Bicycle Translational Acceleration vs Angular Acceleration

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SUMMARY

The discussion centers on calculating the minimum force required on a bicycle pedal to initiate both angular and translational acceleration of the rear wheel. Given a wheel radius of 20 cm, gear radius of 5 cm, a coefficient of static friction of 0.2, and a weight on the rear wheel of 50 N, the maximum static friction force is determined to be 10 N, leading to a maximum torque of 2.5 N*m. The confusion arises from the relationship between pedal force and the required torque, with participants clarifying that the pedal force must exceed the static friction torque to achieve motion, and that the pedal force is not directly equal to the tension in the chain.

PREREQUISITES
  • Understanding of Newtonian mechanics, specifically torque and force relationships.
  • Familiarity with static friction and its role in motion initiation.
  • Knowledge of bicycle mechanics, including gear ratios and torque transmission.
  • Ability to perform free body diagram (FBD) analysis for mechanical systems.
NEXT STEPS
  • Study the relationship between torque and force in mechanical systems, focusing on bicycle dynamics.
  • Learn how to calculate the effective torque from pedal to rear wheel using gear ratios.
  • Explore the principles of static friction and its implications in rolling motion.
  • Practice drawing and analyzing free body diagrams for complex mechanical systems.
USEFUL FOR

Physics students, mechanical engineers, and bicycle enthusiasts seeking to understand the dynamics of bicycle acceleration and the interplay between angular and translational motion.

  • #91
Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached
 
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  • #92
jbriggs444 said:
If you recall, we assumed that the wheel has negligible moment of inertia in order to simplify the calculations. Accordingly, it takes negligible net torque for it to accelerate.

If you want to re-do the calculations with a non-negligible moment of inertia in the rear wheel then we can do so.
We've already done that in previous posts.

Chet
 
  • #93
UMath1 said:
Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached
Huh?
 
  • #94
UMath1 said:
Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached
I want to echo Chet's puzzlement. The torques will be negligibly different until maximum static friction is reached. [At which time the angular acceleration of the wheel becomes significant in spite of the wheel's neglibible moment of inertia]
 
  • #95
The equation is based on neglecting moment of inertia and taking the torques to be equal, but they are not. They are negligibly different as you say...but they are still different.

So what equation indicates that the torque of the chain is little higher than the torque of static friction.
 
  • #96
UMath1 said:
The equation is based on neglecting moment of inertia and taking the torques to be equal, but they are not. They are negligibly different as you say...but they are still different.

So what equation indicates that the torque of the chain is little higher than the torque of static friction.
Your own first equation in post #52.

Chget
 
  • #97
Right but that doesn't express static friction solely in terms of the force of the chain and the two radii. It only arbitrarily shows that there is net torque but it doesn't show how a bigger gear would cause more static friction.
 
  • #98
UMath1 said:
Right but that doesn't express static friction solely in terms of the force of the chain and the two radii. It only arbitrarily shows that there is net torque but it doesn't show how a bigger gear would cause more static friction.
It leads to a set of simultaneous equations involving the mass of rider and bicycle and the moment of inertia of the front and rear wheels which can be solved to obtain a formula for the force of static friction in terms of the force of the chain, the two radii, the mass of the rider and bicycle and the moment of inertia of the front and rear wheels.
 
  • #99
How can I eliminate acceleration?

Screenshot_2015-08-18-17-01-48.png
 
  • #100
UMath1 said:
How can I eliminate acceleration?
Can you write an equation for the angular velocity of the wheel in terms of the linear velocity of the bicycle?
 
  • #101
W=v/r
 
  • #102
UMath1 said:
W=v/r
With some type setting, that's ##\omega = \frac{v}{r_{wheel}}## where ##\omega## is the angular velocity of the wheel, ##v## is the velocity of the bicycle and ##r_{wheel}## is the radius of the rear wheel.

Now can you use that to write down a formula for angular acceleration of the wheel in terms of the linear acceleration of the bicycle?
 
  • #103
Right..but I don't want angular acceleration or acceleration in the equation. It would alpha=a/r.
 
  • #104
UMath1 said:
Right..but I don't want angular acceleration or acceleration in the equation. It would alpha=a/r.
I am trying to lead you there. We'll get rid of the angular acceleration and the linear acceleration in due course. First I want you to write down the equation.
 
  • #105
Ok so did that
Screenshot_2015-08-18-17-32-52.png
 
  • #106
JBriggs444: I just wanted to let you know that I've already gone through all this stuff with UMath1 in posts #47 and #55. Please check it out to save yourself some effort.

Chet
 
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  • #107
UMath1 said:
Ok so did thatView attachment 87477
I'm not sure what your last line represents, since it is not an equation.
The middle two lines are equations, each involving the same two unknowns, FSF and a. Decide which unknown you are interested in determining. If it is the frictional force, use one of your equations to replace a in the other equation.
 
  • #108
The last line is a simplification of the middle line. Jbriggs444 was trying to help me eliminate "a" by using alpa=a/r. But I am not sure how to progress.
 
  • #109
UMath1 said:
The last line is a simplification of the middle line. Jbriggs444 was trying to help me eliminate "a" by using alpa=a/r. But I am not sure how to progress.
You could do worse than what I suggested in post #107. (107!)
 
  • #110
Worse?
 
  • #111
haruspex said:
You could do worse than what I suggested in post #107. (107!)
Hi haruspex,

I've been through all this already with the OP. Please see my post to jbriggs444 in post #106.

Chet
 
  • #112
UMath1 said:
Worse?
Perhaps English is not your first language?
Anyway, try what I suggested in post #107. I see Chet has already tried to help you through this. I don't understand why you are so reluctant to try it.
 
  • #113
So I did try that.
Screenshot_2015-08-19-14-45-43.png

But now I have staticfriction on both sides of the equation.
 
  • #114
Am I doing something wrong?
 
  • #115
UMath1 said:
So I did try that.
It would be much friendlier if you could transcribe that equation.

##F_{sf} = \frac{(T_u - T_l)R_g - M_{w+g}R_w\frac{F_{sf}}{M_{total}}}{R_w}##

But now I have staticfriction on both sides of the equation.

That equation looks complicated. But almost everything there is a constant. If you roll all of the constants together, the equation takes the general form:

##F_{sf} = k F_{sf} + c##

Could you solve that equation for ##F_{sf}##
 
  • #116
Would I need to integrate both sides?
 
  • #117
UMath1 said:
Would I need to integrate both sides?
Huh? Try subtracting ##k F_{sf}## from both sides to start.
 
  • #118
FSF=c/1-k
 
  • #119
UMath1 said:
FSF=c/1-k
That is correct -- or would be if we fix up the notation slightly. When written on one line in text form, the conventional interpretation of what you have written would be to divide c by 1 and then subtract k. i.e.

##F_{sf} = \frac{c}{1} - k##

What you want is

##F_{sf} = \frac{c}{1-k}##

Which you would write on one line in text form as Fsf = c/(1-k).

Now you just have to go back to your original equation...

##F_{sf} = \frac{(T_u - T_l)R_g - M_{w+g}R_w\frac{F_{sf}}{M_{total}}}{R_w}##

and perform the same manipulation you just did on the simplified version.
 
  • #120
So it would be

FSF= [(Tu−Tl)Rg]/ [Rw(1+(Mw+g/RwMtotal))]

Correct?
 
Last edited:

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