Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

I = 2 MR²

Explanation:

Given that

Radius of the hollow ring (hoop) = R

The mass of the hoop = M

We know that mass moment of inertia of a hoop about its center is given as

Io = M R²

By using theorem, mass moment of inertia at distance d from center is given as

I = Io + m d²

Here, M = m, d = R

Now by putting the values in the above equation we get

I = M R² + M R²

I = 2 MR²

Therefore the mass moment of inertia will be 2 M R².