Pure Rolling | JEE Notes

A rolling body translates with its centre of mass while rotating about that centre. These are simultaneous descriptions of the same motion. The most useful strategy is to keep their equations separate until the contact condition connects them. Throughout this section, the support is stationary and the body is rigid, with outer radius $R$ and $I_{C} = β M R^{2}$ .

The signed convention that prevents most errors

Along a horizontal surface, take rightward $v$ , $a$ , and friction $f$ as positive. Take clockwise $ω$ and $α$ as positive. The contact-point velocity relative to the ground is

u = v - R ω .

Pure rolling means $u = 0$ , hence $v = R ω$ and, while that constraint persists, $a = R α$ . These are signed equations. They work for leftward rolling too: both $v$ and $ω$ are then negative.

Every point has its own velocity

Use the vector relation $𝐯_{P} = 𝐯_{C} + 𝛚 \times 𝐫_{P / C}$ . For a wheel rolling right at speed $v$ , the top moves right at $2 v$ ; the centre moves right at $v$ ; the instantaneous bottom point has zero velocity. The rightmost point moves right and downward, with speed $\sqrt{2} v$ . The leftmost point moves right and upward with the same speed.

The instantaneous contact point can therefore be treated as an instantaneous centre of rotation for velocity and kinetic-energy calculations:

K = \frac{12}{M} v^{2} + {\frac{12}{I}}_{C} ω^{2} = \frac{12}{(} I_{C} + M R^{2}) ω^{2} .

Zero velocity is not zero acceleration. The material point touching the ground changes continuously. Its instantaneous rest does not make it a permanently fixed hinge.

Why the contact point accelerates upward

𝐚_{P} = 𝐚_{C} + 𝛂 \times 𝐫_{P / C} + 𝛚 \times (𝛚 \times 𝐫_{P / C}) .

On a straight horizontal track, the first two terms cancel at the bottom during pure rolling. The remaining centripetal term points toward the centre:

𝐚_{bottom} = \frac{v^{2}}{R} \hat{𝐲} .

This result also holds when the wheel speeds up, provided $a = R α$ . At the top, by comparison, $𝐚_{top} = 2 a \hat{𝐱} - (v^{2} / R) \hat{𝐲}$ . Never obtain a material-point acceleration merely by differentiating the zero velocity of a succession of different contact points.

The cycloid: the path a rim marker actually follows

For a wheel starting with a marked rim point on the ground, let $ϕ$ be its clockwise angular displacement. Pure rolling gives $x_{C} = R ϕ$ . Combining the centre's displacement with the marker's position relative to it produces

Watch the animation: The cycloid and the contact point

x = R (ϕ - \sin ϕ), y = R (1 - \cos ϕ) .

The path is a cycloid, with a cusp whenever the marker touches the ground. For constant $ω$ , differentiating gives $\dot{x} = R ω (1 - \cos ϕ)$ and $\dot{y} = R ω \sin ϕ$ . The cusp has zero instantaneous speed but nonzero upward acceleration. The full curve repeats after the centre advances $2 π R$ .

Example 18 · A point that stops without having zero acceleration

A wheel of radius $0.25 m$ rolls at constant centre speed $2 m s^{- 1}$ . Find its angular speed, the top-point speed, the rightmost-point velocity, and the bottom-point acceleration.

Hint

Add translation and rotation as vectors. For acceleration, retain the centripetal term.

Complete solution

$ω = v / R = 8 r a d s^{- 1}$ , clockwise. The top speed is $2 v = 4 m s^{- 1}$ . At the rightmost point, rotation contributes $2 m s^{- 1}$ downward, so $𝐯 = 2 \hat{𝐱} - 2 \hat{𝐲} m s^{- 1}$ , with magnitude $2 \sqrt{2} m s^{- 1}$ . At the bottom, $𝐚 = (v^{2} / R) \hat{𝐲} = 16 \hat{𝐲} m s^{- 2}$ .

Check: constant centre speed removes tangential acceleration, not the centripetal acceleration of rotating material points.

Rolling down an incline: derive once, adapt everywhere

Take downslope as positive. Gravity accelerates the centre; uphill static friction supplies the torque that spins an initially nonrotating body. Write $f_{s} \geq 0$ for the uphill friction magnitude:

Explore in 3D: Rolling race

M g \sin θ - f_{s} = M a, f_{s} R = I_{C} α, a = R α .

a = \frac{g \sin θ}{1 + β}, f_{s} = \frac{β}{1 + β} M g \sin θ .

These equations assume sufficient static friction. Since $N = M g \cos θ$ , the consistency condition is

μ_{s} \geq \frac{β}{1 + β} \tan θ .

If the condition fails, discard $a = R α$ and solve translation and rotation using kinetic friction. Increasing $μ_{s}$ above the threshold does not increase the rolling acceleration: static friction takes the required value, rather than automatically equalling $μ_{s} N$ .

Body about symmetry axis	$β$	$a / (g \sin θ)$	Minimum $μ_{s}$
Thin ring / hoop	$1$	$1 / 2$	$\frac{12}{} θ$
Uniform solid cylinder / disc	$1 / 2$	$2 / 3$	$\frac{13}{} θ$
Uniform solid sphere	$2 / 5$	$5 / 7$	$\frac{27}{} θ$
Thin spherical shell	$2 / 3$	$3 / 5$	$\frac{25}{} θ$

Try the simulator: Rolling race and when rolling fails

Where does the gravitational energy go?

