Rolling

Learning Goals

After this lesson, you should be able to:

• describe rolling as simultaneous translation and rotation;
• apply the no-slip relations among displacement, speed, and acceleration;
• find the velocities of the top, center, and contact point of a rolling wheel;
• write total kinetic energy as translational plus rotational energy;
• derive the acceleration of a rigid object rolling down an incline;
• compare rolling objects using the dimensionless ratio \(I/(MR^2)\);
• explain the role and direction of static friction;
• distinguish ideal rolling from slipping without applying no-slip equations incorrectly;
• interpret graphs that test the rolling constraint; and
• design an experiment that checks a rolling model.

1. Two Motions Occur at Once

For a rigid object rolling along a surface, the center of mass has translational kinetic energy and the object has rotational kinetic energy about its center of mass:

The first term depends on the motion of the whole object; the second depends on how its mass is distributed around the rotation axis.

3. Rolling Without Slipping: The Three Constraints

When the surface contact does not slide, the distance traveled by the center of mass equals the arc length rotated through:

Differentiating with respect to time gives the corresponding speed and tangential-acceleration relations:

Use radians for \(\Delta\theta\). These equations are constraints for no-slip rolling, not universal equations for every rotating wheel.

4. Why the Contact Point Has Zero Instantaneous Speed

The velocity of a point on a rolling wheel is the vector sum of the center-of-mass velocity and the point's rotational velocity about the center:

For a wheel rolling right without slipping:

• at the top, the two velocity contributions point right, so \(v_{\mathrm{top}}=2v_{\mathrm{cm}}\);
• at the center, \(v=v_{\mathrm{cm}}\);
• at the bottom, equal opposite contributions cancel, so \(v_{\mathrm{contact}}=0\) at that instant.

The contact point is not the same material point over time; a new point arrives at the surface as the wheel turns.

5. Static Friction Enforces the Constraint

On an incline, the component of gravity through the center of mass produces no torque about the center. Static friction at the contact point supplies the torque that changes \(\omega\). For an object released from rest and rolling down a fixed incline, static friction points up the incline in the ideal model.

On a horizontal surface, an already rolling rigid object can move at constant speed with zero friction if no other horizontal force or dissipative effect acts. Rolling does not automatically mean that a nonzero friction force must be present.

6. Deriving the Incline Acceleration

Choose down the incline as positive. Translation of the center of mass gives

Taking torques about the center gives

Using \(a_{\mathrm{cm}}=R\alpha\),

Substitution produces

This is smaller than \(g\sin\theta\) because some of the gravitational effect increases rotational motion.

7. Use the Shape Factor \(\beta\)

Write the rotational inertia in the form

Then the incline results become

For objects with the same mass and radius, the smaller value of \(\beta\) gives the larger acceleration. The mass and radius cancel from the ideal acceleration expression.

8. Which Object Reaches the Bottom First?

With equal starting conditions and no slipping, the solid sphere arrives first, followed by the disk, then the hoop. The decisive quantity is not mass alone but how far the mass lies from the axis.

9. Energy Derivation for a Vertical Drop

If static friction does not dissipate energy and other losses are negligible, an object released from rest through vertical height \(h\) obeys

Apply \(\omega=v_{\mathrm{cm}}/R\):

The final speed depends on vertical height and mass distribution, not on the detailed shape or angle of an ideal path.

10. Worked Example: Solid Disk Descending a Ramp

A solid disk with \(R=0.25\,\mathrm m\) rolls without slipping from rest through a vertical drop of \(1.80\,\mathrm m\). Use \(I=\frac12MR^2\) and \(g=9.8\,\mathrm{m/s^2}\).

The energy fractions are

Answer: \(v_{\mathrm{cm}}=4.85\,\mathrm{m/s}\), \(\omega=19.4\,\mathrm{rad/s}\); two thirds of the kinetic energy is translational.

11. Compare the Disk with a Hoop

A thin hoop descending through the same height has \(\beta=1\):

The disk is faster because less of its mass lies far from the axis. The total energy at the bottom is the same \(Mgh\) for equal masses, but the hoop stores a larger fraction as rotational kinetic energy.

