Rolling without slipping: friction, forces and energy

1. State the model before choosing an equation

We model a rigid circular body on a stationary, straight surface. Its centre of mass is at its geometric centre; its radius R and mass distribution are constant. The rotation axis passes through the centre and is perpendicular to the plane of motion. Neglect deformation, air drag and rolling resistance. The normal force acts through the centre, so it produces no torque about that centre.

Write the central moment of inertia as $I_{\mathrm{CM}}=kmR^2$. The dimensionless shape factor k is 1/2 for a uniform solid cylinder or disk about its symmetry axis, 2/5 for a uniform solid sphere, and 1 for a thin hoop. These values are not interchangeable: a hollow cylinder with thick walls needs its own inertia.

You need a free-body diagram, force components, $\sum F=ma$, $\sum\tau_{\mathrm{CM}}=I_{\mathrm{CM}}\alpha$, and kinetic energy. Review torque and lever arms or rotational inertia and pulley tensions first if needed. Pulley no-slip relates a rope to a rotating rim; rolling no-slip relates the body’s contact point to the ground.

In each calculation, positive translation is down the slope or to the right. Positive rotation is the corresponding rolling direction, clockwise for rightward rolling or a surface descending to the right. Thus the signed rolling relation is $v=R\omega$. If counterclockwise were chosen positive instead, a right-moving wheel would have $v=-R\omega$. Choose one convention and retain it.

2. A moving centre can have an instantaneously stationary contact point

The contact point’s velocity along the surface is the sum of translation and rotation:

Rolling without slipping on this stationary surface means $v_{\mathrm{contact}}=0$. Therefore, while the constraint holds,

The acceleration here is the centre’s acceleration along the straight surface, not the total acceleration of the material point at contact. That material point can have a nonzero acceleration even when its instantaneous velocity is zero.

Static friction is a response, not a preset value. First solve for the force needed by no-slip motion. Then test $|f_s|\leq\mu_sN$. Equality describes the limiting case, not every rolling body. If the calculated requirement exceeds the limit, discard the no-slip solution. Use a slipping model with kinetic friction and calculate v and ω independently.

Friction opposes relative sliding, or its tendency, at the contact. It need not oppose the centre’s velocity. A freely rolling ideal wheel on a level surface at constant speed needs no friction at all. A wheel driven by an axle torque can need forward static friction. The direction follows from the applied forces and torques, not the word “rolling.”

Where the energy goes

For no slip, substitution gives

The translation fraction is $1/(1+k)$; the rotation fraction is $k/(1+k)$. Both energies belong to the same body. Do not add a separate $\tfrac12I_{\mathrm{contact}}\omega^2$ to this expression: the parallel-axis form would already include translation.

In our ideal no-slip model on stationary ground, static friction supplies zero net instantaneous power to the rigid body because its point of application has zero velocity. It can nevertheless change the translation–rotation split: its contribution to the centre-of-mass kinetic-energy equation is balanced by its rotational contribution. This is not a claim that static friction never does work in other settings, such as a moving belt.

3. Derive the incline result and check its permission to apply

Release a body from rest on a slope of angle θ measured from the horizontal, with $0<\theta<90^\circ$. Gravity acts at the centre; its component along the slope is $mg\sin\theta$. With no friction, gravity would accelerate the centre but would provide no torque about it: the contact would tend to slide downhill. Static friction therefore acts uphill and supplies the torque that increases the downhill rolling rotation.

Let f be the nonnegative uphill friction magnitude. With positive translation downhill and positive rotation in the downhill rolling direction,

Apply $\alpha=a/R$ and $I_{\mathrm{CM}}=kmR^2$. The torque equation becomes $f=kma$. Substitution into translation yields

The no-slip solution exists in the static-friction model only if

Mass cancels from this threshold, but shape and slope do not. A larger available coefficient does not increase a once there is enough grip; it increases the margin before slipping.

For a vertical descent h from rest, ideal mechanical-energy conservation gives

This is a faster route to speed, not to the required friction. It is valid only after the rolling condition is justified, with no dissipative losses or additional energy input. The distance along the slope is $s=h/\sin\theta$, not h. Checking $v^2=2as$ reproduces the energy result.

