A tennis ball connected to a string is spun around in a vertical, circular path

Question

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m= 0.15 kg and moves at v=4.89 m/s. The circular path has a radius of R=0.94 m What is the magnitude of the tension in the string when the ball is at the bottom of the circle? Shat is the magnitude of the tension in the string when the ball is at the side of the circle? What is the magnitude of the tension in the string when the ball is at the top of the circle? What is the minimum velocity so the string will not go slack as the ball moves around the circle?

Explanation / Answer

1.

In this problem, the velocity is constant. This means the centripetal force is constant.
Fc = m * v^2 / r = 0.15 * 4.89^2/0.94 3.815 N

2)
When the ball is at the side of the circle, the tension and centripetal forces and the weight is a vertical force. So, the weight of the ball does not affect the tension.
Tension =weight =Fc = 0.15*9.81 = Fc = 1.4715 N

3)
When the ball is at the bottom, the tension in the rope must support the weight of the ball and cause the ball to move at a constant speed at that position on the circle. The tension in the rope, the weight of the ball, and the centripetal force are all vertical forces.
Tension = weight + Fc
Tension = 1.4715 + 3.815 = 5.2865 N
This is the maximum tension

4)
At the Top, the tension at the top of the circle is the minimum tension.
Tension = Weight – Fc = 3.815 – 1.4715 = 2.3435 N

4)

PE at top = mass * g * diameter = 0.15 * 9.8 * 2 * 0.94
Set KE at bottom = 0.15 * 9.8 * 2 * 0.94
½ * 0.15 * v^2 = 0.15 * 9.8 * 2 * 0.94
Solve for v

v = 6.07 m/s
This is the minimum “constant” velocity so the string will not go slack as the ball moves around the circle

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