The Physics of a Bicycle

We will neglect friction between the axel, the wheel, and air resistance. The friction at any one moment is static friction.

This is because the x component of the gravity is also a centripetal force, and is not factored by the coeffient of friction (less than 1) and has a greater value, allowing Lance to take a turn faster on a banked track compared with a flat track.

We can conclude that Lance can ride his bicycle faster on a banked track than on a flat track of the same radius.

v=sq. rt. of (F frict + F grav. x)(radius)/mass= 15.35 m/s

Centripetal Force in this circumstance is the friction between the tires and the road that points toward the center of the arc traveled. Thus, we can find the maximum instantaneous velocity possible for Lance to travel without landing on his rear end.

v=Square root of (F centripetal) (radius) / mass = 11.7 m/s

This gives us a centripetal acceleration of 6.6 m/s² found by ac=v²/r

Knowing the Friction Force... How fast can Lance go into a turn (riding vertically) with a radius of 20 meters and not end up like this amateur.

But wait! There is more!

FBD

F normal=804.58N

F frict=563.21N

F gravity=804.58N

Friction Force=(Mu)(Normal Force) or (.7)(82.1kgx9.8m/s²)=563.21N

Again note: This is when Lance is riding vertically, and is a "very short"

Let's say that Lance has just won the Tour de France and is riding on a completely flat surface, what is his friction with the ground?

How can we apply physics to a

bike that is making a turn on a flat track verus a banked track?

82.1 kg

Well let's start with an FBD! (The rider is riding vertically)

And Friction force=(Mu) (Normal Force)

We know the masses

• The mass of Lance Armstrong's Trek Madone 6.9 is 7.26kg
• Lance's mass is 165 lbs or 74.84kg
• The total mass of the system is
• For this analysis, assume that Lance has no height, he is an infinite "shrimp"

Gravity (9.8m/s²)

Mass

The Mu between the tires and the road is .7

Mu

Where is the F friction from?

But how can we find this?

Maybe something like this?

Friction Force

Do you remember ?

"Mu"- The coefficient of friction

The sum of Fy= Fn-Fg = 0

• Fn=Fg
• = mg (82.1kg)(9.8m/s²) = 804.58N

Fn=804.58N

FBD

Fg=804.58N

What would happen if Lance Armstrong entered a banked turn that has an angle of 45 degrees, again assuming that Lance's bike is vertical and he and his bike have no height?

Disclaimer: Ignoring tourque on Lance and the bicycle

Fn

y axis

FBD

45°

+

x axis

45°

F gravity y comp.

F gravity x comp.

Ff=398.23N

Fg is not straight Down,

There is an x and y component

How fast can Lance enter a turn of 20 meters now?

They y component can be found by

Fg(804.58N) cos(45°)=568.9N

45°

sin(x°)=opposite/hypotenuse

The x component can be found by Fg(804.58N) sin(45°)=568.9N

Ajacent

F gravity (y)

hypotenuse

F gravity=804.58N

45°

hypotenuse

F gravity804.58N

opposite

F gravity (x)

The Sum of Fy=Fn-Fg=0... therefore Fn=Fgy and Fn=568.9 N

Mu(Fn)=Ff