Topic 6.5 Rolling: state the rolling-without-slipping constraints on velocity and acceleration, analyze the role of friction in rolling, and apply energy and dynamics methods to rolling bodies.

What this topic is asking

The College Board (Topic 6.5) wants you to state the rolling-without-slipping constraints on velocity and acceleration, to analyze the role of static friction in rolling, and to apply both energy and force-torque methods to rolling bodies. Rolling combines the translational and rotational ideas of the course, and the no-slip constraint is what ties them together.

The rolling-without-slipping constraints

These two constraints are the heart of rolling problems. They couple the translational motion of the center to the rotational motion about the center, so a rolling body has one fewer independent variable than a body that is free to slip. Whenever you write energy or dynamics equations for a rolling body, you use $v = R\omega$ (energy) or $a = R\alpha$ (dynamics) to close the system. Without the no-slip constraint, the body would be sliding and these relations would not hold.

The velocity distribution

A neat way to see rolling is through the velocity of different points. The motion is the superposition of translation at $v$ and rotation at $\omega = v/R$:

• The contact point has the forward translation $v$ exactly cancelled by the backward rim velocity $R\omega = v$, so its velocity is zero.
• The center moves at $v$.
• The top has translation $v$ plus rim velocity $R\omega = v$ in the same direction, so it moves at $2v$.

Because the contact point is momentarily at rest, you can even treat the whole rolling body as rotating instantaneously about that contact point. This picture explains why rolling is so efficient: the part touching the ground is not sliding.

The role of static friction

The friction in rolling without slipping is static, not kinetic, because the contact point does not slide. This has two important consequences. First, static friction supplies the torque that produces the angular acceleration (without it, a body placed on a frictionless incline would slide, not roll). Second, because the contact point has zero velocity, static friction does no work, so mechanical energy is conserved for a body rolling without slipping. This is why you can use energy conservation freely for rolling problems, unlike sliding-with-friction problems where energy is dissipated.

Two methods for a rolling body on an incline

You can analyze a body rolling down an incline two ways. The energy method: $Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2$ with $\omega = v/R$, solved for $v$ at the bottom. The force-torque method: along the incline $Mg\sin\theta - f = Ma$, and about the center $fR = I\alpha$, with $a = R\alpha$, solved for the acceleration $a$. Both give the same physics; the energy method finds speeds and the dynamics method finds accelerations and the friction force. For a solid cylinder the acceleration works out to $a = \tfrac{2}{3}g\sin\theta$, less than the $g\sin\theta$ of a frictionless slider, because some "drive" goes into rotation.

Try this

Q1. A wheel rolls without slipping with its center moving at $4.0$ m/s. State the speed of the top and the contact point. [2 points]

• Cue. Top: $2v = 8.0$ m/s; contact point: zero.

Q2. Explain why mechanical energy is conserved for a body rolling without slipping even though friction acts. [2 points]

• Cue. The friction is static and acts at the contact point, which is instantaneously at rest, so it does no work.