We have three objects, a solid disk, a ring, and a solid sphere. If we release them from rest at the top of an incline, which object will win the race? Assume the objects roll down the ramp without slipping.

1. The sphere
2. The ring
3. The disk
4. Three-way tie
5. Can't tell - it depends on mass and/or radius.

We can take the winner of this race, if there is one, and race it against a slippery block that slides down the ramp with negligible friction and see which one wins that race.

To analyze the rolling race, let's take an object with a mass M and a radius R, and a moment of inertia of cMR². Consider the free-body diagram of such an object.

If there was no friction the object would slide down the ramp without rotating. Friction opposes this motion, so it must be directed up the slope. It's static friction because the object rolls without slipping.

Because the force of gravity and the normal force pass through the center of the object, the frictional force is the only force producing a torque about the center of the object - that's why the object rotates.

We'll analyze the problem it from two different perspectives.

Perspective 1 - Energy Conservation

Because the object does not slip as it rolls, there is no loss of mechanical energy. As it rolls, the object is experiencing a combination of straight-line and rotational motion. Applying conservation of mechanical energy:

U_i + K_i = U_f + K_f

Take the bottom of the ramp to be the zero for the gravitational potential energy. If the object is released from a height h, the initial potential energy is mgh.

The initial kinetic energy is zero. The final kinetic energy is made up of translational and rotational kinetic energies.

Mgh = ½ Mv² + ½ Iω²

Plugging in I = cMR²:

Mgh = ½ Mv² + ½ cMR² ω²

Multiply both sides by 2, and cancel the mass (so mass doesn't matter):

2gh = v² + cR² ω²

For rolling without slipping, the relationship between the velocity of the center-of-mass and the angular velocity is:

ω = v/R, so:

2gh = v² + cR² v²/R²

The factors of R cancel, so the size of the object doesn't matter.

This gives:

2gh = v² + c v²

Solving for v, v = [2gh/(1+c)]^½

So, the larger the value of c, the smaller the speed is. For our different objects we have:

So, the sphere wins the race.

Perspective 2 - Forces and Torques

The free-body diagram of the object shows two forces parallel to the slope. If the angle of the incline is φ, the component of the force of gravity acting down the slope is mgsin(φ). f_s, the force of static friction, acts up the slope.

Summing forces in this direction, with positive down the slope, gives:

ΣF = Ma

Mgsin(φ) - f_s = Ma

With both the normal force and the force of gravity passing through the center-of-mass of the object, summing torques about the center of mass gives:

Στ = Iα

f_sR = Iα

I = cMR² and, for rolling without slipping, α = a/R

The torque equation becomes:

f_s R= cMR² a/R

The factors of R cancel, leaving:

f_s = cMa

Substituting this into the force equation gives:

Mgsin(φ) - cMa = Ma

The factors of mass cancel. Solving the equation for acceleration gives:

a = gsin(φ)/(1+c)

Remember that, when starting from rest, v = at, so both the acceleration and the velocity of the object are reduced by a factor of 1+c from what they would be if the object simply slid down the ramp with zero friction. This is why the object that slides without friction wins the race against any of the rolling objects.

The bottom line is that, whether you analyze it using energy or with forces and torques, the object with the smallest value of c in I = cMR² wins the race between the rolling objects. The mass and radius don't make any difference - we saw that they canceled out in the equations.