Solve the physical pendulum: period of oscillation + example with a thin rod physical pendulum.

We derive the formula for the period of oscillation of a physical pendulum, then we apply the formula to find the period of oscillation of a thin rod hinged about one end.

Derivation of the physical pendulum period of oscillation: we start by viewing the force of gravity attached to the center of mass, because the torque exerted by gravity is computed by acting as if all the mass in the rigid body is located at the center of mass.

Next, we decompose the force of gravity into its radial and tangential components, and we only care about the tangential component, since that's the part exerting a torque on the rigid body.

Now we apply the rotational analog to Newton's second law: tau=I*alpha, where tau is the torque, I is the moment of inertia and alpha is the angular acceleration. We solve for alpha and express it as the second time derivative of angle, and we realize that we're looking at a non-linear second order differential equation.

Now we make the small angle assumption and express sin(theta) as a power series. Provided the angle of oscillation is small, this series can be truncated to get the approximation sin(theta)~theta, which linearizes the differential equation into a guessable form.

We guess the general solution of the differential equation (a linear combination of sines and cosines), then use the fact that period is 2pi divided by the coefficient of t, and we find the period of oscillation of the physical pendulum!

Finally, we apply our formula for the period of a physical pendulum to the case of a thin rod hinged at one end.