M g h = \frac{12}{M} v^{2} + \frac{12}{β} M R^{2} (v / R)^{2}, v^{2} = \frac{2 g h}{1 + β} .

The rotational fraction of kinetic energy is $β / (1 + β)$ ; the translational fraction is $1 / (1 + β)$ . With identical release conditions and adequate friction, a smaller $β$ means faster descent. Mass and radius cancel when the bodies have the same shape factor.

Why static friction does no work here: the instantaneous contact-point velocity is zero in the stationary-ground frame, so its power is

𝐟 \cdot 𝐯_{c o n t a c t} = 0

. Its translational contribution and rotational contribution cancel. This conclusion assumes ideal rigid rolling on a stationary support; a moving belt, deformation, or rolling resistance changes the energy accounting.

Example 19 · Is the standard rolling answer even allowed?

A solid sphere is released on a $30^{\circ}$ incline with $μ_{s} = 0.20$ . Take $g = 10 m s^{- 2}$ . Determine whether it rolls without slipping, its acceleration, and its speed after a vertical drop of $1.4 m$ .

Hint

Test the friction inequality before using either rolling formula.

Complete solution

For a solid sphere, $β = 2 / 5$ . Required $μ_{s} = (2 / 7) \tan 30^{\circ} = 2 / (7 \sqrt{3}) \approx 0.165 < 0.20$ , so pure rolling is consistent. Then $a = (5 / 7) g \sin 30^{\circ} = 25 / 7 m s^{- 2}$ . Energy gives $v^{2} = 2 (10) (1.4) / (7 / 5) = 20$ , hence $v = 2 \sqrt{5} m s^{- 1}$ .

Check: $f_{s} / N \approx 0.165$ , not $0.20$ . A larger available friction coefficient is unused capacity.

Friction can point along the motion

A horizontal force $F$ acts to the right at height $y$ above the centre. Let $f$ be signed positive rightward. Its clockwise torque is $F y$ ; rightward contact friction produces anticlockwise torque $R f$ . Thus

Watch the animation: Which way does friction point?

Try the simulator: Where friction points

F + f = M a, F y - R f = I_{C} (a / R),

a = \frac{F (1 + y / R)}{M (1 + β)}, f = \frac{F (y / R - β)}{1 + β} .

Force at the centre makes friction point left. Force at the top can make it point right. At $y = β R$ , the applied force alone gives the precise translation-to-rotation ratio for rolling, and no friction is needed. The criterion is the tendency to slip at contact, not the direction of centre motion.

Example 20 · Same cylinder, opposite friction

A $3 k g$ solid cylinder is acted on by a $9 N$ horizontal force. Compare application at its centre and at its top. The cylinder rolls without slipping; use $g = 10 m s^{- 2}$ .

Hint

Use $β = 1 / 2$ , first with $y = 0$ , then with $y = R$ .

Complete solution

At the centre: $a = 9 / [3 (3 / 2)] = 2 m s^{- 2}$ and $f = - 3 N$ , leftward. At the top: $a = 18 / [3 (3 / 2)] = 4 m s^{- 2}$ and $f = + 3 N$ , rightward. Both require $μ_{s} \geq 3 / 30 = 0.10$ .

At centre speed $v$ , force power is $F v$ in the first case and $2 F v$ in the second. This explains the larger energy growth for top application without assigning positive work to stationary-contact friction.

Rolling on a moving surface

$a = R α$ is only true on fixed ground. If the surface accelerates at $A$ , no slip means $A = a + R α$ . A solid cylinder on a pulled plank has $a = A / 3$ , and a solid sphere has $a = \frac{27}{A}$ . Friction is what drives the body forward here, while the body turns backward.

The sweet spot

Strike a ball with a horizontal impulse at height $y$ above its centre. It rolls immediately, with no friction needed, when $y = β R$ . For a billiard ball, that is $\frac{25}{R}$ above the centre, or $\frac{75}{R}$ above the table. Below it, the ball starts by sliding; above it, by over-spinning.

Rolling off a curve

Combine energy with $N = 0$ . A small sphere rolling from rest off the top of a large fixed sphere: $\frac{7}{10} v^{2} = g ρ (1 - \cos θ)$ and $v^{2} = g ρ \cos θ$ , so $\cos θ = \frac{10}{17}$ . A block sliding off would leave at $\cos θ = \frac{23}{}$ , so rolling stays on longer. JEE Advanced 2026 used this idea at a table edge.

Loop-the-loop

With no slip and $r ≪ R$ : minimum release height $h = 2 R + \frac{12}{(} 1 + β) R$ , which is $2.7 R$ for a solid sphere and $3 R$ for a ring.

Rolling: one body, two motions, one constraint

In 60 seconds

The signed convention that prevents most errors

Every point has its own velocity

Why the contact point accelerates upward

The cycloid: the path a rim marker actually follows

Example 18 · A point that stops without having zero acceleration

Rolling down an incline: derive once, adapt everywhere

Where does the gravitational energy go?

Example 19 · Is the standard rolling answer even allowed?

Friction can point along the motion

Example 20 · Same cylinder, opposite friction

Rolling on a moving surface

The sweet spot

Rolling off a curve

Loop-the-loop