Slipping, friction and the move to rolling

A ball that skids before it rolls

A bowling ball leaves a player's hand and touches the lane. It does not roll at once. It skids for a few metres, and then it rolls smoothly. What changed in between? And how far does the skid go?

In the free courses you used the rule $v = \omega R$ for a body that rolls without slipping. Here we study what comes before that rule holds. Friction does all the work, so we must be careful about its direction.

The contact point decides

Take a round body of radius $R$. Its centre moves forward with speed $v$. It spins with angular speed $\omega$ in the sense of forward rolling. Because of the spin alone, the rim moves past the centre with the speed $u = \omega R$.

The point that touches the floor moves forward with speed $v$ because of the centre, and backward with speed $u$ because of the spin. Its speed along the floor is therefore

• If $v - u = 0$, the contact point does not slide. The body rolls without slipping.
• If $v - u > 0$, the contact point slides forward on the floor. The body is moving faster than its spin allows.
• If $v - u < 0$, the contact point slides backward. The body is spinning faster than it moves.

Kinetic friction opposes the slide of the contact point. Its size is $f = \mu mg$, where $\mu$ is the coefficient of kinetic friction. Its direction on the body is opposite to the slide of the contact point.

Two equations and a shape number

Let $I$ be the moment of inertia about the centre. Write it as $I = k\,mR^2$. The number $k$ depends on the shape only.

• Thin ring or hollow cylinder: $k = 1$.
• Solid disc or solid cylinder: $k = \tfrac{1}{2}$.
• Solid sphere: $k = \tfrac{2}{5}$.
• Thin hollow sphere: $k = \tfrac{2}{3}$.

Suppose the contact point slides forward, so the friction on the body acts backward. The friction slows the centre, and it also turns the body in the forward sense, because it acts at the bottom.

The speed $v$ falls at a steady rate and the rim speed $u$ rises at a steady rate. They meet when $v - u = 0$. The gap closes at the rate $\mu g\left(1 + \dfrac{1}{k}\right) = \dfrac{\mu g\,(1+k)}{k}$. So for a start with $v_0$ and $u_0$:

There is a shortcut for the final speed. The friction acts at the floor, so it has no torque about any point on the floor. The weight and the normal force lie on one vertical line, so they have no torque about such a point either. The angular momentum $mvR + I\omega$ about a point on the floor therefore stays constant.

The friction coefficient does not appear in $v_f$. It only decides how long the skid lasts. If the body starts with no spin, $u_0 = 0$, and $v_f = \dfrac{v_0}{1+k}$.

A worked example. A solid sphere slides onto a floor with $\mu = 0.20$, at a speed of $6.0\ \text{m/s}$ and with no spin. Find the time until it rolls, its final speed, the distance its centre moves, and the share of the kinetic energy that is lost.

1. The slide. The contact point moves forward at $6.0\ \text{m/s}$, so the friction acts backward.
2. The time. $t = \dfrac{k\,v_0}{\mu g\,(1+k)} = \dfrac{0.4 \times 6.0\ \text{m/s}}{0.20 \times 9.8\ \text{m/s}^2 \times 1.4} = 0.875\ \text{s}$.
3. The final speed. $v_f = \dfrac{6.0\ \text{m/s}}{1.4} = 4.29\ \text{m/s}$.
4. The distance. $s = v_0t - \tfrac{1}{2}\mu g\,t^2 = 6.0\ \text{m/s} \times 0.875\ \text{s} - \tfrac{1}{2} \times 0.20 \times 9.8\ \text{m/s}^2 \times (0.875\ \text{s})^2 = 4.50\ \text{m}$.
5. The energy. Before: $\tfrac{1}{2}mv_0^2$. After: $\tfrac{1}{2}(1+k)\,m\,v_f^2 = \dfrac{\tfrac{1}{2}mv_0^2}{1+k}$. The share lost is $\dfrac{k}{1+k} = \dfrac{2}{7} = 28.6\%$.

