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physics

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We are learning about circular motion, uniform and non uniform. how do you find tension in a string, when the string is attached to a object with mass m and length r? What about if you are swinging it around in a horizontal circle and the string makes a angle theata with the horizontal?

  • physics - ,

    If we ignore gravity, and calculate the tension in the string due solely to the circular motion, then
    Tc=mrω²
    where
    Tc=tension due to circular motion
    m=mass attached to the end of the string
    r=radius
    &omega=angular velocity, in radians / second
    For a horizontal circular motion, the above value of Tc should be vectorially added to the weight, i.e.
    Total Tension, T
    = &radic(T²+(mg)^2)
    The angle
    tan-1(mg/Tc)
    is the angle with the horizontal.

    For a vertical circular motion, Tc must be added to the vertical component of weight, -mg (positive upwards)
    The vertical component can be obtained by -mg*sin(θ),
    where θ=angle above horizontal, equals zero when mass is horizontal and going up.

    Tension T
    =Tc - mg*sin(θ)

  • physics - ,

    The only thing I would add to the above, is that the r in the formula for centripetal force is the radius of the horizonal circle, not the length of the string. if l is the length of the string, then r=lcosTheta. This makes Tc = mw^2 l cosTheta

  • physics - ,

    Thank you Bob.

  • physics - ,

    In the total tension, what does &radic mean?
    Total Tension, T
    = &radic(T²+(mg)^2)
    The angle
    tan-1(mg/Tc)
    is the angle with the horizotal

  • physics - ,

    Total Tension, T
    = √(T²+(mg)^2)
    The angle
    tan-1(mg/Tc)
    is the angle with the horizotal

    √ is the equaivalent of square-root.
    The formula is a typical application of Pythagoras theorem while adding the vertical and horizontal components of force. I could rewrite it as:

    Total Tension, T
    = sqrt(T²+(mg)^2)
    The angle
    tan-1(mg/Tc)
    is the angle with the horizotal

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