Question

PERIOD OF THE LEG The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of Equation representing the period of oscillation for a pendulum, where I is the moment of inertia, m is the mass, and h is the distance from the pivot point to the center of mass. The leg can be considered to be a right cylinder of constant density. For a man, the leg constitutes 16 \% of his total mass and 48 \% of his total height [21]. 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:

Explanation:

We can see from the question that

        $T 2\pi\sqrt{\frac{I}{mgh} }$

  and  $I = \frac{ml^2}{3}$

           $m = 16$% of  $67kg$

 =>      $m =10kg$

      from the question  $l =48$% of $1.83m$

 Substituting this into the equation

                 $I = \frac{10.72 * (0.8784)^2}{3}$

        =>   $I = 2.7571 \ kg m^2$

                      $h = 0.5 * L$

                       $h = 0.5 * 0.8784$

                     $h = 0.4392m$

From the equation above

      $T = 2 \pi \sqrt{\frac{2.7571}{10 .72 * 9.81 * 0.4392} }$

       $T = 1,534\ sec$