We release a solid disk, a ring, and a solid sphere from rest at the top of an incline. The objects roll without slipping. Which object will win the race?

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

Consider an object with mass M, radius R, and moment of inertia cMR². We'll analyze the race from two different perspectives.

Perspective 1 - Energy Conservation

Because each object does not slip as it rolls, there is no loss of mechanical energy by friction. Thus:   U_i + K_i = U_f + K_f

Define h = 0 at the bottom of the ramp; thus U_f = 0. If the object is released from a height h, U_i = mgh.

The initial kinetic energy is zero. The final kinetic energy consists of translational and rotational parts:   K_f = ½ Mv² + ½ Iω²

Apply energy conservation:   Mgh = ½ Mv² + ½ Iω²

Using I = c MR² gives:   Mgh =½ Mv² + ½ c MR² ω²
2gh = v² + cR² ω²
Here c is an arbitrary constant.

For rolling without slipping, the relation between center-of-mass velocity and angular velocity is   ω = v/R   so:

2gh

=

v2

+

cR2 v2 R2

The factors of R cancel, so size doesn't matter. We obtain:   2gh = v² + c v²

Solving for v,

v =

(

2 g h 1 + c

)

½

So smaller c gives a larger speed. For our objects we have:

The sphere wins!

Perspective 2 - Forces and Torques

The force of gravity and the normal force pass through the center of the object and produce no torque about the center-of-mass. The frictional force is the only force that produces a torque about the center-of-mass.

Apply DID TASC:

1. Diagram and coordinate system
2. Isolate the system
3. Draw all forces acting
4. Take components
5. Apply F=ma   and/or   τ = I α and constraints, if needed.
6. Solve
7. Check!

Step 3 -- Draw all forces: There are two forces parallel to the slope.
the component of gravity acting downslope mg sin θ;
the static friction force, f_s, up the slope.

Step 5 -- Apply Newton's 2nd Law:   ΣF = Ma    Σ τ = I α

The force equation is (+x = downslope):   Mg sin θ - f_s = Ma
The torque equation about the CM is (+z = out):      f_sR = Iα

For rolling without slipping, α = a/R, and we also use I = c M R² so that 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:   Mg sin θ - cMa = Ma

Solving for the acceleration gives:

a

=

g sin θ 1 + c

Both the acceleration and the velocity are reduced by a factor of 1+c compared to the case of no friction. Thus an object that slides without friction beats any rolling object. For the rolling race, the object with the smallest c in I = cMR² wins, independent of the mass and the radius.

Question: what attribute produces the smallest value of c in I = c M r² for fixed mass and fixed radius?