 Rotational Dynamics: Rolling Spheres/Cylinders Printer Friendly Version
Rolling Spheres and Cylinders

Consider the following three diagrams. The first one shows the velocity vectors for an object experiencing only pure translation. The center diagram is for an object that is experiencing only rotation. The final diagram is a combination of the two - both translation and rotation.   Notice how the vectors add together. At the bottom of the object that is both rotating and translating, the contact point is instantaneously at rest. This is why static friction is used to calculate the torque that produces rotational motion. Also notice that the very top point on the wheel is moving with speed 2vCM - faster than any other point on the wheel.

Rotational Dynamics

Now consider an object rolling down an incline plane. The first thing that we are going to do is draw a freebody diagram of the forces acting on this mass and then resolve those forces into their components which act parallel and perpendicular to the plane.   Let's consider the axis of rotation passing through the disk's center of mass, cm. Notice in this case that only the instantaneous static friction force will supply a torque since the lines of action of the other two forces (normal and weight) act through the center of mass and cannot produce a torque. As long as the mass rolls without slipping, we can use the relationships: v = rω and a = rα.

Rotationally,

net τ = ICMα
τ =  fsr
fs = ICM(α/r)

Remember linearly,

net F = ma
mg sinθ - fs = mrα  where a = rα

Simultaneously eliminating fs and solving for α yields:

mg sinθ - ICM(α/r) = mrα
α = g sinθ /(r + ICM/mr)

The moment of inertia for a disk, or solid cylinder (see chart provided below), equals ½mr2. Substituting in this value and simplifying gives us

α = 2g sinθ/(3r)

Since this angular acceleration is uniform, you would be free to use any of the rotational kinematics equations to solve for final angular velocity, time to travel down the incline, or the number of rotations it completes as it rolls along the incline's surface.

Moments of Inertia

Below you will find a chart of the three most popular "rolling objects." Notice that their rotational inertia increases from left to right as the mass distribution gets farther from the axis of rotation that passes through their center of mass.

 solid sphereI = 2/5 MR2 solid cylinder (about central axis) I = 1/2 MR2 thin-walled cylinder/hoop/ring(about central axis) I = MR2    Based on this chart, which object in the following picture would you predict would reach the bottom of the incline first: the solid cylinder or the thin ring? To decide, consider the rotational inertia of each object and how inertia affects motion. Both objects have the same mass and equal diameters.

Using Energy Methods

As a solid sphere rolls without slipping down an incline, its initial gravitational potential energy is being converted into two types of kinetic energy: translational KE and rotational KE. If the sphere were to both roll and slip, then conservation of energy could not be used to determine its velocity at the base of the incline. The slipping would result in kinetic friction doing work on the sphere and dissipating energy in the form of heat.

Remember that when an object rolls down an incline, it's linear (also called translational) kinetic energy is always less than what it would have been had it was only slid down a frictionless incline.

An Application of the Parallel-Axis Theorem

The principle that can unite a rolling object's rotational and translational kinetic energy into one expression of total KE is called the Parallel Axis Theorem.

I P = ICM + Mh2

where
• I P represents the object's moment of inertia from any location, P
• I CM represents the object's moment of inertia about its center of mass
• h represents the perpendicular distance from P to the center of mass

For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface.
 Total KE = ½IPω2 Total KE = ½(ICM + mh2)ω2 Total KE = ½ICMω2 + ½mh2ω2 Total KE = ½ICMω2 + ½m(r2ω2) Total KE = ½ICMω2 + ½mv2 Total KE = KErotational + KEtranslational This procedure can apply to any rolling object - just substitute in its correct moment of inertia. Related Documents