2 years ago

Note: Use g=9.8 N/kg in this assignment.

Physics

1 answer:

Karolina [17]2 years ago

5 0

Answer:

16.12

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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 a

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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

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

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