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

`\mathbf{v_(max) = √(gL)}`

Step-by-step explanation:

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

`\mathbf {F = (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 = (mv^2)/(L)`

Dividing both sides by "m"; we have :

`g cos \theta = (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) = √(gL \ cos \theta)`

`v_(max) = √(gL \ cos (0))`

`v_(max) = √(gL * 1)`

`\mathbf{v_(max) = √(gL)}`

Hence; the maximum walking speed for the person is
`\mathbf{v_(max) = √(gL)}`