Angular acceleration in terms of angular velocity

[Flipmeister]

I've been working on problems that deal with pendulums and I've noticed that a few of my answers require me to find the angular velocity, frequency, period of a pendulum. I managed to get the answer right every time, but there's a step that I didn't understand, namely converting angular acceleration to angular velocity by what seems to be using this relationship:

\alpha=\omega^2

Where in the heck does this come from? The book kind of skips this step in the examples and doesn't prove it at all, even though it proves just about every other equation it mentions.

 

[AJ Bentley]

That looks like a conversion factor for radial acceleration.

F=m * v²/r = m * a(rad)

so a(rad) is proportional to w².

You'd need to show the example in detail for anyone to say more than that.

 

[Flipmeister]

Here's how the book derives $w=\sqrt{\frac{Mgl}{I}}$, the angular frequency of a physical pendulum:

\tau=-Mgd=Mglsin\theta

If we restrict the angle to being small (<10 degrees), as we did for the simple pendulum, we can use the small-angle approximation to write

\tau=-Mgl\theta (eq 14.49)

From Chapter 12, Newton's second law for rotational motion is $\alpha=\frac{d^2\theta}{dt^2}=\frac{\tau}{I}$ where I is the object's moment of inertia about the pivot point. Using eq. 14.49, we find

\frac{d^2\theta}{dt^2}=\frac{-Mgl}{I}\theta

Comparison with eq. 14.32 ($\frac{d^2x}{dt^2}=-\frac{k}{m}x$) shows that this is again the SHM equation of motion, this time with angular frequency

\omega=2\pi f=\sqrt\frac{Mgl}{I}

Something happens to get to that last step that doesn't add up for me... Why is Mgl/I square rooted?

 

[AJ Bentley]

It's justified here by comparison to an earlier expression for SHM Eq 14:32. In that section you will probably find a proof that f=1/2*pi sqrt(k/m)

There's a fair explanation in http://en.wikipedia.org/wiki/Simple_harmonic_motion

It's just that the solution of that differential equation is a cosine with √k/m (you can prove that to yourself by substituting it back)