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Rolling problems usually split the motion into two parts: the centre moves forward and the body spins about the centre. That works, but adding velocity vectors point by point is slow and error-prone. There is a shortcut. At any instant a rolling body is purely rotating about the point touching the ground. This post explains why, uses it to find velocities and kinetic energy in one line, applies it to a sliding ladder, and marks where the shortcut breaks.
What is the instantaneous axis of rotation?
When a wheel rolls without slipping, its centre moves at v and it spins at ω = v/R. The bottom point moves backward at Rω relative to the centre, and the centre moves forward at v, so the bottom point's velocity is v − Rω = 0. OpenStax University Physics derives exactly this and notes that the bottom of a rolling tyre is at rest relative to the road.
A rigid body with one point at rest is, at that instant, turning about that point. So every point moves as if the wheel were hinged at the contact point P:
Speed = ω × (distance from P).
Direction = perpendicular to the line joining the point to P.
The JEE (Advanced) 2026 brochure lists "rolling without slipping of rings, cylinders and spheres". Check your exam year's bulletin.
Worked example 1: speeds of points on a rolling wheel
A wheel of radius 0.30 m rolls without slipping at 3.0 m/s, so ω = 3.0/0.30 = 10 rad/s.
Top point. Distance from P is 2R = 0.60 m, so speed = 10 × 0.60 = 6.0 m/s, twice the centre's speed.
Front point level with the centre. Distance from P is √2 × R ≈ 0.42 m, so speed = 3√2 ≈ 4.24 m/s, directed 45° below the horizontal, forward. Check by vectors: forward 3.0 m/s from the centre's motion plus downward 3.0 m/s from the spin gives the same answer.
General point at angle φ around the rim from P: distance 2R sin(φ/2), so speed 2v sin(φ/2).
Worked example 2: kinetic energy in one term
Since the body rotates about P, KE = ½I_Pω², and the parallel axis theorem gives I_P = I_cm + MR².
Take a uniform disc of mass 2.0 kg with the same radius and speed. I_cm = ½MR², so I_P = 1.5 × 2.0 × 0.09 = 0.27 kg m². KE = ½ × 0.27 × 10² = 13.5 J.
Check the long way: ½Mv² + ½I_cmω² = ½ × 2.0 × 9 + ½ × 0.09 × 100 = 9 + 4.5 = 13.5 J. Same answer, one step instead of two.
The same idea gives the acceleration down an incline in one line. Gravity's torque about P is MgR sin θ, so MgR sin θ = I_Pα, and a = g sin θ/(1 + I_cm/MR²). Our post on rolling without slipping reaches this result through forces and friction.
Worked example 3: a ladder sliding down a wall
The axis does not have to be on the body. A 2.0 m rod leans against a smooth wall with its foot on a smooth floor. The rod makes an angle with the floor whose sine is 0.6 (about 37°), so the foot is 1.6 m from the wall and the top is 1.2 m up. The foot slides away at 1.5 m/s. How fast is the top falling?
The foot moves horizontally, so the axis lies on the vertical line through the foot. The top moves vertically, so the axis lies on the horizontal line through the top. They meet at a point I, 1.2 m above the foot and 1.6 m out from the top.
ω = 1.5/1.2 = 1.25 rad/s.
Top speed = 1.25 × 1.6 = 2.0 m/s, downward.
Midpoint speed: I is the far corner of a rectangle whose diagonal is the rod, so the midpoint is L/2 = 1.0 m from I, and moves at 1.25 m/s.
Check with the constraint x² + y² = L²: differentiating, x(dx/dt) = −y(dy/dt), so 1.6 × 1.5 = 1.2 × v_top, giving 2.0 m/s.
Why do students find rolling hard?
Rimoldini and Singh (2005) compared introductory students with physics juniors who had studied rolling in a higher mechanics course. All groups showed similar difficulties, and greater mathematical sophistication did not seem to help. A picture of the contact point at rest is one way to make the motion concrete.
Which mistakes cost marks?
Saying the bottom of a rolling wheel moves at v. It is momentarily at rest.
Writing KE = ½I_cmω² alone for a rolling body. That misses the translation.
Using the instantaneous axis to find accelerations of points. Its own acceleration is not zero.
Applying it to a wheel that is slipping. Then the contact point is moving.
A practice routine
For a rolling ring, find the speeds of the top, front and a point 60° from the bottom.
Compute KE of a rolling ring, disc and sphere at the same speed using I_P only.
Solve the sliding ladder for three different angles and check each with the constraint.
Write one sentence on why friction does no work in rolling without slipping.
Rolling is easier to picture when you can watch the contact point. The Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
Take a coin, roll it slowly along a ruler, and watch which point stays still at each instant.
