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:

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

no

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