Question

A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person’s step length, A? (Assume that the legs are rigid rods of length L.) (b) If the person takes one step every 1.20 s, what is the period T of the gait cycle? (c) Is the period T in this case longer or shorter than the period for most effortless walking? (d) What is the person’s walking speed?

Answer 1

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 $\frac{m}{s}$

t=0.475s V=1.95 $\frac{m}{s}$

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

$sin(\alpha) =\frac{op}{h}$

$op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m$

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

$T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz$

c).

$T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz$

The period is the inverse of the time of the motion so, the T1 is faster that the T because

$t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s$

d).

The speed is the relation between the distance with time so:

$Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}$