Rolling Motion
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Direct answer
A body rolls without slipping when the contact point is instantaneously at rest, imposing v_cm = R omega (and a_cm = R alpha). The friction involved is static and does no work in ideal rolling; the kinetic energy is KE = (1/2) M v^2 (1 + k^2/R^2), with k the radius of gyration, and the motion may be analysed about the centre of mass or about the instantaneous axis through the contact point.
What you must remember
- Constraint of pure rolling: v_cm = R omega, a_cm = R alpha; the topmost point moves at 2 v_cm while the contact point is at rest.
- Rolling kinetic energy: KE = (1/2) M v^2 (1 + k^2/R^2); k^2 = 2R^2/5 for a solid sphere, R^2/2 for a disc, R^2 for a ring.
- Acceleration down an incline: a = g sin(theta)/(1 + k^2/R^2) — solid sphere (5/7) g sin(theta), disc (2/3) g sin(theta), hollow sphere (3/5) g sin(theta), ring (1/2) g sin(theta).
- The incline race depends only on k^2/R^2, never on mass or radius: sphere first, then disc, then hollow sphere, ring last.
- Friction acts up the incline on a body rolling down (it alone supplies the spin torque) and, being static, does no work — mechanical energy is conserved.
- Minimum mu for pure rolling down an incline: tan(theta)/(1 + R^2/k^2); below it, the body slips and rolls together.
- On a frictionless incline the body slides without rotating, with a = g sin(theta) — larger than any pure-rolling value.
Common confusion
Friction's direction is the perpetual trap. Rolling down, gravity acts at the centre and cannot spin the body, so friction acting up the slope must supply the torque — students mark it downhill because friction "opposes motion". Equally unsettling, this friction does no work: the contact point is instantaneously at rest, so the force acts on a point with zero velocity. Only slipping friction dissipates energy.
Exam-focused takeaway
JEE Main asks the energy split, the incline accelerations and the body ranking as numerical-value questions — dependable marks once v = R omega and the energy formula are automatic. JEE Advanced prefers the reasoning side: friction direction under applied forces, a spool or yo-yo pulled by a string at different angles, a slipping body settling into pure rolling, and the minimum-coefficient condition. The method is uniform: translation and rotation equations about the centre of mass plus the rolling constraint, solved together.
Frequently asked questions
Which body reaches the bottom of an incline first?
The one with the smallest k^2/R^2: a solid sphere beats a disc, which beats a hollow sphere, which beats a ring — mass and radius never enter.
Why does friction point up the incline for a rolling body?
Gravity acting at the centre gives no torque about it; only up-slope friction can produce the torque that increases the angular speed.
Is energy conserved while rolling down an incline?
Yes — static friction does no work in pure rolling, so the loss in potential energy equals the gain in translational plus rotational kinetic energy.
What is the topmost point's speed on a rolling wheel?
Twice the centre's speed, since translation v_cm and rotation R omega = v_cm add in the same horizontal direction there.
What happens on a perfectly frictionless incline?
The body slides with a = g sin(theta) without rotating, because no force can exert a torque about its centre of mass.
What is the instantaneous axis of rotation?
The line through the contact point, about which the body is momentarily in pure rotation with omega = v_cm/R — often the quickest route to its kinetic energy.