6.5 Rolling

Learning objectives

6.5A—Describe the kinetic energy of a system that has translational and rotational motionDescribe the kinetic energy of a system that has translational and rotational motion.• The total kinetic energy of a system is the sum of the system’s translational and rotational kinetic energies. Relevant equation:. KKtott =+ rans Krot 6.5B—Describe the motion of a system that is rolling without slippingDescribe the motion of a system that is rolling without slipping.• While rolling without slipping, the translational motion of a system’s center of mass is related to the rotational motion of the system itself with the equations:• For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.6.5C—Describe the motion of a system that is rolling while slipping. BOUNDARY STATEMENT Rolling friction is beyond…Describe the motion of a system that is rolling while slipping. BOUNDARY STATEMENT Rolling friction is beyond the scope of AP Physics 1. BOUNDARY STATEMENT The precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping. AP Physics 1: Algebra-Based Course and Exam Description Energy and Momentum of Rotating Systems UNIT 6 TOPIC 6.6 Motion of Orbiting Satellites• When slipping, the motion of a system’s center of mass and the system’s rotational motion cannot be directly related.• When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.

Recognize both motions

A rolling rigid system can translate through the motion of its center of mass and rotate about its center of mass at the same time. Its total kinetic energy includes both motions.

Add the energy contributions

$K_{\mathrm{tot}}=K_{\mathrm{trans}}+K_{\mathrm{rot}}=\frac12 Mv_{\mathrm{cm}}^2+\frac12 I_{\mathrm{cm}}\omega^2$

Contribution	Quantities used
Translation	Total mass $M$ and center-of-mass speed $v_{\mathrm{cm}}$
Rotation	Inertia $I_{\mathrm{cm}}$ about the center of mass and angular speed $\omega$

Calculate each part before totaling

Example: $M=2.0\,\text{kg}$ , $v_{\mathrm{cm}}=3.0\,\text{m}\,\text{s}^{-1}$ , $I_{\mathrm{cm}}=0.18\,\text{kg}\,\text{m}^2$ , and $\omega=10\,\text{rad}\,\text{s}^{-1}$ .

$K_{\mathrm{trans}}=\tfrac12(2.0)(3.0)^2=9.0\,\text{J}$

$K_{\mathrm{rot}}=\tfrac12(0.18)(10)^2=9.0\,\text{J}$

$K_{\mathrm{tot}}=9.0+9.0=18\,\text{J}$ . The energy is split equally in this case; that is not a general rule.

Keep the decomposition consistent

Do not count only $\tfrac12Mv^2$ or only $\tfrac12I\omega^2$ when the system both translates and rotates. Use the center-of-mass axis consistently so the two terms form one valid decomposition.

Start from the contact condition

Rolling without slipping means the point touching the surface is instantaneously at rest relative to that surface. Translation and rotation are then locked by the rolling radius $r$ .

Match linear and angular quantities

Center-of-mass quantity	Rotational quantity	No-slip relationship
Displacement $\Delta x_{\mathrm{cm}}$	Angular displacement $\Delta\theta$	$\Delta x_{\mathrm{cm}}=r\Delta\theta$
Speed $v_{\mathrm{cm}}$	Angular speed $\omega$	$v_{\mathrm{cm}}=r\omega$
Acceleration $a_{\mathrm{cm}}$	Angular acceleration $\alpha$	$a_{\mathrm{cm}}=r\alpha$

Calculate center-of-mass speed

Example: a wheel of radius $0.25\,\text{m}$ rolls without slipping at $\omega=8.0\,\text{rad}\,\text{s}^{-1}$ .

$v_{\mathrm{cm}}=r\omega=(0.25)(8.0)=2.0\,\text{m}\,\text{s}^{-1}$ .

Interpret ideal friction

In the ideal no-slip case, the contact point has no displacement relative to the surface. Static friction may still set the required translation and rotation, but it does not dissipate mechanical energy at that contact.

Apply the equations only without slip

These relationships apply only while there is no slipping. Use the same physical radius and one consistent sign convention; the equations shown here express magnitudes when direction is already understood.

Recognize that the coupling has failed

While an object is slipping, its contact point moves relative to the surface. The no-slip coupling fails, so $v_{\mathrm{cm}}$ and $r\omega$ cannot be set equal.

Compare translation-too-fast and spin-too-fast

Forward-rolling state	Contact point slips	Kinetic friction on object	Qualitative change
$v_{\mathrm{cm}}>r\omega$	Forward	Backward	$v_{\mathrm{cm}}$ decreases; spin rate increases
$v_{\mathrm{cm}}<r\omega$	Backward	Forward	$v_{\mathrm{cm}}$ increases; spin rate decreases

Follow force and torque together

Kinetic friction opposes the relative slipping at the contact point. Its force changes the center-of-mass motion, and its torque changes the rotation, tending to reduce the mismatch between $v_{\mathrm{cm}}$ and $r\omega$ .

Account for dissipation

Because the point where kinetic friction acts moves relative to the surface, mechanical energy is dissipated. Momentum and angular-momentum changes must still be analyzed using the chosen system and external forces or torques.

Keep the analysis qualitative

AP Physics 1 expects a qualitative explanation of linear and angular changes while slipping, not a precise general mathematical relationship between them. Rolling friction is also outside this Topic's scope.

Next Topic

—6.6 Motion of Orbiting SatellitesRead