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Basic Angular Motion question

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    g.JPG

    2. Relevant equations
    ω=ω0+αt

    θ=θ0+ω0t+12αt2 I think?



    3. The attempt at a solution

    For part a, I know that from the graph, the angular velocity in this case would be constant since the graph illustrates a position in rad vs time graph. The thing is, I'm really sure on how to approach this problem. Could somebody hint me on how to work this problem?
     
  2. jcsd
  3. Nov 2, 2014 #2
    For a, aren't we just subtracting the final angle and initial?
     
  4. Nov 2, 2014 #3

    BvU

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    Yep !

    Your relevant equations are applicable for uniformly accelerated angular motion (##\alpha## constant). Here, we clearly have something else at hand, so you will have to fall back to the definitions
    (##\omega = \dot \phi = {d\phi\over dt}## and (##\alpha = \dot \omega = {d\omega\over dt}## )
    [edit] -- however, you don't need these for this exercise. Basically it's a math exercise :)
     
  5. Nov 2, 2014 #4
    Ah I see. Oh and for angular displacement, do you typically consider them as vectors? Not sure if I should include the notation for thetha final - thetha initial...
     
  6. Nov 2, 2014 #5

    BvU

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    Angles are scalars, not vectors. ##\theta## is in radians. It's a bit strange to put "Position in radians" on a graph (but understandable: angular displacement in radians is a mouthful).


    ##\Delta \theta## already has some sense of direction. It comes close to being a vector.

    and from that ##\omega## and ##\alpha## are pseudo vectors (also this link, and this).

    Easier to follow when you know about cross products and polar coordinates. Is that familiar already ?
     
  7. Nov 2, 2014 #6
    Yup! Just found about that, much thanks!
    Last question was for Part B, do you think I'm on the right track in finding the exponential function of the log-log graph scale (e.g: slope of line of log-log: n= logx2-logx1/logt2-logt1 andetc) to find the equation of the angular position of the object?
     
  8. Nov 2, 2014 #7

    BvU

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    For part b) you indeed make good use of the observation that the relationship between log(theta) and log(time) is linear. I understand what you mean, but to make it unambiguous, better use some brackets: n= (logx2-logx1)/(logt2-logt1) :)

    For c) you use the expression and check with the plot. Don't answer in five digits: the accuracy allows two, max three.
     
  9. Nov 2, 2014 #8
    That's the problem, I tried finding the equation for the log-log plot, however ended up getting Ꝋ(t)=(3 rad/s^2) t^2. Whenever I use some arbitrary point of time from the graph, I don't get a corresponding result that matches up with the y-axis (position in rad).

    The way I solved this was n= log(200,000)-log(3) / (log 42)-log(11) = 2.972
    Then knowing that the position function is parabolic in shape: x(t)=Ct^2. So we need to find proportionality constant we solve log x = log C + 3 log(t). So to make it log(x)=log(c), I use t=1s (because log1=0), and got log(3)=logC + 0 ==> C=3.

    So once I found 3, i then get the answer Ꝋ(t)= (3)t^2...am I doing something wrong here?
     
  10. Nov 2, 2014 #9

    BvU

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    is not right. That is only for constant angular acceleration, which is clearly not the case here.

    You found a linear relationship ##\ln(\theta) - ln(3) = A \;( ln (t) - ln(1) )## with A = 2.972 as the slope of the line. So now they want you to rewrite ##\theta = ## ?
     
  11. Nov 2, 2014 #10
    Oh okay, it works with
    θ(t) = (3 rad/s^2) t^(2.972). In some sense correct now?
     
  12. Nov 2, 2014 #11

    BvU

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    In a physics sense: excellent. And:
    Easy to check: fill in a few t and look at the picture !

    But (math,nitpicking, ...) : I would crank up the 2.972 a little bit. root cause: your 200000 looks to me already achieved at t = 40.

    In part c) notice how small changes in the slope and the intercept give big changes in theta. That's what you get for using log-log plots....
     
  13. Nov 2, 2014 #12

    Yeah I rounded the 2.972 to 3.
    For c) I just took the derivative of Θ(t) to get w(t)= 138 rad/s
     
  14. Nov 2, 2014 #13

    BvU

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    Must be an empty battery in the calculator. ?:)
    You can clearly see in the figure that ##\theta## increases a lot more every
    second than that at t=23.
    How did you calculate this 138 rad/s ?
     
  15. Nov 2, 2014 #14
    Derivative of Θ(t) and got w(t)=6(t) and so for 23 seconds, got 138 rad/s
     
  16. Nov 3, 2014 #15

    BvU

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    Derivative of which expression for ##\theta## ? The wrong one, or the one you found in b) ? what is the derivative (not the number, the expression) ?
     
  17. Nov 3, 2014 #16
    the one I found for b.
     
  18. Nov 3, 2014 #17
    apologies what I meant was w(t)=9t^2 which I got from theta(t)=3t^3. :)
     
  19. Nov 3, 2014 #18
    Correct now?
     
  20. Nov 3, 2014 #19

    BvU

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    Yep :)
    (around 4800 rad/s, I hope ?)
     
  21. Nov 3, 2014 #20
    Yup what I got! I really appreciate your help bvu :D
     
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