Why $\omega = \sqrt{\frac{mgl_{cm}}{I_s}}$ for a physical pendulum?

Question

I'm trying to derive the equations of motion for the physical pendulum, and I'm confused to why $\omega = \sqrt{\frac{mgl_{cm}}{I_s}} $ ? Is it just a definition or can it be derived? I'm pretty sure I heard someone talk about a derivation, but I've had no luck in finding one myself.

My question extends to why $\omega = \sqrt{\frac{k}{m}}$ for a simple pendulum, as I presume the logic behind is the same as in the physical pendulum. My school's textbook derives it from $x(t) = A\sin(\omega t)$, but as far as I know, you need $\omega = \sqrt{\frac{k}{m}}$ to even show (which the book doesn't) that $x(t) = A\sin(\omega t)$ is a solution.

Symbols:

• $I_s$: Moment of inerti about pivot point
• $l_{cm}$: distance from pivot point to center of mass

Answer 1

...you need $\omega = \sqrt{\frac{k}{m}}$ to even show (which the book doesn't) that $x(t) = A\sin(\omega t)$ is a solution.

Actually, you don't. You can go through the whole process of deriving the differential equation that governs a simple (or physical) pendulum, approximating it, and solving it without ever using the symbol $\omega$ or talking about a frequency.

Specifically, if you write the tangential component of Newton's second law for a simple pendulum, you get $$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} = -\frac{g}{\ell}\sin\theta \approx -\frac{g}{\ell}\theta$$ Using standard techniques for solving differential equations, you can find that solutions (to the approximate equation) take the form $$\theta(t) = A\sin\biggl(\sqrt{\frac{g}{\ell}}(t - t_0)\biggr)$$ That tells you that the quantity $\sqrt{\frac{g}{l}}$ is important for the motion of that pendulum. So you can see why it might make sense to invent a name for it. Now, you could call it, say, $Q$ or $\zeta$ or whatever, but if you think about it one step further, you'll notice that the motion of the pendulum is periodic with a period of $T = 2\pi/\sqrt{g/\ell}$, and that that equation bears a striking resemblance to the definition of angular frequency that we use in other contexts, namely $T = 2\pi/\omega$. Therefore it makes sense to use $\omega$ to represent $\sqrt{\frac{g}{\ell}}$ for a simple pendulum, because that quantity is playing the role of an angular frequency for this physical system.

So, in a sense, the equation you're asking about is a definition, but also something that can be derived. What you can derive is the fact that the quantity $\sqrt{\frac{g}{\ell}}$ is important for a simple pendulum, or that $\sqrt{\frac{mg\,l_{cm}}{I_s}}$ is important for a physical pendulum. You can even derive that this quantity is important in the same way that other things we choose to call "angular frequency" are important. But the fact that we choose to call it "angular frequency" and write it as $\omega$ is a definition; one that is a very common convention, but still a definition.

Answer 2

The solution for this kind of differential equation

$${\frac {d^{2}}{d{t}^{2}}}q(t) +{\omega}^{2} q(t) =0$$ is: $$q(t)=A\,\sin(\omega\,t+\varphi)$$

where:

• $\omega$ eigen angular velocity
• $\varphi$ phase
• $A$ Amplitude

the frequency of the sine wave is $f=\frac{\omega}{2\pi}\quad [\text{hz}]$ and the period of the wave is $T=\frac{2\pi}{\omega}\quad [\text{s}]$

Example

I) mass spring unit

$$m\ddot{q}+k\,q=0\quad, \ddot{q}+\frac{k}{m}\,q=0$$ $$\Rightarrow\quad \omega^2=\frac{k}{m}\quad, \omega=\sqrt{\frac{k}{m}}$$

II) Mathematics Pendulum

$$m\,L^2\ddot{q}+m\,g\,L\sin(q)=0$$ with $\sin(q)=q$ $$\ddot{q}+\frac{g}{L}\,q=0$$

$$\Rightarrow\quad \omega^2=\frac{g}{L}\quad, \omega=\sqrt{\frac{g}{L}}$$

III) Physical Pendulum

$$I_s\,\ddot{q}+m\,g\,L_{\text{CM}}\sin(q)=0$$

with $\sin(q)=q$ $$\ddot{q}+\frac{m\,g\,L_{\text{CM}}}{I_s}\,q=0$$

$$\Rightarrow\quad \omega^2=\frac{m\,g\,L_{\text{CM}}}{I_s}\quad, \omega=\sqrt{\frac{m\,g\,L_{\text{CM}}}{I_s}}$$

IV) LC oscillating circuit

$$LC\ddot{q}+q=0$$ $\Rightarrow\quad \omega=\frac{1}{\sqrt{LC}}$