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Math Help - Ferris Wheel Sinusoidal problem

  1. #1
    Junior Member CONFUSED_ONE's Avatar
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    Ferris Wheel Sinusoidal problem

    Jacob rides a Ferris wheel at a carnival. the wheel has a 16-m diameter and turns at 3rpm with its lowest point 1 m above the ground. Assume that Jacob's height h above the ground is a sinusoidal function of time t [in seconds], where t=0 represents the lowest point of the wheel.
    -Write an equation for h.
    - draw a graph of h for 0< [or equal to] t < [or equal to] 30


    can someone help me with this problem?



    thanks .
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CONFUSED_ONE View Post
    Jacob rides a Ferris wheel at a carnival. the wheel has a 16-m diameter and turns at 3rpm with its lowest point 1 m above the ground. Assume that Jacob's height h above the ground is a sinusoidal function of time t [in seconds], where t=0 represents the lowest point of the wheel.
    -Write an equation for h.
    - draw a graph of h for 0< [or equal to] t < [or equal to] 30


    can someone help me with this problem?



    thanks .
    Let's break this into pieces. First I'm going to assume the center of the Ferris wheel is on the ground. (Yes, I realize that half the wheel is now buried. (Dang it Jim, I'm a theoretical physicist not an engineer!)) I'm also going to assume the Ferris wheel is rotating counterclockwise and we are going to measure Jacob's position by an angle theta measured CCW from the ground to his position.

    The angular speed of the Ferris wheel is constant and is
    \omega = \frac{\theta}{t}
    so
    \theta = \omega t

    At this stage Jacob's height is:
    h = 16sin(\omega t)

    Obviously the center of the Ferris wheel isn't on the ground. Since the lowest point of the wheel is 1 m above the ground we know the center is 17 m (1 m + 16 m) above the ground. So Jacob's height is:
    h = 17 + 16sin(\omega t)

    What's \omega? Well, the wheel is turning at 3 rpm, so the angular speed in rad/s will be:
    \omega = \frac{3 \, rpm}{1} \cdot \frac{2\pi \, rad}{1 \, rev} \cdot \frac{1 \, min}{60 \, s} = \frac{\pi}{10} rad/s

    Thus
    h = 17 + 16sin \left ( \frac{\pi t}{10} \right )

    -Dan
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  3. #3
    Junior Member CONFUSED_ONE's Avatar
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    Thank you Dan . :]
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Thus
    h = 17 + 16sin \left ( \frac{\pi t}{10} \right )

    -Dan
    t=0 is supposed to correspond to the lowest point; while here the height
    at t=0 is 17 m. To correct the phase you need to change this to:

    h = 17 + 16sin \left ( \frac{\pi t}{10} - \pi/2\right )

    RonL
    Last edited by CaptainBlack; January 19th 2007 at 01:53 AM. Reason: add a semi-colon
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  5. #5
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    Quote Originally Posted by CONFUSED_ONE View Post
    Jacob rides a Ferris wheel at a carnival. the wheel has a 16-m diameter and turns at 3rpm with its lowest point 1 m above the ground. Assume that Jacob's height h above the ground is a sinusoidal function of time t [in seconds], where t=0 represents the lowest point of the wheel.
    -Write an equation for h.
    - draw a graph of h for 0< [or equal to] t < [or equal to] 30

    can someone help me with this problem?

    thanks .
    Here is one way.

    3 rpm = 3 revolutions per min
    So, 1 revolution = (1/3) minute.
    That means the period of the h is 1/3 minute.
    Its equivalent in degrees (one revolution = 360 degrees) is:
    0 deg ---> 0 min.
    90 deg ---> (1/4)(1/3) = 1/12 min.
    180 deg ---> (2/4)(1/3)= 1/6 min.
    270 deg ---> (3/4)(1/3) = 1/4 min.
    360 deg ---> (4/4)(1/3) = 1/3 min.
    Meaning,
    At t=0 min, Jacob is at the lowest point, or h = 1 m.
    At t=1/12 min, he is at 90 degrees counterclockwise fromm the bottom, or h = 1 +16/2 = 9 m.
    At t=1/6 min, he is at 180 degrees counterclockwise fromm the bottom, or h = 1 +16 = 17 m.
    At t=1/14 min, he is at 270 degrees counterclockwise fromm the bottom, or h = 1 +16/2 = 9 m. again.
    At t=1/3 min, he is at 360 degrees counterclockwise fromm the bottom, or h = 1 m. again

    Graph those on a (theta,h) rectangular axes.
    The graph is like an inverted cosine curve. It is a normal cosine curve that is shifted at 180 degrees to the right, because at 180deg, h is maximum.

    So the sinusoidal curve may be in the form
    h(t) = A*cos[Bt +C] +D --------------(i)
    where
    A = amplitude
    B = frequency
    C = horizontal shift
    D = vertical shift

    A = 16/2 = 8 m.

    Period = 360/B
    1/3 = 360/B
    B = 3*360 = 1080

    C = 180 deg, to the right.

    D = 1 +16/2 = 9 m.

    Hence,
    h(t) = 8cos[1080t -180] +9 -----------answer.

    where
    [1080t -180] is in degrees.
    Horizontal shift C is negative because the shift is to the right of the supposed to be origin.
    h(t) is in meters.
    t is in minutes.

    ----------------
    Check.

    When t = 0, h = 1m.
    h(0) = 8cos[1080(0) -180] +9
    h(0) = 8cos[-180deg] +9
    h(0) = 8[-1] +9
    h(0) = 1 m. ------------OK.

    When t = 1/12 min, h=9m.
    h(1/12) = 8cos[1080(1/12) -180] +9
    h(1/12) = 8cos[90 -180] +9
    h(1/12) = 8cos[-90deg] +9
    h(1/12) = 8[0] +9
    h(1/12) = 9 m. ------------OK.

    When t = 1/6 min, h=17m.
    h(1/6) = 8cos[1080(1/6) -180] +9
    h(1/6) = 8cos[180 -180] +9
    h(1/6) = 8cos[0deg] +9
    h(1/6) = 8[1] +9
    h(1/6) = 17 m. ------------OK.

    When t = 1/4 min, h=9m again.
    h(1/4) = 8cos[1080(1/4) -180] +9
    h(1/4) = 8cos[270 -180] +9
    h(1/4) = 8cos[90deg] +9
    h(1/4) = 8[0] +9
    h(1/4) = 9 m. ------------OK.

    When t = 1/3 min, h=1m again.
    h(1/3) = 8cos[1080(1/3) -180] +9
    h(1/3) = 8cos[360 -180] +9
    h(1/3) = 8cos[180deg] +9
    h(1/3) = 8[-1] +9
    h(1/3) = 1 m. ------------OK.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    t=0 is supposed to correspond to the lowest point; while here the height
    at t=0 is 17 m. To correct the phase you need to change this to:

    h = 17 + 16sin \left ( \frac{\pi t}{10} - \pi/2\right )

    RonL
    Whoops! Sorry about that. Thanks for the catch.

    -Dan
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