[Physics] Why does a leaning bike not fall over

centrifugal forcenewtonian-mechanicsrotational-dynamicsstabilitytorque

This question has been bothering me for a while now. Everywhere I look, everyone talks about 'fictitious forces' and how they apparently explain the bike being in equilibrium. However, if we just look at a simple force diagram, we can see that turning moments around certain points are unbalanced:Force diagram

If we take moments around, for example, the point where the wheel touches the ground, we obtain that a resultant moment of $amg\cos \theta$, where $a$ is some length, $m$ the mass of the combined system of the driver and the bike, and $\theta$ the angle between the bike and the ground, is acting so as to make the system fall down in the direction of the centre.

This moment will always be 'towards the centre' – irrespective of the bike's position. That means it should fall down. So why doesn't it? Clearly, either there should be no resultant moment, or the resultant moment throughout the duration of one lap made by the bike should be $0$, but none is evidently the case here. What am I missing?

Best Answer

In the inertial frame, the torques are unbalanced and the rider's rotational momentum about a point on the ground changes. It's just that this change does not result in the rider toppling.

If instead you consider the frame where the rider is at rest, then this (accelerating) frame will have fictitious forces opposite the acceleration appear.

These forces will act through the center of mass of the rider, and will be in the opposite direction of the acceleration. Since the rider is accelerating to the left, there should be a fictitious frame force to the right. This balances the torques about the tire contact.

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