**whirling pendulum**

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,

T_{V} = |T|cosθ
and
T_{R} = |T|sinθ .

The vertical component of the tension must balance the weight of the mass, so
T_{V} =|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
ω^{2}R and R = H tan θ, so
T_{R} =
|T|sinθ = mω^{2}Htanθ, or

|T| = mω^{2}H/cosθ.

Taking these two values for |T| to be equal, we get
mg/cosθ = mω^{2}H/cosθ.
The mass and cosine terms cancel, and we are left with

ω^{2} = g/H.

Since ω = 2π /P,

P = 2π √(H/g) .