Analyzing the Movement of a Bicycle (requesting feedback)

  • #1
As I watched a few cyclists zoom by today, I started thinking about the mechanics behind the motion of a bicycle. I will now run through the work I have done in analyzing it. Please tell me if I have made any glaring mistakes in my logic and let me know what you think.

Purpose
Evaluate the relationship between the force exerted downward on the pedal gear to the change in momentum of the bicycle's center of mass.

Assumptions
- No air resistance
- Interactions between the tires and Earth are elastic
- The chain connecting the pedal gear to the back wheel absorbs no energy; energy is completely transfered from the pedal gear to the back wheel
- The bicycle is on level surface so gravitational force contributes nothing to change in momentum
- The bicycle starts at rest

Picture
A bicycle starts in place at rest. The pedals are attached to a gear which rotates clockwise. As the pedal gear rotates clockwise, its momentum is transfered by a chain to the back tire. The back tire begins to rotate pushing the bicycle forward.

Values
Radius of pedal gear - Rg = r (m)
Radius of back wheel - Rb = 5r (m)
Mass of gear - mg = m (kg)
Mass of back wheel - mb = 10m (kg)
Force acting on pedal gear - [tex]\vec{}F[/tex] = <0, -F, 0> (N)
Initial angular velocity of pedal gear - [tex]\omega[/tex]g, i = 0 (rad/s)
Time that [tex]\vec{}F[/tex] acts upon the pedal gear - [tex]\Delta[/tex]t = t (s) = 5s (s)
Mass of the section of back tire that interacts with the Earth at a time - Msec = 10m/7 (kg)
Time that a section of back tire interacts with Earth - [tex]\Delta[/tex]t = s (s)
Angular momentum of pedal gear - [tex]\vec{}L[/tex]g
Angular momentum of back wheel- [tex]\vec{}L[/tex]b
Angular velocity of back wheel- [tex]\omega[/tex]b

Calculations
1) First step is to calculate the angular momentum that the pedal gear acquires over [tex]\Delta[/tex]t and use that to find the final angular velocity of the back wheel.

[tex]\vec{}L[/tex]g = Ig[tex]\omega[/tex]g + [tex]\vec{}\tau[/tex][tex]\Delta[/tex]t

I[tex]\omega[/tex]g = 0
[tex]\vec{}\tau[/tex] = [tex]\vec{}r[/tex]g X [tex]\vec{}F[/tex] = <0, 0, -Fr> (kg m2/s2)

[tex]\vec{}L[/tex]g = [tex]\vec{}\tau[/tex][tex]\Delta[/tex]t = <0, 0, -Frt> (kg m2/s)

This value above is the angular momentum that the pedal gear attains over the time interval. Since it is transfered completely to the back wheel, we have:

[tex]\vec{}L[/tex]g = [tex]\vec{}L[/tex]b = <0, 0, -Frt> (kg m2/s)

To calculate the angular velocity of the back wheel we have:

[tex]\vec{}\omega[/tex]b = [tex]\vec{}L[/tex]b/Ib

Ib = mbRb2 = 10m x (5r)2 = 250mr2 (kg m2)

[tex]\vec{}\omega[/tex]b = <0, 0, -Ft/250mr> (rad/s)


Thus we have: when a constant downward force F is applied to the pedal gear over a time interval t, the back wheel gains angular speed of |[tex]\omega[/tex]| = Ft/250mr (rad/s)


2) Use angular speed of the back tire to analyze the collision between a section of the back tire and the Earth. Use this information to find the constant net force acting on the center of mass.

(focusing only on the momentum of the section in contact with the earth. Moving in the -x direction)
[tex]\omega[/tex] = v/Rb = Ft/250mr (rad/s)
[tex]\vec{}v[/tex]sec = <-Ft/50m, 0, 0> (m/s)
[tex]\vec{}p[/tex]sec = [tex]\vec{}v[/tex]sec x Msec = <-Ft/35, 0, 0> (kg m/s)

Section is in contact with the Earth for [tex]\Delta[/tex]t = s (s)

Since the collision between the section of the bike and the Earth is elastic, and mearth >> Msec, we have a reflection of the center of mass in the opposite direction. This means that:

[tex]\Delta[/tex][tex]\vec{}p[/tex]cm = -2[tex]\vec{}p[/tex]sec = <2Ft/35, 0, 0> (kg m/s)

Since
[tex]\Delta[/tex][tex]\vec{}p[/tex]cm = [tex]\vec{}F[/tex]net, cm[tex]\Delta[/tex]t
we have that:
[tex]\vec{}F[/tex]net, cm = [tex]\Delta[/tex][tex]\vec{}p[/tex]cm/[tex]\Delta[/tex]t = <2Ft/35s, 0, 0> (N)


In the time interval [tex]\Delta[/tex]t = s (s) that the section of the tire is in contact with the Earth, the constant force [tex]\vec{}F[/tex]net, cm = <2Ft/35s, 0, 0> (N) acts on the center of mass.


3) Using information of the time intervals, find the total change in momentum of the bicycle's center of mass of [tex]\Delta[/tex]t = t (s)

Since
t = 5s
then
[tex]\Delta\vec{}p[/tex]cm, total = <2Ft/35s, 0, 0> x 5s = <2Ft/7, 0, 0> (kg m/s)


While applying a constant downward torque on the pedal gear over the time interval [tex]\Delta[/tex]t = t (s), the center of mass of the bicycle gains 2Ft/7 (kg m/s) of translational momentum.





That's all I've got. Please let me know if you find any errors or if you have general advice for me in any way. This is the first time I've ever tried something like this on my own.

Thanks a ton,
keep on doing work, haha! :biggrin:
 

Answers and Replies

  • #2
167
0
It looks like you are forgetting to take into account the torque on the tire caused by friction, the one that actually applies the force on the bike to move it.
I looked through most of your work and it seems OK, but you must superimpose the effect of friction on the angular momentum of the bike tire.
 
  • #3
Hmmm, do you think you could explain in detail the forces at play during the rolling of a wheel? I know that the motion of the center of mass is due to friction, but I'm having a difficult time wrapping my mind around the details involved. If you could upload a free-body diagram as well, that would be great.

Thanks!
 

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