Rolling Without Slipping

Put a solid ball and a hollow ball of the same mass and the same radius at the top of a ramp and let them go together. They do not tie. The solid one wins, every time, and the gap does not depend on how heavy they are or how big they are. Nothing about the ramp cares which is which: same gravity, same slope, same height. So the difference has to come from inside the balls.

That is the whole subject of this lesson. When something rolls, the energy you give it gets split between moving forward and spinning, and the split is decided by how the mass is arranged. In the previous lesson you treated rotation on its own: a wheel on a fixed axle, spinning in place. Now the axle moves too, and the two motions are locked together by the fact that the object is touching the ground.

The constraint

Roll a wheel of radius $r$ along the floor without letting it slip. Mark the rim with chalk. After one full turn the chalk has laid down the whole circumference onto the floor, once, without stretching or skidding, so the wheel has advanced exactly $2\pi r$. Turn it through a smaller angle $\theta$ (in radians) and it advances the corresponding arc:

$x = r\theta$

Here $x$ is how far the centre has moved and $\theta$ is how far the wheel has turned. Both are measured from the same starting moment. Now differentiate with respect to time. The rate of change of $x$ is the speed of the centre, $v_{cm}$, and the rate of change of $\theta$ is the angular speed $\omega$:

$v_{cm} = \omega r$

Differentiate once more and you get the same link for the accelerations:

$a_{cm} = \alpha r$

where $\alpha$ is the angular acceleration. These two equations are the rolling constraint, and almost every problem here uses one of them. Notice what kind of statement this is. It is not a law of physics. It is geometry, and it is true only while the contact grips. Kick a ball hard along the floor and for a while it slides: the centre is moving faster than $\omega r$ and the constraint is simply false. You will handle that case later in the lesson.

There is a second way to see rolling that is worth as much as the algebra. The point of the object touching the floor is instantaneously at rest. It has to be: if it were moving, it would be sliding. So at each instant the whole body is rotating about the contact point, like a door about its hinge. The centre is a distance $r$ from that pivot, so it moves at $\omega r$. The top of the wheel is a distance $2r$ from it, so the top moves at $2\omega r$, which is twice the speed of the car it belongs to. Look at a photograph of a moving car: the bottom of the wheel is sharp and the top is a blur.

Where the energy goes

A rolling object has two kinds of kinetic energy at once. The centre of mass is moving, which is worth $\tfrac12 M v_{cm}^2$, and the body is spinning about the centre of mass, which is worth $\tfrac12 I\omega^2$, where $I$ is the moment of inertia about the centre. Those add:

$K = \tfrac12 M v_{cm}^2 + \tfrac12 I \omega^2$

On its own that has two unknowns. The constraint fixes it. Write $\omega = v_{cm}/r$ and substitute:

$K = \tfrac12 M v^2 + \tfrac12 I \frac{v^2}{r^2} = \tfrac12 M v^2\left(1 + \frac{I}{Mr^2}\right)$

dropping the subscript on $v$ from here on. The combination $I/Mr^2$ appears constantly, so give it a name. Call it $\beta$:

$\beta = \frac{I}{Mr^2}, \qquad K = \tfrac12 M v^2 (1 + \beta)$

$\beta$ is a pure number that depends only on the shape, because the $M$ and the $r^2$ have been divided out. A hoop or a thin ring has all its mass at radius $r$, so $I = Mr^2$ and $\beta = 1$. A solid disk or cylinder has $I = \tfrac12 Mr^2$, so $\beta = \tfrac12$. A solid sphere has $I = \tfrac25 Mr^2$, so $\beta = \tfrac25$. A hollow spherical shell has $\beta = \tfrac23$. Mass pushed out towards the rim means a bigger $\beta$.

Read the formula in words: a rolling object carries $(1+\beta)$ times as much kinetic energy as a sliding one at the same speed. Equivalently, the fraction of the total that is rotational is

$\frac{K_{rot}}{K} = \frac{\tfrac12 I\omega^2}{\tfrac12 Mv^2(1+\beta)} = \frac{\beta}{1+\beta}$

For a solid sphere that is $(2/5)/(7/5) = 2/7 \approx 0.286$. For a hoop it is $1/2$: exactly half of a rolling hoop's energy is spin.