Worked example 1: a solid cylinder descends

A uniform solid cylinder has $m=3.00\,\mathrm{kg}$, $R=0.200\,\mathrm m$ and $k=1/2$. It starts from rest on a 20.0° slope with $\mu_s=0.200$. Its centre descends vertically by $h=0.800\,\mathrm m$. Find the acceleration, friction and final speed. Use $g=9.81\,\mathrm{m/s^2}$.

Step 1: solve the force and torque equations

The acceleration is downhill; friction is uphill. The angular acceleration is $\alpha=a/R=11.18\,\mathrm{rad/s^2}$ in the downhill rolling direction. The friction torque about the centre increases the spin even though the friction force reduces the centre’s acceleration relative to a frictionless sliding object.

Step 2: verify static friction can provide that force

Since 3.355 N is less than 5.531 N, no-slip rolling is feasible. Equivalently, $\mu_{s,\min}=\tfrac13\tan20^\circ=0.1213$. The actual friction is 3.355 N, not 5.531 N.

Step 3: use both kinetic-energy terms

The lost potential energy is 23.544 J. Translation receives 15.696 J, rotation receives 7.848 J, and their sum is 23.544 J. A calculation using $mgh=\tfrac12mv^2$ alone would predict too large a speed.

Step 4: check with the distance along the slope

Here $s=0.800/\sin20^\circ=2.339\,\mathrm m$. The force result gives $v=\sqrt{2as}=3.235\,\mathrm{m/s}$, agreeing with energy. Radius affects ω and α, but cancels from this model’s a and v when k is fixed.

Worked example 2: a horizontal pull through the centre

A uniform solid cylinder of mass 4.00 kg and radius 0.250 m starts at rest on level ground. A constant horizontal force of 12.0 N acts through its centre to the right, without applying a couple. Take $\mu_s=0.150$. Find its motion and its speed after the centre travels 2.00 m.

The central pull has zero torque about the centre. Without friction it would translate the cylinder without spinning it, making the bottom tend to slide right. Friction must act left to produce clockwise rotation. Let f denote its leftward magnitude.

Thus $f=kma$, giving

Now check: $N=mg=39.24\,\mathrm N$, so $\mu_sN=5.886\,\mathrm N$. The required 4.00 N is available. The minimum coefficient is $4.00/39.24=0.1019$.

The applied force does work $Fs=(12.0)(2.00)=24.0\,\mathrm J$. Static friction has zero net work in this ideal model. Starting from rest,

The translation energy is 16.0 J and rotation energy is 8.0 J. The check $v^2=2as=8.00\,\mathrm{m^2/s^2}$ agrees.

Why “friction always points backward” is not a rule

Replace the central pull by a clockwise driving couple of magnitude τ, with no horizontal applied force. The only horizontal force is static friction. In this different model, it points right: $f=ma$ and $\tau-fR=I_{\mathrm{CM}}\alpha$. With no slip, $a=\tau/[mR(1+k)]$. Friction gives forward translation and opposes part of the driving torque. Its required magnitude still must satisfy $f\leq\mu_smg$. Direction depends on how the wheel is driven.

Four graduated problems

Attempt each setup before opening its solution. Use $g=9.81\,\mathrm{m/s^2}$ and the ideal rigid-body assumptions above.

A final decision checklist

1. State the inertia about the centre, the contact model and the positive directions.
2. Use translation and torque to find the friction required by no slip.
3. Check the static bound before accepting the no-slip result or using its energy formula.
4. Use both kinetic-energy terms; distinguish vertical height from slope distance.
5. If slipping occurs, determine its direction and solve translation and rotation separately.

Scope and references

This original lesson supplements the rotational-inertia lesson and work and energy. For another treatment of the physical model, see OpenStax University Physics, Rolling Motion. Course topic coverage and assessment requirements must be checked against the student’s current section outline; a legacy mechanics guide is not evidence of a current-term requirement.

Vanier’s legacy NYA rotation study guide includes rolling as enrichment. That supports offering supplemental practice, not declaring it a current SN1 requirement. Use this topic only when it matches your current section.