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Angular displacement and bicyclist

  1. Oct 24, 2009 #1
    1.

    An exhausted bicyclist pedals somewhat erratically, so that the angular velocity of his tires follows the equation

    omega(t) = 0.5t - 0.25sin(2t), t greater than/equal to zero.

    where t represents time (measured in seconds).




    2. omega is defined as the first derivative of theta.



    3. I integrated omega(t), obtaining 0.25t^2 + (1/8)cos(2t), plugged in 2 and got my incorrect answer.
     
  2. jcsd
  3. Oct 24, 2009 #2

    cepheid

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    Hey, can you post the actual question? Is it to find the angular displacement after 2 seconds?
     
  4. Oct 24, 2009 #3
    What angular displacement theta has the spot of paint undergone between time 0 and 2 seconds?

    That is the question. Apparently I integrated incorrectly, but I do not see where I made a mistake.......
     
  5. Oct 24, 2009 #4

    cepheid

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    I'm really not sure what the problem is either. Is your calculator in the right mode?
     
  6. Oct 24, 2009 #5
    Yes, it is in radians. So, cos(4) = -.6536

    (1/8)cos(4) = -.0817

    + 1 = 0.918


    ??????
     
  7. Oct 24, 2009 #6
    BTW, it says that I did my integral wrong. In other words, the angular displacement does NOT equal:

    0.25t^2 + (1/8)cos(2t)

    ???????????
     
  8. Oct 24, 2009 #7

    cepheid

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    You calculated the *indefinite* integral:

    [tex] \theta(t) = \int \omega(t) \, dt [/tex]

    [tex] = \int \left(\frac{1}{2}t - \frac{1}{4}\sin(2t) \right) \, dt [/tex]

    [tex] = \frac{1}{4}t^2 + \frac{1}{8}\cos(2t) + C [/tex] ​

    where C = [itex]\theta(0) [/itex]

    The question is asking you for the angular *displacement* between t = 0 and t = 2, which is given by [itex]\theta(2) - \theta(0) [/itex], which is given by the *definite* integral:

    [tex] \theta(2) - \theta(0) = \int_0^2 \omega(t) \, dt [/tex]

    [tex] = \int_0^2 \left(\frac{1}{2}t - \frac{1}{4}\sin(2t) \right) \, dt [/tex]

    [tex] = \left[\frac{1}{4}t^2 + \frac{1}{8}\cos(2t)\right]_0^2 [/tex] ​
     
  9. Oct 24, 2009 #8
    Damn it, I forgot the blood 1/8 from taking cos(0).

    Thank you.
     
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