Rolling Down an Incline

Three forces on a slope

Place a body of mass \(M\), rolling radius \(R\) and radius of gyration \(k\) on a rough slope of angle \(\theta\). Three forces act on it. The weight \(Mg\) acts at the centre. The normal force \(N\) acts at the contact, perpendicular to the slope, and points through the centre. The friction \(f\) acts at the contact along the slope.

Which way does the friction point? Neither the weight nor the normal force has any torque about the centre. So without friction the body would slide down without turning at all. A body that rolls must spin faster and faster, and only the friction can supply that torque. To spin the body forward, the friction must point up the slope.

Read the result slowly. The mass has cancelled, and so has the radius; only the shape number \(k^2/R^2\) survives. A ring, with \(k^2/R^2 = 1\), rolls down at half of \(g\sin\theta\). A solid sphere rolls down at \(\tfrac57 g\sin\theta\). Every roller is slower than a block sliding down a smooth slope at \(g\sin\theta\). Part of the pull of gravity is used to spin the roller.

How much grip rolling needs

The friction a rolling body needs is fixed by the formula above. The slope can supply at most \(\mu_s N\), with \(N = Mg\cos\theta\). Rolling survives only while the friction needed stays within that limit. Divide one by the other, and the mass cancels once more.

A steeper slope needs more grip, through \(\tan\theta\). A body with its mass far from the axis needs more grip too. On a \(30°\) slope a ring needs \(\mu_s \ge \tfrac12\tan 30° \approx 0.29\), while a solid sphere needs only \(\tfrac27\tan 30° \approx 0.165\). For the stone roller of the example, \(10\,\text{N}\) out of \(N = 6 \times 10 \times \cos 30° \approx 52\,\text{N}\) gives \(\mu_s \ge 0.19\). A rougher ramp changes nothing; the roller still takes only the \(10\,\text{N}\) it needs.

If the slope is perfectly smooth, there is no friction at all. Then nothing has a torque about the centre, so the angular velocity cannot change. A body released without spin slides down at the full \(g\sin\theta\) and never starts to turn. If \(\mu_s\) is too small but not zero, the body rolls and slides at once; the chapter's fourth lesson handles that case.

The race

Now return to the bangle, the wheel and the marble. Before you press play, decide the order in which they will arrive.

The slopes in the figure are at \(30°\) and the race is \(2.8\,\text{m}\) long, shown three times slower than life. The solid sphere accelerates at \(\tfrac57 \times 5 \approx 3.57\,\text{m/s}^2\) and arrives in \(1.25\,\text{s}\). The disc, at \(3.33\,\text{m/s}^2\), arrives in \(1.30\,\text{s}\). The ring, at \(2.5\,\text{m/s}^2\), needs \(1.50\,\text{s}\). When the sphere arrives, the disc is still \(0.19\,\text{m}\) behind and the ring \(0.84\,\text{m}\) behind.

A steel marble and a solid iron ball the size of a football arrive together, because both are solid spheres. A small coin and a heavy cart wheel, both close to discs, arrive together. The angle of the slope changes how long the race lasts, but never the order, because \(\sin\theta\) multiplies every acceleration alike.

The energy route

The chapter on Rolling Motion showed that a rolling body carries kinetic energy in two parts. With \(\omega = v/R\) the total is \(K = \tfrac12 Mv^2 + \tfrac12 Mk^2\omega^2 = \tfrac12 Mv^2(1 + k^2/R^2)\). The first lesson showed that friction on a rolling body does no work. The normal force does none either. So only gravity does work, and mechanical energy is conserved.

For the stone roller, the \(2.4\,\text{m}\) along the ramp is a drop of \(h = 2.4\sin 30° = 1.2\,\text{m}\). Then \(v = \sqrt{2 \times 10 \times 1.2/1.5} = \sqrt{16} = 4\,\text{m/s}\), the same answer as before. The energy route never needed the friction, the acceleration or the time. It also works on a curved rough track, where the acceleration changes at every point.

Rolling up a rough slope and a smooth one

Run the energy route backward. A solid ball rolling at \(5\,\text{m/s}\) meets a rough slope. Friction keeps it rolling all the way up, so its spin slows with its speed, and it stops with \(v = 0\) and \(\omega = 0\) together. All of its kinetic energy becomes height: \(h = v^2(1 + k^2/R^2)/2g = 25 \times 1.4/20 = 1.75\,\text{m}\).

Now make the slope perfectly smooth. Gravity and the normal force both pass through the centre, so nothing has a torque about it. The spin cannot change. Only the translational energy \(\tfrac12 Mv^2\) can become height, so the ball rises only \(v^2/2g = 1.25\,\text{m}\). Watch the smooth chip in the figure. At the top the ball has stopped moving, yet it is still spinning.

The smoother slope takes the ball lower. That surprises almost everyone the first time. The energy is not lost on the smooth slope; it is locked in the spin, which nothing can reach. When the ball slides back to level ground, it arrives at \(5\,\text{m/s}\) again, still spinning forward at \(20\,\text{rad/s}\) while it moves backward.

Now the race of the opening has its answer. The marble wins, the wooden wheel comes second, and the bangle comes last. The heavy wheel gains nothing from its weight, and the small marble gains nothing from its size. The one number that decides the order is the shape number \(k^2/R^2\). The less of a body's mass lies far from its axis, the faster it rolls down.