Question

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 mm at the top of a leg of length LL.Find an expression for the person's maximum walking speed vmaxvmax.

Answer 1

Answer:

$\mathbf{v_{max} = \sqrt{gL}}$

Explanation:

Considering an object that moving about in a circular path,  the equation for such centripetal force can be computed as:

$\mathbf {F = \dfrac{mv^2}{2}}$

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

$mg cos \theta = \dfrac{mv^2}{L}$

Dividing both sides by "m"; we have :

$g cos \theta = \dfrac{v^2}{L}$

Making "v" the subject of the formula: we have:

$v^2 = g Lcos \theta$

$v=\sqrt{ gL cos \theta$

So, when $\theta$ = 0; the velocity is maximum

∴

$v_{max} = \sqrt{gL \ cos \theta}$

$v_{max} = \sqrt{gL \ cos (0)}$

$v_{max} = \sqrt{gL \times 1}$

$\mathbf{v_{max} = \sqrt{gL}}$

Hence; the maximum walking speed for the person  is $\mathbf{v_{max} = \sqrt{gL}}$