In a snooker club in Kolkata a player strikes the white ball level, through its centre. Watch it closely. For a short distance it slides across the cloth without turning much. Then it starts to roll, and from there it runs smoothly to the far cushion. Nobody made it roll. The cloth did, and the cloth also decided how fast the ball would be rolling when it began.
When the contact slides
Rolling without slipping needs \(v = \omega R\), where \(v\) is the speed of the centre, \(\omega\) the angular velocity and \(R\) the radius. When a ball is struck through its centre, it starts with \(v\) large and \(\omega = 0\). Then \(v > \omega R\), and the bottom point moves forward over the cloth at \(v - \omega R\). The contact slides, so the friction is kinetic friction, of size \(\mu_k N\).
Kinetic friction opposes the sliding of the contact. Here the contact slides forward, so the friction points backward. That single force does two jobs at once. As a force on the ball, it slows the centre, so \(v\) falls. As a torque about the centre, it spins the ball forward, so \(\omega R\) rises. The two approach each other.
When \(v \neq \omega R\), kinetic friction acts against the sliding of the contact point. It changes \(v\) and \(\omega R\) towards each other. When they become equal the sliding stops, and on level ground the friction then drops to zero and the body rolls on by itself.
The opposite case is a wheel that spins too fast, as when a spinning disc is set down on the floor with no forward speed. Then \(\omega R > v\), and the bottom point slides backward over the floor. The friction points forward. It pushes the centre forward and slows the spin. Again the two meet. Kinetic friction always opposes the sliding of the contact, not the motion of the body.
A disc is spun fast about its own axis and then lowered gently onto a rough floor, with no forward speed. Which way does the friction on it point at first?
The struck ball, in numbers
Take a solid ball of mass \(m\) and radius \(R\), struck at \(v_0\) with no spin, on a surface with coefficient of kinetic friction \(\mu\). The friction \(\mu mg\) slows the centre at \(\mu g\), so \(v = v_0 - \mu g t\). Its torque \(\mu mgR\) about the centre gives \(\alpha = \mu mgR/(\tfrac25 mR^2)\), so \(\omega R = \tfrac52\mu g t\). Rolling starts when the two agree.
\[ v_0 - \mu g t_0 = \tfrac52 \mu g t_0 \]
Why. rolling starts at the instant \(t_0\) when \(v = \omega R\).
\[ t_0 = \dfrac{2v_0}{7\mu g} \]
Why. collect the terms in \(t_0\): \(v_0 = \tfrac72\mu g t_0\).
\[ v_f = v_0 - \mu g\,\dfrac{2v_0}{7\mu g} \]
Why. put \(t_0\) into \(v = v_0 - \mu g t\).
\[ v_f = \tfrac57 v_0 \]
Read it. \(\mu\) and \(g\) cancel, leaving \(v_0(1 - \tfrac27)\).
The final speed contains no \(\mu\) at all. A rougher cloth only brings the moment of rolling sooner. The ball always ends up rolling at five sevenths of the speed it was struck with. In Fig 4.1 the ball struck at \(3.5\,\text{m/s}\) on a cloth with \(\mu = 0.2\) skids for \(0.5\,\text{s}\) over \(1.5\,\text{m}\), and then rolls at \(2.5\,\text{m/s}\).
A bowling ball of mass \(6\,\text{kg}\) is released along the lane at \(7\,\text{m/s}\) with no spin. The coefficient of kinetic friction between the ball and the oiled lane is \(0.1\). Take \(g = 10\,\text{m/s}^2\). When and where does the ball start to roll, how fast is it then, and how much kinetic energy has the skid cost?
Treat the ball as a uniform solid sphere, \(I = \tfrac25 mR^2\). The friction is \(\mu mg = 6\,\text{N}\), backward.
\(t_0 = 2v_0/(7\mu g)\), \(v_f = \tfrac57 v_0\) and \(s_0 = 12v_0^2/(49\mu g)\).
\(t_0 = 14/7 = 2\,\text{s}\), \(v_f = 5\,\text{m/s}\) and \(s_0 = 12 \times 49/49 = 12\,\text{m}\). The energy at the start is \(\tfrac12 \times 6 \times 49 = 147\,\text{J}\). Rolling at \(5\,\text{m/s}\), the ball has \(\tfrac12 mv^2(1 + \tfrac25) = 0.7 \times 6 \times 25 = 105\,\text{J}\).
The ball starts to roll after \(2\,\text{s}\), \(12\,\text{m}\) down the lane, at \(5\,\text{m/s}\). The skid has turned \(42\,\text{J}\), two sevenths of the starting energy, into heat in the ball and the lane.
