6.2 Physical Pendula

Let’s now analyse the motion of a pendulum that is not a point mass. In this case, the we will look at the torques to find the motion. The weight force acts on the centre of mass, and the lever arm for the torque is the distance from the pivot point to the centre of mass. We call this $h$ , then the torque is

\tau=I\alpha=mgh\sin\theta\approx mgh\theta.

(6.28)

If we write this as a differential equation using the small-angle approximation, we will get equation 6.20 for the pendulum above just without the moment of inertia specified. A physical pendulum therefore also obeys SHM (under the small-angle approximation) with frequency and period

\omega=\sqrt{\frac{mgh}{I}},\quad T=2\pi\sqrt{\frac{I}{mgh}}.

(6.29)

Here we use the small-angle approximation because it allows us to solve for the motion analytically for lots of physical pendula. Of course, if we had some object with a moment of inertia that cannot be expressed analytically, we might as well do away with the small-angle approximation since we are going to have to solve the equation of motion numerically anyway!

Let’s do some examples of physical pendula.

Example 6.3.

Consider a long thin rod swinging about a pivot at one end. If $m$ is the mass of the rod and $L$ is its length, what is its resonant frequency?

As we have seen, the moment of inertia of a thin rod rotating about its end is $\frac{1}{3}mL^{2}$ . The distance between the pivot and the centre of mass is $\frac{L}{2}$ , so by equation 6.29 the resonant frequency is

\omega=\sqrt{\frac{mg\frac{L}{2}}{\frac{1}{3}mL^{2}}}=\sqrt{\frac{3g}{2L}}.

(6.30)

Example 6.4.

Consider a thin hoop pendulum of mass $m$ and radius $R$ . What is its resonant frequency?

The moment of inertia of a thin ring rotating about its centre is $mR^{2}$ , so by the parallel axis theorem the moment of inertia about the pivot is $2mR^{2}$ . Therefore, the resonant frequency is

\omega=\sqrt{\frac{mgR}{2mR^{2}}}=\sqrt{\frac{g}{2R}}.

(6.31)