# Where does ω=sqrt(MgL/I) come from?

## Homework Statement

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

## 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 Helper

## Homework Statement

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

## 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

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
Homework Helper
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

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