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.

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.