Slipping to Rolling | JEE Notes

Suppose a circular body is placed on a rough, horizontal, stationary plane with arbitrary signed initial $v_{0}$ and clockwise-positive $ω_{0}$ . During this transition, gravity and the normal balance; kinetic friction is the only horizontal force. Ignore rolling resistance and take a constant kinetic coefficient $μ_{k} > 0$ .

One signed derivation covers forward spin, backspin, and reversal

Define $Δ = v_{0} - R ω_{0}$ . For $Δ \neq 0$ , let $s = sgn Δ$ . While slipping persists,

f = - s μ_{k} M g, \dot{v} = - s μ_{k} g, \dot{ω} = \frac{s μ_{k} g}{β R} .

u (t) = Δ - s μ_{k} g (1 + 1 / β) t .

The slip reaches zero at

t_{*} = \frac{| Δ |}{μ_{k} g (1 + 1 / β)} .

Substitute this time into $v (t)$ , then apply $v_{f} = R ω_{f}$ :

v_{f} = \frac{v_{0} + β R ω_{0}}{1 + β}, ω_{f} = \frac{v_{f}}{R} .

The final rolling speed is independent of $μ_{k}$ . Friction strength determines how quickly the transition happens. If $Δ = 0$ , the body already rolls and $t_{*} = 0$ . If $μ_{k} = 0$ while $Δ \neq 0$ , no transition occurs.

Watch the animation: Sliding settles into rolling

After $t_{*}$ , kinetic friction must stop. With no other horizontal force or torque, the required static friction is zero and rolling remains uniform. Extending the original constant-friction acceleration past $t_{*}$ invents an unphysical overshoot.

A useful invariant, with a precise reason

Multiply the rotational acceleration equation by $β R$ and add it to the translational equation. The friction terms cancel, giving

\frac{d}{d t} (v + β R ω) = 0 .

Thus $v + β R ω = v_{0} + β R ω_{0}$ throughout the transition. Combining this with the final rolling condition immediately reproduces $v_{f}$ . This is a consequence of the same contact friction producing linked linear and angular impulses. It is not conservation of the body's linear momentum: the ground supplies an external horizontal impulse. Nor is it conservation of kinetic energy: the contact slides and produces heat.

This invariant is especially valuable in a question that asks only the final state. You can find that state before calculating the transition duration. If an extra applied force or torque is present, differentiate the combination again and retain those new contributions; the invariant then generally disappears.

Try the simulator: Slip to roll, any initial state

Energy loss depends on slip, not simply on travel distance

The total frictional power delivered to the rigid body is $f v - R f ω = f u = - μ_{k} M g | u |$ . The graph of $| u |$ against time is triangular. Therefore

Q = K_{i} - K_{f} = \frac{β M}{2 (1 + β)} (v_{0} - R ω_{0})^{2} .

This is nonnegative for every signed initial condition. It vanishes only when the initial state already satisfies rolling. Multiplying friction by centre displacement alone calculates only the change in translational energy, not the total energy loss. Friction can increase translational kinetic energy while decreasing rotational energy by an even larger amount.

Example 21 · A ball projected without spin

A solid sphere of mass $0.50 k g$ and radius $0.10 m$ enters a rough horizontal surface at $7 m s^{- 1}$ without spin. If $μ_{k} = 0.20$ and $g = 10 m s^{- 2}$ , find the transition time, final motion, centre displacement during slipping, and energy lost.

Hint

Here $Δ = 7 m s^{- 1}$ and $β = 2 / 5$ . Do not set $v = R ω$ before the transition ends.

Complete solution

$t_{*} = 7 / [2 (1 + 5 / 2)] = 1 s$ . The final speed is $v_{f} = 7 / (7 / 5) = 5 m s^{- 1}$ , giving clockwise $ω_{f} = 50 r a d s^{- 1}$ . Centre displacement is $x = 7 (1) - \frac{12}{(} 2) (1)^{2} = 6 m$ .

The loss is $Q = [(2 / 5) (0.5) / (2 \cdot 7 / 5)] 7^{2} = 3.5 J$ . Initially $K_{i} = 12.25 J$ ; finally $K_{f} = \frac{12}{(} 0.5) (7 / 5) (5^{2}) = 8.75 J$ , confirming the difference.

Does friction reverse when the centre reverses?

No. Friction follows $- sgn (v - R ω)$ . With no extra horizontal force, this slip keeps its original sign until it becomes zero. The centre may reverse earlier while friction continues in the same direction. In a driven problem, recompute contact slip whenever circumstances change; a reversed slip requires reversed kinetic friction.

Initial condition	Friction during free slipping	Effect
$v_{0} > R ω_{0}$	Left	$v$ decreases; clockwise $ω$ increases
$v_{0} < R ω_{0}$	Right	$v$ increases; clockwise $ω$ decreases
$v_{0} = R ω_{0}$	Zero in this unforced problem	Uniform rolling already established

Example 22 · Strong backspin sends the cylinder back

A solid cylinder has $R = 0.20 m$ , $v_{0} = + 2 m s^{- 1}$ , and $ω_{0} = - 30 r a d s^{- 1}$ . With $μ_{k} = 0.25$ and $g = 10 m s^{- 2}$ , determine when the centre stops, when rolling begins, and the final direction.

Hint

Negative clockwise-positive angular speed means anticlockwise backspin. Stopping the centre and eliminating contact slip are different events.

Complete solution

$Δ = 2 - 0.2 (- 30) = 8 m s^{- 1}$ , so friction points left. During slipping, $v = 2 - 2.5 t$ and $ω = - 30 + 25 t$ . The centre stops at $t = 0.80 s$ , but $u = 8 - 7.5 (0.80) = 2 m s^{- 1}$ , so slipping continues.

Rolling begins at $t_{*} = 8 / 7.5 = 16 / 15 s$ . Then $v_{f} = [2 + (1 / 2) (0.2) (- 30)] / (3 / 2) = - 2 / 3 m s^{- 1}$ and $ω_{f} = - 10 / 3 r a d s^{- 1}$ . The cylinder rolls left. Friction never reversed during the transition.

General reversal test: with $v_{0} > 0$ , eventual leftward rolling requires $β R ω_{0} < - v_{0}$ . Equality makes the body finally come completely to rest.

Why the invariant works

Take angular momentum about any point on the floor. Friction acts along the floor, so it has no moment there, and gravity balances the normal reaction. So $M v R + I ω$ stays constant. Dividing by $M R$ gives the conserved $v + β R ω$ . This is the fastest route when a question asks only for the final state.

Sliding plus spinning: track the contact slip

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