Which item gets to the bottom first?

Question

A Hoop (a car tire), and a solid sphere (a bowling ball) and a solid cylinder (a round fence post) all roll down a hill at once.  Which item gets to the bottom first?

m g h = ½ m v2 + ½ I w2    w = vT / r

m g h = ½ m v2 + ½ I vT2 / r2

m g h = ½ m v2 + ½ I vT2 / r2 for Hoop I = m r2

m g h = ½ m v2 + ½ m r2 vT2 / r2

g h = ½ v2 + ½ v2

V = SQRT (g h)

Answer 1

Since the radii cancel out in the end, all radii will be used with the letter r.

You are correct in stating that the moment of inertia for a hoop is:

I = m * r²

It should also be noted that the moment of inertia for a sphere is:

I = 2 / 5 * m * r²

And for a cylinder about its center axis (the rounder part that it will actually roll with) is:

I = 1 / 2 * m * r²

Intuition at this point can at this point may begin to say they will arrive at the bottom of the hill at different times, and it will be proven mathematically soon. This is due in part that some of the energy from the potential energy they all had before going down the hill is being imparted to rotational kinetic energy and not exclusively to translational kinetic energy ( 1 / 2 m v² ). If all of them freely slipped down the hill without spinning and just moved down the hill, they'd all arrive at the same time. Kind of weird to think about.

As you've also stated, all initial energy equals final energy:

m g h = 1 / 2 m v² + 1 / 2 I ω²

Since all obey the rolling without slipping condition: v = r ω ω = v / r

m g h = [ 1 / 2 m + 1 / 2 ( I / r² ) ] * v²

For the "hoop": For the "sphere":

m g h = [ 1 / 2 m + 1 / 2 m ] * v² m g h = [ 1 / 2 m + 1 / 2 ( 2 / 5 m ) ] * v²

m g h = m v² m g h = 7 / 10 m v2

v = 1 * √( g h ) v = √( 10 / 7 * g h ) ≈ 1.195 * √( g h )

For the "cylinder":

m g h = [ 1 / 2 m + 1 / 2 ( 1 / 2 m ) ] * v²

m g h = 3 / 4 m v2

v = √( 4 / 3 * g h ) ≈ 1.155 * √( g h )

1.195 > 1.155 > 1

Thus:

The sphere will make down the fastest, followed by the cylinder, and in last place, the hoop.

I hope this helps!