Bicycle physics - Flying over the handlebars

[SysResearch]

It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. Skilled cyclists use the front brake alone probably 95% of the time.

However, improperly modulating the force applied on the front brake will bring the bike to a sudden stop, rider and bicycle's weight shifts to the front thus generating a momentum (force) that will send the rider "sailing" forward over the handlebars and eventually the bike will flip.

The second case scenario a bicycle hits a end parking concrete barriere at 15km/h.
Chances are the bicycle and rider will flip over.

Q1): Does anyone know what is the name of the generated force ( that throws you over the handlebars).

Q2): Based on speed, bike/rider mass-height how can this force be calculated.


Any links?



Thank you anticipated

S.R.

 

[FredGarvin]

q1) There doesn't need to be a shift in weight. The rider has momentum. Newton's first law tells you that an object wants to keep going in a straight line until acted on by an external force. When you stop normally, your arms and legs exert a force to help keep you upright on the bike and to slow you down with the bike. The rider's momentum is what carries them over the handlebars. There is no force, that is the problem. The rider needs a force to stop with the bike. So your statement that "...rider and bicycle's weight shifts to the front thus generating a momentum (force)..." is not really what is happening. If you are wondering about the flip part, that is simply a pivot point being introduced and the rider's arms or whatever providing the centripetal force required to introduce circular motion. Think of a situation where the rider let's go completely when they hit a stationary object. The rider will, most likely, just keep going in a straight line and not flip over.

q2) Essentially, worst case, the rider will keep moving at the same rate they were before they hit the brake. There will be a slight deceleration because their arms should exert some force to try to keep themselves on the bike.

This is the same exact scenario as a person riding in a car without their seat belt. If the car is moving 50 mph and suddenly stops, the person without the seatbelt to exert a force to decelerate, will keep moving at 50 mph.

 

[pixel01]

The difference between a car and a bike is that a bike has the centre of mass compared to the base higher than that of a car (a man is much heavier than the bike itself). So if you brake the front wheel of a bike, the braking force is much stronger than the back wheel.
High centre of mass also explains for the fact that one can easily fly over the handlebar when he/she used the front brake. It creates a torque with the point that the front wheel contacts with the road as the pivot.

 

[dst]

Q1.) Not exactly a force. Angular momentum, a quantity that likes to be conserved and will take any opportunity that arises to lob you off the front. And I only have a back brake... Well, the force would be centripetal force.

Q2.) Use the Schrodinger equation: Fast speed + wheel lock = Faceplant. Let's assume that the wheel locks up and does not slide whatsoever, or very little (as is the case in real life). Then, all your momentum is transferred into a rotational movement about the wheel, the center of the virtual circle would be the point where the wheel contacts the ground. If the cyclist is only thrown off and the bike is stationary and remains so, then mass (of combined bike & human) * velocity = mass of human * new velocity.

So a cyclist of mass 70kg at say, 15m/s, with a bike of mass 10kg, will have a momentum of [(70kg+10kg)*15ms], which means he'll be moving at 17m/s, and will have some 10kJ of energy to dissipate, which I imagine is not going to be pretty.

 

[SysResearch]

Super nice! Thank you!

Q1.) Not exactly a force. Angular momentum, a quantity that likes to be conserved and will take any opportunity that arises to lob you off the front. And I only have a back brake... Well, the force would be centripetal force.

Q2.) Use the Schrodinger equation: Fast speed + wheel lock = Faceplant. Let's assume that the wheel locks up and does not slide whatsoever, or very little (as is the case in real life). Then, all your momentum is transferred into a rotational movement about the wheel, the center of the virtual circle would be the point where the wheel contacts the ground. If the cyclist is only thrown off and the bike is stationary and remains so, then mass (of combined bike & human) * velocity = mass of human * new velocity.

So a cyclist of mass 70kg at say, 15m/s, with a bike of mass 10kg, will have a momentum of [(70kg+10kg)*15ms], which means he'll be moving at 17m/s, and will have some 10kJ of energy to dissipate, which I imagine is not going to be pretty.

That was nice! Thanks a lot