Formula for maximum angular frequency and velocity

[MenchiKatsu]

TL;DR

I can't seem to find it anywhere

What does this formula mean ? I can't find it anywhere. Vmax=Wmax x L. And Wmax=Wθmax. It came up in the pendulum chapter. L is string length.

 

[kuruman]

I can imagine that Vmax=Wmax x L is actually $V_{\text{max}}=\omega_{\text{max}}L$ where
$V_{\text{max}}=~$ maximum speed of the pendulum bob
$\omega_{\text{max}}=~$ maximum angular speed of the pendulum relative to the point of support
$L=~$ the length of the pendulum.

I cannot imagine what Wmax=Wθmax could be.

 

[jbriggs444]

And Wmax=Wθmax. It came up in the pendulum chapter. L is string length.

$$\omega_\text{max} = \omega \theta_\text{max}$$I will make a guess at this one. The notation is confusing because there are two different omegas ($\omega$) being considered here.

On the left hand side of the equality we have $\omega_\text{max}$ which is the rotation rate (in radians per unit of time) for the pendulum at the bottom of its swing.

On the right hand side of the equality we have $\omega$ which is the angular frequency of the oscillations of the pendulum.

Finally we have $\theta_\text{max}$ which is the maximum deflection angle of the pendulum from the vertical.

For instance...

If we have a one meter pendulum under earth gravity, its period is about two seconds. This is an angular frequency ($\omega$) of about $2 \pi$ radians per $2$ seconds. Or about one radian per second.

If this pendulum is launched from rest at a deflection angle ($\theta_\text{max}$) of $0.1$ radians from the vertical, the equation asserts that the rotation rate ($\omega_\text{max}$) of the pendulum at the bottom of its arc will be $0.1$ radians per second.

 

[kuruman]

I will make a guess at this one. The notation is confusing because there are two different omegas (ω) being considered here.

I agree. It should be something like $$\omega_{\text{max}}=\theta_{\text{max}}\sqrt{\frac{g}{L}}$$ where $\omega$ is a reserved symbol for the angular speed of the pendulum bob.