Question

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.

Answer 1

Answer:

 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.

Answer 2

The moment of inertia should be I = 2 MR².

Calculation of the moment of inertia:

Since

The radius of the hollow ring ( hoop ) = R

The mass of the hoop = M

Now the mass moment of the inertia should be

Io= M R²

Now

I= Io + m d²

Here ,M= m  ,d =R

So, the equation be

I =  M R² +  M R²

I = 2 MR²

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

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