Calculate the natural period of the swinging motion of a human leg. Treat the leg as a rigid physical pendulum with an axis at the hip joint. Pretend that the mass distribution of the leg can be approximated as two rods joined rigidly end to end. The upper rod (thigh) has a mass of 6.8 kg and a length of 43 cm; the lower rod (shin plus foot) has a mass of 4.1 kg and a length of 46 cm.
let's start finding the center of mass:
$$CM = \frac{6.8*0.215+4.1*0.66}{10.9} =0.38$$
Now let's find the moment of inertia of the object: $$I = I_1 +I_2 $$ the moment of inertia of the two cylinders: note($L_t = L_1+L_2$ ): $$ I_1 = \frac{1}{3}M_1L_1^2$$ $$I_2 = \frac{1}{3}M_2(\frac{L_2}{2})^2 + M(L_t)^2$$
$$I = 1.28$$
now let's find the Period from SHO ($sin\theta = \theta)$ small angle aproximation: $$\tau = I\alpha = (CM)MGsin\theta$$ $$T = 2\pi \sqrt{\frac{I}{(CM)MG}} = 1.1$$
which is not the desired answer, what did I get wrong?
I think you calculated the moment of inertia wrong for the shin-plus-foot incorrectly. The formula for the parallel axis theorem is $I_{new} = I_{cm} + md^2$ where $m$ is the mass of the object and $d$ is the distance from the axis about the center of mass to the new axis. In our case, the new axis is the pivot about the hip, so when we apply the parallel axis theorem we get that $I_{2} = \frac{1}{12} M_2 L_2 ^2 + M_2 (L_2/2 + L_1)^2$, giving us a moment of inertia of $1.858 \ kg\cdot m^2$. Note that $(L_2/2 + L_1)$ is the distance from the center of mass of the foot-plus-shin to the hip pivot. Using this value, you should get an answer of 1.5 seconds.