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# Spinning Out

While riding a bike in a straight line, I suddenly squeeze the rear brake. This locks the rear wheel and keeps the front wheel rolling freely. Will I continue in a straight line until the bike stops? Let us assume that I keep the front wheel straight at all times (so, strictly speaking, I should have training wheels to avoid falling over).

It seems quite natural to argue that the friction on the locked rear wheel will hold the rear of the bike backwards and act as a stabilizer, ensuring that the bike continues going straight. Interestingly, this argument is wrong; the bike will actually spin out.

Figure 1. The front wheel, free to spin, tracks the rear (locked) wheel.

Where is the mistake in the no-spinout argument? It comes from ignoring the front wheel and thus missing the fact that the front wheel acts as a destabilizer. This destabilization beats the stabilizing effect of the rear wheel. Indeed, the front wheel’s velocity aligns with the bike frame, assuming that no slipping occurs. The front wheel therefore acts like an arrow’s fin, and our “arrow” is initially moving backwards (see Figure 1). This arrow will consequently turn around and the fin/front will follow the rear, resulting in a complete spinout. Regardless of the rear wheel’s trajectory (and we do not address its shape here), the front wheel tracks the rear wheel due to the no-sideslip assumption. In our “proof” of stability, we got it exactly backwards (pun intended): it is the front wheel and not the rear wheel that acts as a stabilizer.

Figure 2. If both wheels slip, the perpendicular component of friction is greater on the front wheel, causing a spinout.
So far we have assumed that the front wheel holds its grip on the road, i.e., that no slippage occurs. But what if the front wheel sideslips (as could happen when riding on ice)? The key is to observe that the frictional force acting on a sideslipping wheel in the absence of rolling friction is perpendicular to the direction of rolling (see Figure 2). Indeed, the projection of the frictional force in the direction of rolling vanishes by assumption, and the force is hence purely perpendicular to the wheel’s direction (i.e., aligned with the wheel’s axle).

As a result, the front friction is spent entirely on trying to spin the bike out while the rear friction is partly “wasted” on the longitudinal component; more formally (and assuming, perhaps unrealistically, that the two frictions are the same), we have

$f>f \: \sin \: \theta.$

The sideways frictional push on the front wheel exceeds that on the rear wheel, again resulting in a spinout.