Example 1. A hoop of mass $2.0\ \mathrm{kg}$ rolls without slipping at $3.0\ \mathrm{m/s}$. Find its total kinetic energy.

A hoop has $\beta = 1$. So

$K = \tfrac12 (2.0)(3.0)^2 (1 + 1) = \tfrac12 (2.0)(9.0)(2) = 18\ \mathrm{J}$

Half of that, $9.0\ \mathrm{J}$, is rotational. Had it been a solid disk ($\beta = \tfrac12$) at the same speed, you would have got $\tfrac12(2.0)(9.0) (1.5) = 13.5\ \mathrm{J}$ instead.

Down a ramp: who wins

Now release an object from rest and let it roll down a slope, dropping a height $h$. Static friction is what makes it roll rather than slide, and here is the key fact about it: the contact point is not moving, and work is force times the displacement of the point where the force acts, so static friction does zero work. No energy is lost. Mechanical energy is conserved, even though friction is essential to the motion.

So all the lost height turns into that two-part kinetic energy:

$Mgh = \tfrac12 M v^2 (1 + \beta)$

The $M$ cancels from both sides, which is why mass never appears in the answer. Solve for $v$:

$v = \sqrt{\frac{2gh}{1+\beta}}$

Compare that with a frictionless slider, which would arrive at $\sqrt{2gh}$. Every rolling object is slower, and the bigger $\beta$ is, the slower it is, because more of the same energy budget has been diverted into spin. That settles the race: sphere ($\beta=0.4$) beats disk ($0.5$) beats shell ($0.667$) beats hoop ($1$), regardless of mass, radius, slope, or height.

Example 2. A solid sphere rolls from rest down a $1.5\ \mathrm{m}$ drop. Find the speed of its centre at the bottom.

$\beta = 2/5$, so $1 + \beta = 1.4$.

$v = \sqrt{\frac{2(9.8)(1.5)}{1.4}} = \sqrt{\frac{29.4}{1.4}} = \sqrt{21.0} = 4.58\ \mathrm{m/s}$

A solid cylinder on the same ramp gets $\sqrt{29.4/1.5} = \sqrt{19.6} = 4.43\ \mathrm{m/s}$. The sphere wins by about three percent.

Because $h$ appears inside the same square root for both, a ratio of arrival speeds loses the ramp entirely:

$\frac{v_1}{v_2} = \sqrt{\frac{1+\beta_2}{1+\beta_1}}$

Solid sphere against hoop: $\sqrt{2/1.4} = \sqrt{1.4286} = 1.195$. Solid sphere against hollow shell: $\sqrt{(5/3)/(7/5)} = \sqrt{1.1905} = 1.091$.

Forces and torques: the acceleration

Energy gives you the speed after a drop but not the acceleration along the way, and it never gives you the friction force. For those, go back to Newton.

Three forces act on a ball rolling down a slope of angle $\theta$: gravity $Mg$ straight down, the normal force $N$ perpendicular to the surface, and static friction $f$ along the surface. Take axes along and perpendicular to the slope. Along the slope, with downhill positive:

$Mg\sin\theta - f = Ma$

Friction points uphill. It has to, because it is the only force with a torque about the centre, and that torque must spin the ball up in the forward direction. Take torques about the centre of mass. Gravity acts at the centre so its lever arm is zero; $N$ points straight at the centre so its torque is zero too. Only $f$ is left, at distance $r$:

$f r = I\alpha = (\beta M r^2)\frac{a}{r}$

using $I = \beta Mr^2$ and the constraint $\alpha = a/r$. One $r$ cancels from each side, leaving

$f = \beta M a$

Put that into the force equation:

$Mg\sin\theta - \beta M a = Ma \quad\Longrightarrow\quad a = \frac{g\sin\theta}{1+\beta}$

The same $1+\beta$ as before, which is the sign you did it right. And now you also have the friction force for free: $f = \beta M a = \beta Mg\sin\theta/ (1+\beta)$.

Example 3. A solid cylinder rolls without slipping down a $30^\circ$ incline. What is the smallest coefficient of static friction that allows it?

