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Thread: vectors in circular motion

  1. #1
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    vectors in circular motion

    A pendulum consist of a particle P of mass m, attached to one end of a model string of length l = 2.02 m. The other end of the string is attached to a fixed point O and the particle moves in a vertical circle about the fixed point as shown.

    The angle made by the string with the vertical is measured anticlockwise and we assume that the only forces acting on the particle are its weight W and the tension force T due the string. The particle starts at the lowest point with an initial horizontal velocity of √(7gl/2) in the anticlockwise direction.

    Calculate the speed of the particle, in m s-1 when the tension becomes zero, giving your answer to 3 decimal places. Use g = 9.81 ms-2.


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    Re: vectors in circular motion

    Hey noobpronoobpro.

    Can you please show us what you have tried?
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    Re: vectors in circular motion

    I've gotten |T|= mg cos ϴ+ ml dotϴ^2 when resolved in the er-direction.
    Im stuck there and dont know how to continue from there.
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    Re: vectors in circular motion

    i've somehow managed to figure out how to solve it haha... i've gotten another eqn for mg cos ϴ using conservation of energy and substituted into |T|= mg cos ϴ+ ml dotϴ^2 to get the answer. The answer i gotten was 3.148m/s.
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  5. #5
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    Re: vectors in circular motion

    The pendulum bob has sufficient energy to achieve a height of $\dfrac{7L}{4}$ above its initial position, so the position where $T=0$ and $|v| > 0$ occurs at $L < h < \dfrac{7L}{4}$, where $h=0$ is its initial position.

    At that height, $h$, the radial component of weight is providing the centripetal force required for circular motion.

    Let $\phi$ be the angle between the string of length $L$ and the weight vector, $W=mg$ when the pendulum bob is at that height, $h$.

    $F_c = \dfrac{mv^2}{L} = mg\cos{\phi} \implies v^2 = gL\cos{\phi}$

    using conservation of energy ...

    $mgh + \dfrac{1}{2}mv^2 = \dfrac{1}{2}mv_0^2$

    $v^2 = v_0^2 - 2gh = \dfrac{7gL}{2} - 2g\left(L+L\cos{\phi}\right)$

    setting the two equations for $v^2$ equal ...

    $3gL\cos{\phi} = \dfrac{3gL}{2} \implies \cos{\phi} = \dfrac{1}{2}$

    therefore, $v^2 =gL\cos{\phi}=\dfrac{gL}{2} \implies v = \sqrt{\dfrac{gL}{2}}$
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