A ball with a mass of 0.5 kg is attached to one end of a light rod that is 0.5 m long. The other end of the rod is loosely pinned at a frict

Question

A ball with a mass of 0.5 kg is attached to one end of a light rod that is 0.5 m long. The other end of the rod is loosely pinned at a frictionless pivot. The rod is raised until it is vertical, with the ball above the pivot. The rod is released and the ball moves in a vertical circle. The tension in the rod as the ball moves through the bottom of the circle is closest to:

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Verity 3 years 2021-08-11T14:14:28+00:00 1 Answers 7 views 0

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    2021-08-11T14:15:29+00:00

    Answer:

    The tension in the rod as the ball moves through the bottom circle is 9.8 N

    Explanation:

    When the ball is released from rest, the centripetal force equals the weight of the ball. So mv²/r = mg where m = mass of ball = 0.5 kg, v = speed of ball, r = radius of vertical circle = length of rod = 0.5 m and g = acceleration due to gravity = 9.8 m/s²

    v = √gr = √9.8 m/s² × 0.5 m = √4.9 = 2.21 m/s

    Now at the bottom of the circle T – mg = mv²/r where T = tension in the rod

    T = m(g + v²/r)

    = m(g + (√gr)²/r)

    = m(g+ gr/r)

    = m(g + g)

    = 2mg

    = 2 × 0.5 kg × 9.8 m/s²

    = 9.8 N

    So, the tension in the rod as the ball moves through the bottom circle is 9.8 N

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