12. Does Static Friction Do Work?

For ideal rolling without slipping on a stationary rigid surface, the instantaneous contact point has no displacement relative to that surface. Static friction can therefore redirect the energy into translational and rotational parts without dissipating mechanical energy.

Do not generalize this into “static friction never does work” for every system choice or moving surface. The safe statement here is specific to ideal no-slip rolling on the stationary surface used in this model.

13. Rolling Uphill

For an object rolling uphill without slipping, both translational and rotational kinetic energy are converted to gravitational potential energy. If \(I=\beta MR^2\),

A solid disk entering the incline at \(6.0\,\mathrm{m/s}\) reaches

Ignoring the initial rotational energy would underpredict the height.

14. Slipping: Do Not Use the Rolling Constraint

If the surfaces slide relative to each other at contact,

For a wheel moving right, compare \(v_{\mathrm{cm}}\) with \(R\omega\):

• If \(v_{\mathrm{cm}}>R\omega\), the contact point tends to slide forward; kinetic friction acts backward, reducing translation and increasing the appropriate rotational speed.
• If \(R\omega>v_{\mathrm{cm}}\), the contact point tends to slide backward; kinetic friction acts forward, increasing translation and reducing the excess spin.

Kinetic friction converts mechanical energy into thermal energy. AP Physics 1 emphasizes this qualitative comparison; a precise quantitative relation between linear and angular changes during slipping is beyond this topic's required scope.

15. Graphs That Reveal Rolling

A systematic departure from \(v_{\mathrm{cm}}=R\omega\) is evidence of slipping, measurement bias, or an incorrect effective radius.

16. A Reliable Problem-Solving Workflow

1. Identify the object, surface, center-of-mass axis, and whether no slipping is stated or justified.
2. Write the provided or derived rotational inertia about the center of mass.
3. For force questions, use \(\sum F=Ma_{\mathrm{cm}}\), \(\sum\tau_{\mathrm{cm}}=I_{\mathrm{cm}}\alpha\), and \(a_{\mathrm{cm}}=R\alpha\).
4. For height and speed questions with negligible dissipation, use energy plus \(v_{\mathrm{cm}}=R\omega\).
5. For point velocities, add translational and rotational velocity vectors.
6. If slipping occurs, remove the no-slip constraint and determine kinetic-friction direction from relative motion at contact.
7. Check units, signs, limiting behavior, and whether the answer depends reasonably on mass distribution.

17. Investigation: Test the Rolling Model

Use a ramp, disk or cylinder, hoop, meterstick, angle measurement, video analysis, and marked radii.

1. Measure each object's mass, radius, release height, and ramp angle.
2. Release without pushing and verify visually that the contact point does not slide.
3. Track the center-of-mass position and rotation angle over time.
4. Plot \(x_{\mathrm{cm}}\) versus \(\theta\); compare the fitted slope with the measured radius.
5. Determine acceleration from the position data and compare it with \(g\sin\theta/(1+\beta)\).
6. Race objects with different mass distributions from the same height.
7. Compare predicted and measured final speeds using energy conservation.
8. Discuss rolling resistance, surface deformation, camera perspective, release timing, radius uncertainty, and possible slipping.

18. Common Traps and Key Takeaways

• Rolling combines translation and rotation; include both kinetic-energy terms.
• Use \(v_{\mathrm{cm}}=R\omega\) only for no-slip rolling.
• The contact point is instantaneously at rest, while the top point moves at \(2v_{\mathrm{cm}}\).
• Use \(I_{\mathrm{cm}}\) in the center-of-mass energy and torque models.
• Static friction points up the incline for an object released from rest and rolling down under the standard model.
• Static friction may be zero for constant-speed rolling on an ideal horizontal surface.
• Smaller \(I/(MR^2)\) means larger incline acceleration.
• Mass and radius cancel from the ideal acceleration when \(I=\beta MR^2\).
• Static friction does not dissipate energy in ideal rolling on a fixed surface.
• Slipping uses kinetic friction and \(v_{\mathrm{cm}}\ne R\omega\).