First the acceleration, with $\beta = 1/2$:

$a = \frac{9.8 \sin 30^\circ}{1.5} = \frac{9.8(0.500)}{1.5} = 3.267\ \mathrm{m/s^2}$

The friction actually required is

$f = \beta M a = (0.5)M(3.267) = 1.633M$

The most static friction can supply is $\mu_s N$, and perpendicular to the slope $N = Mg\cos\theta = M(9.8)(0.866) = 8.487M$. Rolling is possible only if $f \le \mu_s N$:

$\mu_s \ge \frac{1.633M}{8.487M} = 0.192$

The $M$ cancels, as it must. In symbols the condition is

$\mu_s \ge \frac{\beta\tan\theta}{1+\beta}$

Worth committing to memory, because it is the check that tells you which of the two worlds a problem lives in.

Example 4. A solid sphere is placed on a $40^\circ$ incline with $\mu_s = 0.20$. Find the acceleration of its centre.

Run the check first. With $\beta = 2/5$, rolling needs

$\mu_s \ge \frac{(0.4)\tan 40^\circ}{1.4} = \frac{(0.4)(0.8391)}{1.4} = 0.240$

You only have $0.20$, so the sphere cannot roll without slipping. Throw away the constraint. Now the contact really slides, kinetic friction acts at its full value $\mu_k N$ up the slope, and the translation no longer knows about $\beta$ at all:

$Ma = Mg\sin\theta - \mu_k Mg\cos\theta$ $a = 9.8\left(\sin 40^\circ - 0.20\cos 40^\circ\right) = 9.8\left(0.6428 - 0.20(0.7660)\right)$ $a = 9.8(0.6428 - 0.1532) = 9.8(0.4896) = 4.80\ \mathrm{m/s^2}$

Faster than the $3.15\ \mathrm{m/s^2}$ it would have had while rolling, because less of gravity is being spent spinning it up. It is also losing energy now, which the rolling version never did.

Skid, then roll

Bowl a ball down a lane and it leaves your hand with a forward speed $v_0$ and essentially no spin. The constraint is badly violated: $v_0 > \omega r = 0$, so the contact point is sliding forward. Kinetic friction acts backwards on the ball, at the contact. Two things happen at once. The backward force slows the centre:

$v(t) = v_0 - \mu_k g t$

And the same force, acting at distance $r$ below the centre, exerts a torque that spins the ball up:

$\mu_k M g\, r = I\alpha = \beta M r^2 \alpha \quad\Longrightarrow\quad \alpha = \frac{\mu_k g}{\beta r}$ $\omega(t) r = \frac{\mu_k g}{\beta} t$

One quantity is falling, the other is rising. When they meet, the contact point stops sliding, kinetic friction switches off, and the ball rolls from then on at constant speed. Set them equal:

$v_0 - \mu_k g t = \frac{\mu_k g}{\beta} t \quad\Longrightarrow\quad t = \frac{\beta v_0}{\mu_k g (1+\beta)}$

Substitute back to get the speed at that moment:

$v = v_0 - \mu_k g \cdot \frac{\beta v_0}{\mu_k g(1+\beta)} = v_0\left(1 - \frac{\beta}{1+\beta}\right) = \frac{v_0}{1+\beta}$

Clean, and $\mu_k$ has dropped out. Friction decides when and where rolling starts, never at what speed.

Example 5. A $6.0\ \mathrm{kg}$ bowling ball ($\beta = 2/5$) is launched at $8.0\ \mathrm{m/s}$ with no spin onto a lane with $\mu_k = 0.10$. Find the rolling speed, the distance it slides first, and the energy lost.

Rolling speed:

$v = \frac{8.0}{1.4} = 5.714\ \mathrm{m/s}$

For the distance, use the constant deceleration $a = \mu_k g = (0.10)(9.8) = 0.98\ \mathrm{m/s^2}$ with $v^2 = v_0^2 - 2ad$:

$d = \frac{v_0^2 - v^2}{2a} = \frac{64.0 - 32.653}{2(0.98)} = \frac{31.347}{1.96} = 16.0\ \mathrm{m}$

For the energy, compare start and finish. At the start there is no spin, so $K_i = \tfrac12(6.0)(8.0)^2 = 192.0\ \mathrm{J}$. At the end it rolls, so use the $(1+\beta)$ form:

$K_f = \tfrac12 (6.0)(5.714)^2(1.4) = \tfrac12(6.0)(32.653)(1.4) = 137.14\ \mathrm{J}$

$\Delta K = 192.0 - 137.14 = 54.9\ \mathrm{J}$

lost to heat at the sliding contact. Do not try to conserve energy here: the contact point moves, so this friction does real work.

