Question

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ml2/3. ________________________ sec The pace of normal walking (3.0 mi/hr) is close to the natural frequency of the leg because the most efficient frequency to "drive" a system is the natural frequency. It takes less effort to walk at this rate.

Answer 1

Answer:

1.54 s

Explanation:

Considering that the legs constitute 16% of the total weight of the man then mass, $m= \frac {16}{100}\times 67=10.72 Kg$

The legs also constitute 48% of his height hence $H=\frac {48}{100}\times 1.83=0.8784 m$

The moment of inertia of a cylinder rotating about a perpendicular axis at one end is $\frac {ml^{2}}{3}$ hence $I=\frac {10.72\times 0.8784^{2}}{3}=2.757135974Kg.m^{2}$

We also know that the period is given by $2\pi \sqrt{\frac {I}{mgh}}$

Here, h=0.5H= 0.5*0.8784=0.4392 m

Taking g as 9.81 kg/m2 then

$T= 2\pi \sqrt{\frac {2.757135974}{10.72\times 9.81\times 0.4392}}\\=1.535132615 s\\\boxed{\approx 1.54 s}$