📂Classical Mechanics

Physical Pendulum

Definition¹

A physical pendulum refers to a rigid body swinging about a fixed horizontal axis due to gravity.

Physical Pendulum

Pendulum motion is a type of harmonic oscillation. The magnitude of the torque acting on the center of mass is as follows:

$$\begin{align*} N &=\left| \mathbf{r} \times \mathbf{F} \right| \\ &= rF\sin\theta \\ &=lmg \sin\theta \end{align*}$$

Expressing the torque in terms of moment of inertia, we get the following equation of motion from $N=I \dot{\omega}=I\ddot{\theta}$:

$$\begin{align*} && I\ddot{\theta} &= mgl\sin \theta \\ \implies&& I\ddot{\theta} -mgl\sin\theta &=0 \\ \implies && \ddot{\theta} -\frac{mgl}{I}\sin\theta &=0 \end{align*}$$

When $\theta$ is sufficiently small, it becomes $\sin\theta \approx \theta$, so the equation of motion can be written as follows:

$$\ddot{\theta} - \frac{mgl}{I}\theta=0$$

The differential equation above is of the same form as a simple harmonic oscillation, and the solution is as follows:

$$\theta (t) = \theta _{0}\cos (2\pi f_{0}t-\delta)$$

Here, $\theta_{0}$ represents the amplitude, $\delta$ the phase angle, and $f_{0}=\frac{1}{2\pi}\sqrt{\frac{mgl }{I}}$ the frequency. Therefore, the period is the reciprocal of the frequency, as follows:

$$\begin{equation} T_{0}=\frac{1}{f_{0}}=2\pi \sqrt{\frac{I}{mgl}}=2\pi \sqrt{\frac{ k^{2}}{gl}} \label{eq1} \end{equation}$$

In this case, $k$ is the radius of gyration. The result above is the same as the period of a simple pendulum movement with length $\frac{k^{2}}{l}$.

Center of Oscillation

Using the parallel axis theorem, the moment of inertia $I$ can be represented by the moment of inertia $I_{cm}$ about the center of mass.

$$I=I_{cm}+ml^{2}$$

Represented as the radius of gyration, it is as follows:

$$mk^{2}=mk_{cm}^{2}+ml^{2}$$

Simplifying $m$ in the equation above yields the following equation:

$$k^{2}=k_{cm}^{2}+l^{2}$$

Substituting this into the period $\eqref{eq1}$ gives the following:

$$T_{0}=2\pi \sqrt{\frac{k_{cm}^{2} +l^{2}}{gl}}$$

Now consider the situation where the axis of rotation changes from $O$ to $O^{\prime}$. Then, the period around the axis of rotation $O^{\prime}$ can be known as follows:

$$T_{0}^{\prime}=2\pi \sqrt{\frac{k_{cm}^{2}+{l^{\prime}}^{2}}{gl^{\prime}}}$$

Therefore, under the condition $$\frac{k_{cm}^{2} +l^{2}}{l}=\frac{k_{cm}^{2}+{l^{\prime}}^{2}}{l^{\prime}}$$

it can be known that the oscillation periods around the axes $O$ and $O^{\prime}$ are the same. The above formula can be simply expressed as follows:

$$\begin{align*} &&l^{\prime}(k_{cm}^{2}+l^{2}) &= l(k_{cm}^{2}+{l^{\prime}}^{2}) \\ \implies && (l^{\prime}-l)k_{cm}^{2}&=ll^{\prime}(l^{\prime}-l) \\ \implies && k_{cm}^{2}=ll^{\prime} \end{align*}$$

At this time, point $O^{\prime}$ is called the center of oscillation with respect to point $O$. Conversely, point $O$ is the center of oscillation with respect to point $O^{\prime}$.