Simple pendulum of a leg

Question

A particular persons lower leg is 56 cm long and weighs 9.5 kg. If allowed to swing freely, with what time period would the leg swing? Is this close to what you would estimate such a persons walking pace is? (Treat the leg as a simple pendulum with length equal to half the leg length and all the leg's mass concentrated at this point.)

Answer 1

The period of a simple pendulum is dependent on its length. Since the leg is 56 cm long, its center of mass, or the length of our pendulum, would be half of that so L=28 cm (.28 m). The equation for the period of a pendulum is:

T = 2π√(L/g)

So by substituting your information you can solve for the period.

T=2*3.14*√(.28/9.8) = 1.06 s

The mass of the leg has no effect on its period.

Answer 2

Our legs swing forward and backward about the hip joint as a pivot when walking, therefore we can assume the leg to be a uniform rod of length, d = 56 cm = 0.56 m. Now let's say that it takes one-half of the period, T for the leg to swing forward. Then the frequency,f for the simple pendulum (which is inversely related to the period) is give by

f = (1/2π)√(mgL/I)

where I = 1/3md² is the moment of inertia for the uniform rod (leg) of length, d rotating about an axis perpendicular to one end. The length, L is the distance between the pivot at the hip and the center of gravity of the leg. this means that L = d/2 = .56/2 m = .28m

==> f = 1/T ==> T = 2π√(I/mgL) = 2π√(1/3md²/mgL) = 2π√(1/3(.56m)²/(9.8m/s²)(.28m)) = 1.23 s

The required time is one-half the period, which is 0.61 s

This number is pretty good approximation to average human walking pace.

[see this link for comparison: https://www.healthline.com/health/exercise-fitness/average-walking-speed]