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.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg