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1 July, 06:21

A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. What are the magnitude and direction of the acceleration of the ball?

What's the tension in the string?

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  1. 1 July, 06:56
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    For circular motion.

    Centripetal acceleration = mv²/r = mω²r

    Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m

    m = mass = 175g = 0.175kg.

    Angular speed, ω = Angle covered / time

    = 2 revolutions / 1 second

    = 2 * 2π radians / 1 second

    = 4π radians / second

    Centripetal Acceleration = mω²r = 0.175 * (4π) ² * 0.5 Use a calculator

    ≈13.817 m/s²

    The magnitude of acceleration ≈13.817 m/s² and it is directed towards the center of rotation.

    Tension in the string = m*a

    = 0.175*13.817

    = 2.418 N
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