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Math Help - Walking on a merry-go-round

  1. #1
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    Walking on a merry-go-round

    Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
    What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?


    R[m] = 3.54;

    Someone help me please
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  2. #2
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    Quote Originally Posted by m_i_k_o View Post
    Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
    What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?


    R[m] = 3.54;

    Someone help me please
    the acceleration is centripetal.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by m_i_k_o View Post
    Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
    What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?


    R[m] = 3.54;

    Someone help me please
    Write the (vector) velocity equation out as a function of time and differentiate it to give the (vector) acceleration as a function of time, then as you are at a radius R at time t=R find the required acceleration and take its absolute value.

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Write the (vector) velocity equation out as a function of time and differentiate it to give the (vector) acceleration as a function of time, then as you are at a radius R at time t=R find the required acceleration and take its absolute value.

    CB
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Xapphire13 View Post
    Like this? (Attached pic)
    Assume that \theta=0 at t=0

    Then at time t (=R) we have the velocity is:

    \bold{v}=[R \omega \sin(\omega t)+\cos(\omega t), R \omega \cos(\omega t)]+\sin(\omega t)] = [t \omega \sin(\omega t)+\cos(\omega t), t \omega \cos(\omega t)]+\sin(\omega t)]

    So:

     <br />
\dot{\bold{v}}=[t \omega^2 \cos(\omega t), 2 \omega \cos(\omega t)+t \omega^2 \sin(\omega t)]<br />

    Now \omega=\frac{2 \pi}{3} radians/s, so when t=3.54 we have:

    \dot{\bold{v}}\approx[6.61158,\ 15.8338],

    so the magnitude of the acceleration is \approx 17.16 m/s^2

    .. check the algebra and the arithmetic


    CB
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