# I Bike going downhill brakes front wheel

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1. Sep 4, 2016

### Quantaliinuxite

I'm an avid bike rider and a first year physics student. A couple days ago a friend of mine told me about how his friend was going downhill on her bike, used her front brake, and went over. I would like to determine what the minimum speed for a given slope angle is for the biker to do a full 180° rotation.

Let's assume the slope is 30°, and the bike is a point-mass. I've given a shot at the problem but I feel like I'm walking in the dark here. Here are a few possible ways to think about the question:

1. In terms of energy conservation principle: I think this is probably the most straightforward way to go about it, since we can draw a very clear picture of what's going on with very basic equation:
Before rotation: Etotal = Elinear kinetic
During rotation: Etotal = Erotational kinetic + Epotential - Elost from friction.
This makes for some easy calculations.
2. In terms of conservation of momentum: Since friction is applied though, I don't think momentum is conserved.
3. In term's of Newton's equations of motions and torque: This seems like a plausible approach, except that for an object to rotate you need to apply torque. I drew force diagrams to try to see where the torque could come from but I simply don't. This hints at the fact that I don't understand why the bike turns over like this (my intuition tells me it does, of course, but I don't understand why).
Could somebody give me some pointers for approach #3? Thank you!

2. Sep 4, 2016

### A.T.

It's not that hard to guess, is it?

3. Sep 4, 2016

### PeroK

Where would the torque come from if she'd hit a low wall?

4. Sep 4, 2016

### Quantaliinuxite

Well, again I intuitively see it in this situation, but my guess is that thinking about it as a point mass isn't enough to understand what's going on isn't it? I think that the collision would only happens at one point on the wheel, while the rest of the bike would want to continue moving forward, and that one point on the wheel where the collision happens would act as the end of a tense string, thus causing the bike to rotate around it. It sort of makes more sense if i imagine myself rollerblading, and then grabbing on to a string thrown from my left. I would start rotating leftwards because of the string. Is that intuition about right? In this case, which equations can I use to figure out the magnitude of the torque?

5. Sep 4, 2016

### PeroK

A point mass can't rotate! You need to consider the bike as a rigid body. I don't understand what you are saying about a tense string.

It's an impact. An impact which is directed through the centre of mass will produce a linear acceleration. An impact directed otherwise will produce both linear and angular acceleration. That's the basis of rigid body motion.

The torque itself is not going to be uniquely defined by these circumstances. What you could do is calculate the maximum braking force before the bicycle rotates, but to do that you'd need the dimensions and mass distribution of the bike and the rider.

6. Sep 4, 2016

### A.T.

Figure out the external forces on the bike using Newtons Laws, and the total torque they produce around the center of mass.

7. Sep 16, 2016

### zanick

think of the bike standing still and not moving. (and the ride on the seat)... grab the brake pads which are attched to the front forks and pull them along the rotaion path of the tire (brakes locked in place) ... the force at which it takes to lift the bike and rotate upwards is the force the front brakes and tires need to generate to create this motion. the force will be reduced as the rear wheel lifts off the ground and progressively be less until the bike is vertical, then, it will require no force but gravity to complete the next 90 degrees of rotation.

so, now apply the above idea to the bike dynamically, and you now know the force needed to be generated at the front tire on the ground to create the rotation of the bike around the front axle.

8. Sep 16, 2016

### lychette

guess !!!!! how?

9. Sep 16, 2016

### jbriggs444

[emphasis mine]

There is a difficulty, of course. The bike does not rotate around its front axle. With the brakes locked and the tire not skidding, the instantaneous center of rotation will be at the contact patch of tire on pavement. Worse, as the bike rolls up over its front wheel, the contact patch will move farther forward, thus further complicating the problem.

A good first cut at the problem would be to assume that the front wheel is about the size of a roller blade. Then that particular discrepancy goes away.

There is another related problem. On a slope, the maximum elevation of the bicycle center of gravity is not attained when the center of mass is directly above the axle. See if you can figure out why.

10. Sep 16, 2016

### Quandry

It is in principle a simple matter of moments. Your bicycle and you are in equilibrium whilst rolling freely on the road. When the brake is applied equilibrium is lost and moment of force come into play. The force is applied through the centre of gravity of the bike/rider combination through the pivot point which is initially where the rubber hits the road but transfers the the axle as the moments take effect. The moments of force are equal to the force applied through the CoG times the perpendicular distance between the CoG and the pivot point (take it as the axle as the transfer is almost instantaneous). Easy to work that out, then modify it for the slope (change in perpendicular distance) and remember the moments change as the the relationship between the pivot point and the CoG changes. There are a dozen other factors to take into account if you are so inclined.

11. Sep 17, 2016

### Quandry

P.s.
In the interest of safety, it is best to advise your friends friend that rear brakes are for stopping and front brakes are for controlling.