Check. At $0.875\ \text{s}$ the rim speed is $\dfrac{\mu g}{k}\,t = \dfrac{0.20 \times 9.8\ \text{m/s}^2}{0.4} \times 0.875\ \text{s} = 4.29\ \text{m/s}$. It equals $v_f$, as it must.

Backspin: a body that comes back

Now give the sphere backspin. The rim speed $u_0$ is negative. A billiard player calls this a draw shot.

A worked example. A solid sphere ($k = 0.4$) slides forward at $v_0 = 3.0\ \text{m/s}$ with a backspin that gives a rim speed of $u_0 = -9.0\ \text{m/s}$. The coefficient is $\mu = 0.25$.

1. The slide. The contact point moves at $v_0 - u_0 = 3.0 - (-9.0) = 12\ \text{m/s}$ forward. The friction acts backward all the time until rolling starts.
2. The time. $t = \dfrac{0.4 \times 12\ \text{m/s}}{0.25 \times 9.8\ \text{m/s}^2 \times 1.4} = 1.40\ \text{s}$.
3. The final speed. $v_f = \dfrac{3.0\ \text{m/s} + 0.4 \times (-9.0\ \text{m/s})}{1.4} = -0.43\ \text{m/s}$. The minus sign means that the sphere is moving backward when it starts to roll.
4. Where it turns. The centre stops at $t = \dfrac{v_0}{\mu g} = 1.22\ \text{s}$, after moving forward by $\dfrac{v_0^2}{2\mu g} = 1.84\ \text{m}$. Then it comes back, and rolling starts $0.18\ \text{s}$ later, with the centre $1.80\ \text{m}$ from the start.

Almost all of the kinetic energy is lost here, since $v_f$ is so small.

Set the spin to the same value as the speed. What do you see? Now give a large backspin. How does the final speed change, and in which direction does the body move at the end?

Where you push decides the friction

Now the body rolls without slipping, and a horizontal force $F$ pushes it at a height $h$ above the centre. The friction $f$ is not known in advance, so we call it positive when it acts forward. Rolling without slipping gives $\alpha = a/R$.

• The linear motion: $F + f = ma$.
• The turning about the centre: the push turns the body in the rolling sense with torque $Fh$, and the friction at the bottom turns it the other way with torque $fR$. So $Fh - fR = I\alpha = k\,mR\,a$.

Take $f = ma - F$ from the first and put it in the second. Then $F(h + R) = (1+k)\,mRa$. We want $a$ and then $f$:

The result is surprising. The friction is not always backward.

• Push through the centre, $h = 0$: $f = -\dfrac{k\,F}{1+k}$. The friction acts backward.
• Push at the top, $h = R$: $f = F\,\dfrac{1-k}{1+k}$. The friction acts forward. For a thin ring ($k = 1$) it is zero.
• The friction is zero at $h = kR$. For a solid disc this is half a radius above the centre. For a solid sphere it is $0.4$ of a radius.

Rolling without slipping is possible only while $|f| \le \mu_s\,mg$. This gives the largest push that the floor can hold.

A worked example. A solid disc of mass $6.0\ \text{kg}$ lies on a floor with $\mu_s = 0.25$. A horizontal force $F$ pulls it at the top. What is the largest $F$ for rolling without slipping, and what is the acceleration then?

1. The friction. With $h = R$ and $k = \tfrac{1}{2}$, $f = F\,\dfrac{1 - 0.5}{1.5} = \dfrac{F}{3}$, forward.
2. The limit. $\dfrac{F}{3} \le \mu_s mg = 0.25 \times 6.0\ \text{kg} \times 9.8\ \text{m/s}^2 = 14.7\ \text{N}$. So $F \le 44.1\ \text{N}$.
3. The acceleration. $a = \dfrac{F}{m}\cdot\dfrac{1 + 1}{1.5} = \dfrac{44.1\ \text{N}}{6.0\ \text{kg}} \times 1.333 = 9.8\ \text{m/s}^2$.

Check. $F + f = 44.1\ \text{N} + 14.7\ \text{N} = 58.8\ \text{N}$ and $ma = 6.0\ \text{kg} \times 9.8\ \text{m/s}^2 = 58.8\ \text{N}$. It fits.

