Where does ω=sqrt(MgL/I) come from?

[jumbogala]

Homework Statement

I'm trying to find the frequency of a hoop pivoting on an axle. The hoop has mass M and radius R.

Homework Equations

The Attempt at a Solution

Obviously f = ω / 2pi

But apparently ω=sqrt(MgL/I), which you plug into the above. I don't understand where this part comes from, though. L is the center of gravity of the hoop (which is just R).

Any help?

 

[Andrew Mason]

Homework Statement

I'm trying to find the frequency of a hoop pivoting on an axle. The hoop has mass M and radius R.

Homework Equations

The Attempt at a Solution

Obviously f = ω / 2pi

But apparently ω=sqrt(MgL/I), which you plug into the above. I don't understand where this part comes from, though. L is the center of gravity of the hoop (which is just R).

Any help?

You will have to explain the problem better. It sounds like the hoop is oscillating like a pendulum but it is unclear why it would do this if the axle is through the centre of the hoop. Is there a diagram?

AM

 

[jumbogala]

Whoops sorry, the hoop pivots through its edge. There was no diagram given, but if the hoop is like a unit circle, then imagine an axle passing through the hoop at the 90 degree position.

 

[Andrew Mason]

Whoops sorry, the hoop pivots through its edge. There was no diagram given, but if the hoop is like a unit circle, then imagine an axle passing through the hoop at the 90 degree position.

Ok. It is similar to a single bob pendulum with the bob located at the centre of mass (ie the centre of the hoop) BUT with a slight difference.

Analyse the forces the same as you would a simple pendulum, BUT you have to let:

\vec{F}\cdot{\vec{R}} = \vec{\tau} = I\vec{\alpha}

Here is a little more help:

For small angles,

\vec{\tau} = -mgR\sin\theta \approx -mgR\theta = I\alpha

which reduces to:

\alpha = \ddot\theta = -\frac{mgR}{I}\theta

[Hint: Find the general solution to that differential equation ]

AM