Question

The physics of circular motion sets an upper limit to the speed of human walking. (If you need to go faster, your gait changes from a walk to a run.) If you take a few steps and watch
what’s happening, you’ll see that your body pivots in circular motion over your forward foot as you bring your rear foot forward for the next step. As you do so, the normal force of the
ground on your foot decreases and your body tries to “lift off” from the ground.
a. A person’s center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass m at the top of a leg of length L. Find an expression for the person’s maximum walking speed v_max.
b. Evaluate your expression for the maximum walking speed of a 70 kg person with a typical leg length of 70 cm. Give your answer in both m/s and mph, then comment, based on your
experience, as to whether this is a reasonable result. A “normal” walking speed is about 3 mph.

Short answer

a) Maximum walking speed is $v_{max}=\sqrt{gL}$

b) V_max for given person is =2.62m/s =5.86mph

Step-by-step solution

Part(a) Step 1 : Given information

Mass of person =70kg
Length of = 70 cm

Part(a) Step 2 : Explanation

Lets consider at the bottom part of a circle, we have

$\frac{m{v}^2}{L}=mg-N$

where N is the normal reaction force.

We know at max speed normal force is zero.
So,

$$\begin{gathered} \frac{m{v}^2}{L}=mg-0 \\ v=\sqrt{gL} \end{gathered}$$

So maximum speed is

role="math" localid="1649080916204" $v_{max}=\sqrt{gL}...........................\left(1\right)$

Part(b) Step 1: Given Information

Mass of person =70kg
Length of = 70 cm =0.7 m

Part(b) Step 2 : Explanation

Substitute values in equation(1) we get

$$\begin{gathered} v_{max}=\sqrt{gL}=\sqrt{(9.81m/s^2)\times (0.7m)}=2.62m/s \\ {v}_{max}=5.86mph \end{gathered}$$