A mass m is suspended from a frictionless pivot at the end of a string of arbitrary length, and is set to whirling in a horizontal circular path whose plane is a distance H below the pivot point. Find the period of revolution of the mass in its orbit.
Solution by Ronald Lovejoy
Let R be the radius of the circular path. Let θ be the angle between the string and the vertical axis. Let P be the period of revolution of the mass, and let ω = 2π / P be its angular velocity.
There are three forces at play in this problem: the gravitational force pulling the mass down vertically, the tension T on the string, and the radial (centripetal) force required to keep the mass moving in a circle at constant speed.
We resolve the tension vector into vertical and radial components,
TV = |T|cosθ and TR = |T|sinθ .
The vertical component of the tension must balance the weight of the mass, so TV =|T|cosθ = mg, or
|T| = mg/cosθ .
The radial component of the tension must balance the centripetal force. The (radial) acceleration of the mass equals ω2R and R = H tan θ, so TR = |T|sinθ = mω2Htanθ, or
|T| = mω2H/cosθ.
Taking these two values for |T| to be equal, we get mg/cosθ = mω2H/cosθ. The mass and cosine terms cancel, and we are left with
ω2 = g/H.
Since ω = 2π /P,
P = 2π √(H/g) .