What causes the back of a bike to lift when the front brake is applied? (Like in an endo.)
Also, if I were to replicate this effect with a wood block with wheels that crashes against a wall (only the wheels touch the wall) would it work?
If we disregard wind resistance, there are three points where external forces apply to the bike/rider combo body: The front and rear wheel contact points with the road, and the center of mass. If these forces generate a torque that is not zero, the rear wheel will lift off while braking (or the front wheel lift off during hard acceleration).
If you want to replicate the system with a block of wood with wheels bumping into a wall, use very small wheels. If the front wheel suddenly stop due to braking on a bike, the bike will rotate around the contact point of the front wheel. On the other side, if you run the front wheel into a wall, the front wheel will stop and the center of the rotation will be the front wheel hub. (Assuming sufficiently stiff wheels.)
I have originally built the page for a tadpole tricycle. For looking at a standard bicycle, select "upright bicycle" on the page and then play a little with the
There is simplified "cheat" method for thinking about this.
The hard way is to draw a free-body diagram of the bicycle, in which all the forces acting on it are made plain: the force of gravity acting through its center of mass, the force of friction from braking, and so on.
The cheat method is to visualize the bicycle in an accelerating frame of reference, and then to pretend that the acceleration is just a form of gravity. Under Newtonian physics, acceleration is not distinguishable from gravity.
For instance if you are standing on a train which accelerates, then you simply tilt forward, as if you were standing on a slope. It feels exactly as if gravity has increased and slanted.
So the reason the rear wheel lifts is because, since you are decelerating, it is as if you are cycling down a steep slope (on a planet where gravity is a little big greater than that of Earth, but that is irrelevant). If you're on a slope facing downhill, your bike is tilted forward, so that its center of gravity is shifted so that it is more above the front wheel, rather than the rear.
If the tilt is sufficiently pronounced, your center of gravity will move ahead of the front wheel, and at that point, the rear wheel will lift.
While that force still acts through the base of support, the rear wheel will not lift. However, there is also the question of how much support comes from the rear versus the front wheel. Even before the rear wheel lifts, "weight transfer" takes place to the front wheel. The bike and rider are supported more by the front than the rear wheel, because the imaginary force acting through the center of mass intersects the imaginary support base closer to the front wheel than to the rear wheel. This weight distribution toward the front limits the effectiveness of brakes on the rear wheel, since brakes can only work to the extent that the wheel can develop friction against the road. Also, it means that you can never brake so hard using the rear brake alone that the bike will tilt. As soon as it is close to the verge of tilting, the braking power is lost: an example of stabilizing negative feedback.
The center of mass of the bike+rider wants to keep moving forward (Newton's first law)
Since the center of mass is higher on the bike - when it goes forward the bike pivots around the point where the front wheel sticks to the road.
Imagine attaching a string to the middle of the wooden block and pulling it forward, while having the front wheel stuck to the ground.
My naive argument would be based on energy conservation. I won't write any equations. This is for you to do.
The bike + rider has both rotational and translational kinetic energy in addition to the potential energy of the system. When you apply the brakes, the translational KE is converted to both rotational KE and PE which manifests as work done against gravity. You lift up, pivoting about a point, which accounts for a change in rotational motion due to an external torque (friction).
A very interesting real world experiment which I do every day (I bike to work) is to stand up if I wish to brake efficiently (in my case it is only marginally efficient, but I do feel the difference). The translational KE is converted to PE and hence the brakes don't have to work as hard to stop my motion. The heavier the person, the more effective the technique.
So, if you are a fat bastard, this may be a useful way to stop quickly.
Applying the brakes makes the wheel stop turning in relation to the bicycle's frame but not in relation to the road. The bike's center of mass (especially with a rider pressing against the handle bars) is higher than the hub of the front wheel.
When the brakes are applied that mass has momentum toward the front of the bike that exerts a force on the front wheel to turn toward the front of the bike. Since the wheel can still turn on the road, it does, but since the wheel and bike are locked together, the bike rotates with the front wheel.