Answer:
[tex]\mathbf{v_{max} = \sqrt{gL}}[/tex]
Explanation:
Considering an object that moving about in a circular path, the equation for such centripetal force can be computed as:
[tex]\mathbf {F = \dfrac{mv^2}{2}}[/tex]
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:
[tex]mg cos \theta = \dfrac{mv^2}{L}[/tex]
Dividing both sides by "m"; we have :
[tex]g cos \theta = \dfrac{v^2}{L}[/tex]
Making "v" the subject of the formula: we have:
[tex]v^2 = g Lcos \theta[/tex]
[tex]v=\sqrt{ gL cos \theta[/tex]
So, when [tex]\theta[/tex] = 0; the velocity is maximum
∴
[tex]v_{max} = \sqrt{gL \ cos \theta}[/tex]
[tex]v_{max} = \sqrt{gL \ cos (0)}[/tex]
[tex]v_{max} = \sqrt{gL \times 1}[/tex]
[tex]\mathbf{v_{max} = \sqrt{gL}}[/tex]
Hence; the maximum walking speed for the person is [tex]\mathbf{v_{max} = \sqrt{gL}}[/tex]
