Energy and angular momentum in bike braking

Question

I have been thinking about the physics of braking in bicycles. We can think about the bike as three rigid bodies: bike, front wheel, back wheel.

First part: What happens when we go from unaccelerated motion to braked motion? Absent any braking force, static friction is what causes the wheel to continue rotating normally (i.e. constant centripetal acceleration). But when the brake is applied to a wheel (let's just assume front), we create a torque (around wheel axis) that causes the wheel to "want" to stop spinning. Does the ground, at the point of contact with the wheel, now increase the static frictional force? (because it wants to "hold" contact with the part of the wheel it was already touching?) Or is there something else going on here? We don't have slipping yet, so it must be that the static friction force increases? But now the static friction provides centripetal acceleration of the wheel (to keep spinning), centripetal acceleration of the whole bike body (around the wheel axis -- the tendency to flip), and backward linear acceleration (bike slowing).

The wheel begins to decelerate. Why does the frictional force not exactly counteract the braking force to keep the wheel moving at constant angular velocity? Possible answer: We also have the entire combination body (bike + 2 wheels) which wants to rotate around the (braked) wheel axis. Is the reason that the wheel decelerates (in angular velocity) because part of the frictional torque (from the road's static friction) resists the wheel-braking torque, while it also creates rotational acceleration of the whole bike body? (Thus only part of the torque can act to stop wheel motion?)

Second part: Finally, what is going on with the tendency of the whole bike to rotate around the front wheel axis: when we brake, we create (stopping) force at the point of contact between road and wheel, which creates angular velocity of full bike-body center of mass around the pivot point (the wheel's axle). The rear wheel generally does not lift off the ground because gravity also exerts a torque in the opposite direction (tending to keep the rear wheel down).

Is there any way to think about energy dissipation in this interaction of gravity with the braking torque? If the rear wheel does lift off the ground, then the road's frictional force does work to accelerate the bike upward off the ground (angular kinetic energy), and then gravity does work to push the bike down if it doesn't flip over.

But what about when the bike does not lift: we're creating a "desire" for the bike body to have angular velocity but gravity prevents it from happening (net zero angular acceleration). It's similar to when we push on a wall that doesn't move. Our muscles get warm trying to move the wall but no external work is done. Is the equivalent warming here in the brake that creates a torque around the front wheel axis that is countered by gravity?

(And assuming we don't care about stability of the bike, the fastest braking could actually occur when we just push the bike to the edge of flipping over? Since then we actually get gravity to do some "work" to help us slow the bike?)

Answer 1

I'll answer the first part of your question by considering a model system with a single wheel rolling without slipping on a perfectly flat surface.

First note that the role of static friction is to apply enough force to ensure that the contact point of the wheel on the ground is zero, i.e. the wheel rolls without slipping. Static friction is only needed to speed up and slow down the wheel without slipping; it isn't needed to keep the wheel in motion at a uniform velocity. If static friction were non-zero and it were the only force acting on the wheel, then the wheel would accelerate over time and the velocity would be non-uniform.

Now in order to slow down the wheel, let's apply a force on the top of the wheel, opposite to the direction of travel. Static friction kicks in to ensure that the wheel doesn't slip. To oppose the slowing of the wheel, the static friction also acts opposite of the direction of the travel of the wheel. Both the applied force and the static frictional force are acting in the same direction, so the wheel velocity must decrease. Rolling without slipping implies that the angular velocity must also decrease. The conclusion is that the static frictional force cannot apply enough torque to cancel out the applied torque, or else the wheel would start to slip (and static friction is no longer applicable).