Caption: A diagram of "a conical pendulum in motion. The pendulum bob moves in a horizontal circle with constant speed v, θ = suspension angle, r = radius of bob's circular motion, h = vertical height of suspension above the plane of the bob's motion, L = length of the wire connecting the bob to the suspension point, T = wire's tension force acting on the bob, and mg=weight of the bob." (Slightly edited.)

The conical pendulum in motion is a bit like a planet. In both cases, there is a body perpetually falling toward the center, but it keeps missing because it has sideways motion or, in other words, angular momentum.

The analysis of the conical pendulum in motion is straightforward:

1. To satisfy Newton's 2nd law of motion (AKA F=ma) for a case of uniform circular motion, the horizontal component of the tension force must equal the centripetal force:

     T*sin(θ) = mv**2/r  ,  

   where the centripetal force is actually NOT a force, but rather "ma" of "F=ma" (i.e., Newton's 2nd law of motion (AKA F=ma)). The T*sin(θ) is the net horizontal force in "F=ma".

   Note the bob is in acceleration in the horizontal direction since it is NOT in straight line motion. The net horizontal force T*sin(θ) is the cause of the acceleration.

2. To satisfy Newton's 2nd law of motion for the vertical direction where there is NO acceleration, there must be balanced forces:

     T*cos(θ) = mg  which gives T = mg/cos(θ)  .  

3. Substituting T = mg/cos(θ) in the first formula above, we find

     mg*tan(θ) = mv**2/r = m(2πr/P)**2/r = m[(4(π**2)r)/P**2]  ,
   
                      where v = 2πr/P and P is period.  

   Now we get the formulae

     θ = arctan[ ( 4(π**2)*r )/(gP**2) ] , where arctan is the function arctangent.
     P = sqrt[ ( 4(π**2)r )/(g*tan(θ)) ]  .

   Formulae for other variables can be derived, of course.

4. We can devise a fiducial value formula that is convenient for analyzing classroom demonstations using fiducial values: θ = 45°, tan(θ = 45°) = 1, r = 1 m, 4(π**2) ≅ 40, and Earth's surface gravitational field fiducial value g = 9.8 m/s**2 ≅ 10 m/s**2. The fiducial value formula is

     P = ( 2.00708992 ... s) * sqrt[ (r/(1 m))/[(g/(9.8 m/s**2))*(tan(θ)/1)] ]  .

   So if the demonstrator chooses r ≅ 1 m and θ ≅ 45°, the period P ≅ 2 s.

   If the demonstrator chooses r ≅ 0.25 m and θ ≅ 45°, the period P ≅ 1 s.

5. The analysis above assumed an ideal case where there are NO resistive forces (e.g., air drag and friction at the suspension point) and NO driving forces.

   In real demonstrations, there are both. In fact, the demonstrator, trying to maintain a uniform motion, adjusts their conical swinging driving force to roughly compensate for the resistive forces.