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Math Help - Uniform Circular Motion

  1. #1
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    Uniform Circular Motion

    A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t(1)=4.00s, it is at point (5.00m, 6.00m) with a velocity (3.00 m/s)j and acceleration in the position x direction. At time t(2) = 10.0s, it has a velocity of (-3.00 m/s)i and acceleration in the positive y direction. What are the x and y coordinates of the center of the circular path if t(2)-t(1) is less than one period?

    I have tried a bunch of stuff on this problem and I keep getting the wrong answer....help please.
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  2. #2
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    Quote Originally Posted by topsquark View Post
    What about r? We are stuck here. I cannot find a way to find r without more information.

    -Dan
    From the first statement you know that k = 6, because the particle at (5,6) is directly to the left of the center of the circle. Also, the equations should be:

    x = h + r \sin(\omega{t} + b)

    y = 6 + r \cos(\omega{t} + b)

    The trig functions are reversed because the particle is moving clockwise.

    Furthermore,  6\omega = \frac{3}{4} \cdot 2\pi.
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  3. #3
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    Ima agree with iceman here, I think you solved for the angular velocity wrong top.


    I think maybe we can use a relation relating the constant velocity around the circular track to the distance traveled to find r


    we know velocity of a particle in uniform circular motion, v = omega*r... we have omega, we have v... vooooooila make sure your units are rad/s
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  4. #4
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    I would suggest this method of solving the problem:

    \frac{dx}{dt} = \frac{\pi}{4}r\cos{(\frac{\pi}{4}t + b)}

    \frac{dy}{dt} = -\frac{\pi}{4}r\sin{(\frac{\pi}{4}t + b)}

    3 = \sqrt{\frac{\pi^2r^2}{16}} = \frac{\pi{r}}{4}

    12 = \pi{r}

    \frac{12}{\pi} = r
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  5. #5
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    iceman,

    I'm not sure you can make the assumption that this person is experienced w/ calculus.... it may or may not be the case

    To Elite, if you can follow the calculus...

    Just note that be broke the velocity vector into x and y components (dx/dt) and (dy/dt). Same way you solve for the magnitude of any right triangle you would take sqrt(dx/dt^2 + dy/dt^2) sorry for not using the script.

    Assuming you don't know calculus, I would

    1) Solve for omega = (change in angle)/(change in time) = 3pi/2 / 6 = pi/4 rad/s

    2) Note that v = omega*r => r = v/omega = 3/pi/4

    Same solution, different method depending on your background
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  6. #6
    Forum Admin topsquark's Avatar
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    Thanks guys. Don't know where my head was at.

    -Dan
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  7. #7
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    Quote Originally Posted by icemanfan View Post
    From the first statement you know that k = 6, because the particle at (5,6) is directly to the left of the center of the circle. Also, the equations should be:

    x = h + r \sin(\omega{t} + b)

    y = 6 + r \cos(\omega{t} + b)

    The trig functions are reversed because the particle is moving clockwise.

    Furthermore,  6\omega = \frac{3}{4} \cdot 2\pi.
    Where did you guys get these equations exactly?
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  8. #8
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    Quote Originally Posted by Elite_Guard89 View Post
    Where did you guys get these equations exactly?
    The equations for a particle in circular motion about the center (h, k) are either going to be

    x = h + sin(\omega{t} + b)
    y = k + cos(\omega{t} + b)
    if the particle is moving clockwise

    or

    x = h + cos(\omega{t} + b)
    y = k + sin(\omega{t} + b)
    if the particle is moving counterclockwise.

    We know from the direction of acceleration, the given fact that between t = 4 and t = 10 we do not have a full cycle, and the directions of velocity at t = 4 and t = 10 that we have moved 3/4 of a revolution clockwise. In radians, a full revolution is 2\pi radians, so 3/4 of a revolution is \frac{3}{4} \cdot 2\pi radians. This is equal to the total change in angle, which is \omega(t_2 - t_1) = (rate of change in angle per time)x(time taken), or in this case \omega(10 - 4) = 6\omega
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