The mirror image works the same way. Spin a ball up to $\omega_0$ in the air and drop it on the floor with no forward speed. Friction now points forward, pushing the centre up to speed while slowing the spin. Setting $\mu_k g t = \omega_0 r - (\mu_k g/\beta)t$ and solving gives

$v = \frac{\beta}{1+\beta}\,\omega_0 r$

For a solid ball spun at $40\ \mathrm{rad/s}$ with $r = 0.10\ \mathrm{m}$: $v = (0.4/1.4)(40)(0.10) = 1.14\ \mathrm{m/s}$. That is the backspin shot that comes back towards you, halfway through its life.

Two harder shapes

A spool has a cord wound round an inner axle of radius $r$ inside a wheel of outer radius $R$ that touches the ground. Pull the cord at angle $\theta$ above the horizontal and the spool can roll towards you, which is the surprise. Torque about the contact point settles it in one line: the normal force and friction both act there, so only the tension $T$ has a torque about it. Its lever arm is the perpendicular distance from the contact point to the line of the cord, which works out to $R\cos\theta - r$. If that is positive the spool rolls forward; if negative it rolls back; if zero it does not turn at all:

$\cos\theta = \frac{r}{R}$

With $R = 0.20\ \mathrm{m}$ and $r = 0.08\ \mathrm{m}$, $\cos\theta = 0.40$ and $\theta = 66.4^\circ$.

A ball rolling inside a bowl is the rolling version of a pendulum. A ball of radius $r$ inside a bowl of radius $R$ has its centre on a circle of radius $R - r$, so if the centre is at angle $\phi$ from the bottom, its speed is $(R-r)\dot\phi$ and its height above the lowest point is $(R-r)(1-\cos\phi)$. Energy conservation, with the rolling factor:

$\tfrac12 M (1+\beta)(R-r)^2\dot\phi^2 + Mg(R-r)(1-\cos\phi) = \text{constant}$

That is exactly a pendulum's energy equation with $g$ replaced by $g/(1+\beta)$, so the small-oscillation angular frequency is

$\omega_{osc} = \sqrt{\frac{g}{(1+\beta)(R-r)}}$

For a solid ball of $r = 0.020\ \mathrm{m}$ in a bowl of $R = 0.50\ \mathrm{m}$: $\omega_{osc} = \sqrt{9.8/(1.4 \times 0.48)} = \sqrt{14.58} = 3.82\ \mathrm{rad/s}$. Note the symbol clash and keep them apart: $\omega_{osc}$ is the rate of the back-and-forth, not the spin rate of the ball. That $1/(1+\beta)$ slowdown is the first thing the next lesson will ask you to recognise.

Recognising a rolling problem

Three questions sort nearly everything.

Is the contact slipping? If the problem says "rolls without slipping", you have $v = \omega r$ and $a = \alpha r$ and static friction does no work, so energy conservation is legal. If it says "launched with no spin", "placed spinning", "the surface is slick", or gives you $\mu_k$, the contact is sliding: the constraint is gone, friction is $\mu_k N$ at full value, and energy is lost. If it gives you $\mu_s$ and an angle, it is asking you to test which case you are in with $\mu_s \ge \beta\tan\theta/(1+\beta)$.

Do you need a force, or just a speed? A question that asks only for the speed after a known drop, or a ratio of speeds, wants energy: one line, $Mgh = \tfrac12 Mv^2(1+\beta)$. A question that asks for acceleration, friction, or a minimum $\mu$ wants $F = Ma$ together with $\tau = I\alpha$ and the constraint linking them. Asking for both means writing both.

What is $\beta$? Every answer comes out in terms of $I/Mr^2$, so identify the shape first and write the number down before anything else: $2/5$ sphere, $1/2$ disk, $2/3$ shell, $1$ hoop. If mass or radius survives to your final answer in a problem that gave you neither, you have made an algebra mistake.

And one habit worth building: if the answer to a ramp question does not contain $1+\beta$ somewhere, look again.