talked/wrote about w. so sorry Mister T.... it was hidden in my notes... i guess he did write an equation relating the 2.
     
  11. Oct 4, 2015 #10

    Mister T

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    Science Advisor

    Right. So looking at the x-components and the y-components of the position vector and the acceleration vector, what can you conclude about the direction of those two vectors?

    Now make the same comparison of the velocity vector and the jerk vector.

    Can you find the magnitude of these vectors from their components?

    These should lead you to being able to describe the magnitude and direction of the jerk vector, which is what your teacher is asking you to do.
     
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