Why is energy lost here, when rolling friction does none? During the skid the contact point slides, so the kinetic friction acts through a real displacement of the surfaces over each other. That rubbing turns mechanical energy into heat. For the snooker ball of Fig 4.1, with mass \(0.16\,\text{kg}\), the contact slides \(0.875\,\text{m}\) against a friction of \(0.32\,\text{N}\), which makes \(0.28\,\text{J}\) of heat. That is exactly the drop from \(0.98\,\text{J}\) to \(0.70\,\text{J}\).
The ball of Fig 4.1 starts to roll at \(2.5\,\text{m/s}\) after \(0.5\,\text{s}\). The same shot is played on a cloth with twice the coefficient of kinetic friction. When and how fast does the ball start to roll now?
A solid ball is struck through its centre at \(7\,\text{m/s}\) on a floor with \(\mu_k = 0.25\). Take \(g = 10\,\text{m/s}^2\). After how many seconds does it start to roll without slipping?
The one-line shortcut
Choose a fixed point \(O\) on the ground, on the line along which the contact travels. Take torques about \(O\). The weight acts down through the centre and the normal force acts up through the contact, directly below it. They are equal, opposite and on one vertical line, so their torques about \(O\) cancel. The friction acts along the ground line itself, which passes through \(O\), so its torque about \(O\) is zero.
With no net torque about \(O\), the angular momentum about \(O\) stays constant through the whole skid. The next lesson shows that the angular momentum of a moving, spinning body about such a point is \(MvR + I_{cm}\omega\). At the start the struck ball has only \(Mv_0R\). When it rolls, it has \(Mv_fR + \tfrac25 MR^2(v_f/R) = \tfrac75 Mv_fR\). Equating the two gives \(v_f = \tfrac57 v_0\) at once, with no time and no friction in sight.
Two things are not conserved here. The friction is an outside horizontal force, so the linear momentum of the ball changes. The skid makes heat, so the kinetic energy falls. Angular momentum about \(O\) is the one quantity that survives, and it is enough.
A uniform disc of radius \(R\) spins at \(\omega_0\) about its own axis and is lowered gently onto a rough floor with no forward speed. At what speed does it finally roll?
About a fixed point on the ground line the angular momentum is conserved. At the start the centre is at rest, so only the spin counts: \(L = \tfrac12 MR^2\omega_0\).
When the disc rolls at \(v_f = \omega_f R\), \(L = Mv_fR + \tfrac12 MR^2\omega_f = \tfrac32 MR^2\omega_f\).
So \(\omega_f = \omega_0/3\) and \(v_f = \omega_0 R/3\). The disc of Fig 4.1, set down with \(\omega_0 R = 3\,\text{m/s}\), rolls off at \(1\,\text{m/s}\). A solid sphere treated the same way rolls off at \(\tfrac27\omega_0 R\).
A ball skids on a level table before it starts to roll. Which quantities stay constant during the skid?
A hoop is pushed along a rough floor so that it starts at \(4\,\text{m/s}\) with no spin. At what speed does it roll once the skid is over, in m/s?
Coriolis, a French engineer, is remembered for the force felt in a rotating frame. In 1835 he also published a whole book on the game of billiards. In it he worked out how a struck ball slides, spins and finally rolls. He also explained why a ball struck high or low curves and follows through. He used exactly the reasoning of this lesson: friction at the sliding contact changes \(v\) and \(\omega\) together until the contact stops sliding.
When a driver brakes too hard, the wheels can lock and the tyres skid over the road. The friction is then kinetic, which is usually smaller than the largest static friction. So a skidding car stops over a longer distance. A skidding front wheel also cannot steer. Anti-lock brakes release and apply the brakes many times a second. Each wheel then keeps rolling, and the road can use its larger static friction.
Now the cue ball of the opening is fully explained. At first it slides, because the cue gave it speed but no spin. The cloth's friction slows it and spins it up together, until \(v = \omega R\). From then on the contact is at rest, friction vanishes and the ball rolls on. It always rolls at five sevenths of its struck speed. A rougher cloth only brings that moment sooner.
- Taking the friction on a skidding body to oppose the motion of the centre.Kinetic friction opposes the sliding of the contact point. Find the velocity of the contact first.
- Expecting a rougher surface to change the final rolling speed.The final speed \(\tfrac57 v_0\) contains no \(\mu\); roughness changes only the time and the distance.
- Conserving momentum or kinetic energy through a skid.Friction is an outside force and sliding makes heat. Conserve angular momentum about a point on the ground line.