Question

A tether ball is attached to a swivel in the ceiling by a light cord of length L. When the ball is hit by a paddle, it swings in a horizontal circle with constant speed v, and the cord makes a constant angle beta with the vertical direction. The ball goes through one revolution in time T. Assume that T, the mass m, and the length L of rope are known, derive algebraic expression for the tension FT in the cord and the angle beta.

Answer 1

As the ball goes through one revolution, it moves a distance that is equal to the circumference of the horizontal circle.

Circumference = 2 * π * r
If you look at the diagram, you will see a right triangle that contains the radius of the circle, the angle, and the length of the cord. Let’s use following equation to determine the radius.

r = L * sin β, r = L * sin β
Let’s use this for r in the circumference equation.

Circumference = 2 * π * L * sin β = 2 * π *L* sin β
To determine the velocity of the ball, divide by T seconds
v = 2 * π * L*sin β / T

Let’s determine the vertical and horizontal component of the tension
Vertical = F * cos β, Horizontal = F * sin β

The vertical component of the tension is equal to the weight of the ball. The horizontal component of the tension is equal to the centripetal force.

F * cos β = m * g
F * sin β = m * v^2/r
Divide the second equation by the first equation.
F^2 = (mg)^2 +( mv^2/R)^2

on substituting and solving. we get

F = m*4*pi^2*L/T^2

We have

F * cos β = m * g

cos β = mg/F = mg/ [  m*4*pi^2*L/T^2]

cos β = [g*T^2 / 4 *pi^2*L]

β = cos^-1 [g*T^2 / 4 *pi^2*L]