Slide the push to $h/R = 0.5$ for a solid disc ($k = 0.5$). What does the friction become? Now try a ring ($k = 1$) and push at the top.

A slope with enough grip, and a slope without

For a body that rolls without slipping down a slope of angle $\theta$, you know that $a = \dfrac{g\sin\theta}{1+k}$. The friction that holds it is

The slope can supply at most $\mu_s\,mg\cos\theta$. Rolling without slipping therefore needs

A thin ring needs the most grip, because it has the largest $k$. When the slope is too smooth, the body slips as it goes down. The friction is then the kinetic friction $\mu mg\cos\theta$. The centre accelerates at $a = g(\sin\theta - \mu\cos\theta)$, and the spin builds up at the rate $\dfrac{\mu g\cos\theta}{k}$ for the rim speed.

A worked example. A thin ring is put on a slope with $\tan\theta = \tfrac{3}{4}$, so $\sin\theta = 0.6$ and $\cos\theta = 0.8$. The coefficient is $\mu = 0.30$.

1. The grip needed. $\dfrac{k\tan\theta}{1+k} = \dfrac{1 \times 0.75}{2} = 0.375$. The slope gives only $0.30$, so the ring slips.
2. The acceleration. $a = g(\sin\theta - \mu\cos\theta) = 9.8\ \text{m/s}^2 \times (0.6 - 0.30 \times 0.8) = 3.53\ \text{m/s}^2$.
3. Compare. A ring that did not slip would have $a = \dfrac{9.8\ \text{m/s}^2 \times 0.6}{2} = 2.94\ \text{m/s}^2$.

The slipping ring is faster, because less of the energy goes into spin. A solid sphere on the same slope needs only $\dfrac{0.4 \times 0.75}{1.4} = 0.214$, so it rolls without slipping.

Practise with new numbers

Try it in Python

The program starts a solid sphere on a rough floor with a forward spin that is slower than the rolling spin. It moves in tiny steps of time, and it stops when the contact point stops sliding. Then it compares the result with the formulas.

Common mistakes

• Friction always acts backward. It acts against the slide of the contact point. When the spin is too fast, it acts forward.
• Using $v = \omega R$ from the start. That holds only after the slipping has stopped.
• Saying that the friction decides the final speed. The final speed depends on $k$ and on the start values. The coefficient only sets the time.
• Taking the friction on a rolling body as $\mu mg$. Static friction is whatever the motion needs, up to $\mu_s\,N$.
• Saying that static friction on a rolling body does heat. The contact point is at rest, so static friction does no work and makes no heat. Heat appears only while the body slips.
• Using angular momentum about the centre of the body. The friction has a torque about the centre. Use a point on the floor, or use the two equations.

Remember these

• The speed of the contact point is $v - \omega R$. Rolling without slipping means that it is zero.
• Kinetic friction $\mu mg$ acts against the slide of the contact point.
• With $I = k\,mR^2$: $v_f = \dfrac{v_0 + k\,\omega_0R}{1+k}$ and $t = \dfrac{k\,|v_0 - \omega_0R|}{\mu g\,(1+k)}$.
• A push at height $h$ above the centre gives $f = F\,\dfrac{h/R - k}{1+k}$, forward when positive.
• A body rolls down a slope without slipping if $\mu_s \ge \dfrac{k\tan\theta}{1+k}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A thin ring is put on a rough floor with no spin and a speed $v_0$. By the time it rolls, what fraction of its kinetic energy is lost?

2. A pool ball slides with no spin on the felt. While it slips, which way does the friction act on it? How does this change its spin?

3. A ring is pulled by a horizontal force at its top. It rolls without slipping on a floor. What is the friction on it?

4. A solid cylinder and a thin ring roll down the same slope without slipping. Which one needs the rougher slope? Why?

References

• OpenStax, University Physics Volume 1, Section 11.1, Rolling Motion (rolling without slipping, static and kinetic friction, rolling down an incline).
• NCERT, Physics Class 11, Part 1, Chapter 6, System of Particles and Rotational Motion, the part on rolling motion.