Check whether rolling without slipping is possible

Check a constraint before using it

A rolling formula can give a plausible number even when the surface cannot support rolling without slipping. Before trusting v = Rω, ask whether the required static friction is within its available range. The investigation shows both quantities on a fixed force scale.

The body starts from rest on a stationary incline, driven only by gravity. Its inertia about the center is I = βmR². Here m = 2 kg, R = 0.5 m, g = 10 m/s² and the static-friction coefficient μs = 0.4. The angle control changes the incline; the inertia-factor control changes mass distribution. Neither control changes the surface coefficient.

Determine the friction direction from the motion

Gravity acts through the center and therefore supplies no torque about that center. Without friction, the body would accelerate downhill without acquiring the rotation required by no-slip motion. The contact would tend to slide downhill. Static friction acts uphill in this passive release, providing the torque that starts the body rotating.

Friction does not always oppose the center’s motion. Its direction responds to the tendency of the surfaces to slip relative to each other. A driven wheel or a pushed wheel can have different force and torque balances. Do not transfer the uphill direction to every rolling problem.

Use force and torque together

Choose downhill as the positive direction for center motion and the corresponding rolling sense as positive rotation. The downslope force equation is mg sin θ − f = ma, where θ is the incline angle and f is the magnitude of uphill static friction. The normal force is N = mg cos θ because there is no acceleration perpendicular to the straight plane.

About the center, friction supplies torque fR = Iα. For no slipping, α = a/R; substituting I = βmR² gives f = βma. Combining this with the force equation produces a = g sin θ/(1+β). We have found the friction required for the assumed motion, not yet proved that it is available.

The condition is f ≤ μsN. Dividing the required force by N gives a minimum coefficient μ_required = [β/(1+β)] tan θ. The OpenStax rolling-motion derivation applies both force and torque equations before testing this inequality.

Static friction is not automatically μsN. That expression is its maximum magnitude under this model. A smaller required force can be supplied without using the full limit. Increasing the available coefficient beyond the threshold does not change the ideal no-slip acceleration for this passive release.

Test a passing case

For a thin hoop, β = 1. At 30 degrees, acceleration is 10 × 0.5/2 = 2.5 m/s² and required uphill friction is 1 × 2 × 2.5 = 5 N. Normal force is 20 cos(30°) ≈ 17.321 N, so the available static-friction limit is 0.4 × 17.321 ≈ 6.928 N. Required friction is below the limit, and the rolling endpoint is supported.

The minimum coefficient is about 0.289, less than the supplied 0.4. Capture this case. The force is 5 N, not 6.928 N; treating the upper limit as the actual force would produce the wrong acceleration.

Find where the rolling model fails

Keep β = 1 and increase the angle through 35, 40 and 45 degrees. At 35 degrees the minimum coefficient is approximately 0.350, so no-slip rolling remains possible. At 40 degrees it is approximately 0.420, exceeding the available 0.4. At 45 degrees it is 0.5.

For the 45-degree hoop, required friction is about 7.071 N while the limit is about 5.657 N. The model withholds its rolling speed, travel time and energy allocation. Those values would rely on an invalid constraint. Insufficient static friction does not stop the body moving; it allows slipping. To calculate its actual subsequent motion, we need a kinetic-friction law and separate translation and rotation equations.

Now keep 45 degrees but reduce β to 0.5. The required coefficient becomes 1/3, below 0.4, and the required force is about 4.714 N. The cylinder model can roll without slipping on the same ideal surface where the hoop fails. This comparison isolates mass distribution rather than attributing the difference to mass or radius.

Why the friction torque does not imply an energy loss

In ideal rigid rolling on a stationary surface, the instantaneous contact point has zero velocity. Static friction therefore does no net work on the whole body. It affects the division between translational and rotational motion while gravity supplies the energy.

Multiplying uphill friction by center displacement alone is not the total friction work on the rigid body. Its contribution to center-of-mass translational energy is balanced by its contribution to rotation. The correct whole-body contact-power calculation uses the contact point’s velocity. Deformation, slipping and rolling resistance would change this energy account; the ideal zero-work statement is not a claim that real tires never dissipate energy.

Transfer to a new surface

A cylinder with β = 0.5 rolls from rest down a 30-degree incline. Surface A has μs = 0.15 and surface B has μs = 0.25. Determine which surface can support no-slip rolling. Explain whether the passing surface’s actual static friction must equal its maximum.

Check the friction transfer

The required coefficient is (0.5/1.5) tan(30°) ≈ 0.192. Surface A fails and surface B passes. On B, friction supplies the required force, which is less than 0.25N; it does not automatically saturate the limit. For A, no actual sliding speed follows without a kinetic-friction model.