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

1.54 s

Step-by-step explanation:

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

The legs also constitute 48% of his height hence
`H=\frac {48}{100}* 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* 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* 9.81* 0.4392}}\\=1.535132615 s\\\boxed{\approx 1.54 s}`