Rolling and Slipping Problems
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Direct answer
A body rolls without slipping when the contact point is instantaneously at rest, which ties translation to rotation through v = ωR. Rolling down a rough incline therefore has acceleration a = g sin θ/(1 + I/mR²), giving (1/2)g sin θ for a ring, (2/3)g sin θ for a disc and (5/7)g sin θ for a solid sphere — so a sphere beats a disc beats a ring down any incline, independent of mass and radius. The friction that produces the spin is static, does no work, and points up the incline while rolling. When v and ωR do not match, the body slips, kinetic friction acts on the contact, and the motion converts between sliding and rolling until v = ωR is reached.
What you must remember
- Rolling constraint: v_cm = ωR at every instant; differentiating gives a_cm = αR, the link between the two equations of motion.
- Incline acceleration: a = g sin θ/(1 + I/mR²); ring 0.50 g sin θ, disc 0.67 g sin θ, solid sphere 0.71 g sin θ, hollow sphere 0.60 g sin θ — the smaller I/mR², the faster the descent.
- Energy split while rolling: kinetic energy divides as translational : rotational = mR² : I; a ring stores 50 percent in spin, a disc 33 percent, a solid sphere 29 percent, so mgh = ½mv²(1 + I/mR²).
- Minimum friction for rolling on an incline: μ ≥ (I/mR²) tan θ/(1 + I/mR²); that is tan θ/2 for a ring, tan θ/3 for a disc, (2/7) tan θ for a solid sphere.
- Friction direction: while rolling down, friction acts up the incline (it must supply the torque that spins the body); it is static and does zero work because the contact point is at rest.
- During slipping: kinetic friction μmg acts opposite the relative slip at the contact; it simultaneously changes v linearly and ω linearly, and pure rolling begins when v = ωR.
- Rolling bodies on a horizontal surface move with constant v and ω (no friction needed in the ideal case) — friction appears only when an external torque or force breaks the constraint.
A billiard ball learns to roll
Strike a solid sphere (I = 2mR²/5) through its centre so it starts with v0 and ω0 = 0 on a rough table. Kinetic friction f = μmg acts backward: translation decelerates as v = v0 − μgt, while the same friction's torque μmgR spins it up, ω = (f R/I) t = (5μg/2R) t. Rolling begins when v = ωR, that is v0 − μgt = (5/2)μgt, giving t = 2v0/(7μg) and a final speed v = v0 − μg × 2v0/(7μg) = 5v0/7. Notice the elegant result: the final rolling speed is 5/7 of the launch speed whatever μ is — friction only sets how fast the transition happens, not where it lands. The energy difference, ½mv0² minus ½(7/10)m(5v0/7)², has gone to heat at the sliding contact.
Contrast this with a ball already rolling down a 30° incline. Here friction is static and self-adjusting; the two equations mg sin θ − f = ma and fR = Ia/R solve to f = (I/(I + mR²)) mg sin θ = (2/7) mg sin θ for the sphere, pointing up the slope. Because the contact point is instantaneously at rest, the friction force acts through zero displacement and does no work — mechanical energy is conserved even though friction is present, which is why the energy method mgh = ½mv²(1 + I/mR²) reproduces the acceleration instantly.
Where the exam probes
JEE Main likes the ranking question (which body reaches the bottom first) and the direct a = g sin θ/(1 + I/mR²) substitution. JEE Advanced builds the slide-to-roll transition: a cylinder given initial spin but no translation, a spool pulled by a string, or asks for the friction direction when v_cm exceeds ωR (friction then acts forward, slowing spin and speeding translation). The evergreen trap is "friction opposes motion" — on a rolling body friction opposes relative slipping at the contact, not the motion of the centre of mass, and on a freely rolling wheel on level ground ideal friction is simply absent. A second trap: using v = ωR with the diameter, or I about the centre when the situation demands the instant centre at the contact point (I_contact = I_cm + mR²).
Frequently asked questions
What is the condition for rolling without slipping?
The velocity of the contact point must be zero, which requires v_cm = ωR at every instant, and equivalently a_cm = αR while accelerating.
Why does friction do no work in pure rolling?
The point of contact is instantaneously at rest, so the static friction force acts through zero displacement even while it supplies torque.
Which body wins a race down an incline?
The one with the smallest I/mR²: a solid sphere edges out a hollow sphere, both beat a disc, and a ring finishes last — mass and radius are irrelevant.
What is the minimum coefficient of friction for rolling down an incline?
μ must be at least (I/mR²) tan θ/(1 + I/mR²), which works out to (2/7) tan θ for a solid sphere and tan θ/2 for a ring.
How does a sliding body transition to rolling?
Kinetic friction decelerates v and builds ω simultaneously until v = ωR; for a solid sphere struck through its centre, pure rolling begins at v = 5v0/7 after time 2v0/(7μg).