Bicycle Translational Acceleration vs Angular Acceleration

[UMath1]

So if I drawn the tension in the bottom chain in opposite direction for both diagrams, that would still be right?

And what is the next step in answering the inital question?

 

[Chestermiller]

So if I drawn the tension in the bottom chain in opposite direction for both diagrams, that would still be right?

Yes.

And what is the next step in answering the inital question?

The next step is to eliminate the term involving the chain tensions between the two equations and express the torque exerted by the ground on the wheel in terms of the torque exerted by the rider on the pedals (and the radii of the pedal sprocket and rear wheel sprocket).

Chet

 

[haruspex]

I think I understand that the tension in the upper chain would pull left in reaction to being pushed downward by the sprocket. But why should the lower chain also pull left? It would seem that it would react to the sprocket's force by pulling right.

Consider a stationary bicycle. The lower part of the chain must hang in a catenary between the sprockets. It is under tension. The derailleur spring takes up slack, increasing the tension so that the catenary is almost flat.
Now start turning the pedals. The front sprocket pushes out links of chain below it, but they don't fall straight down because of the tension in the lower chain. Meanwhile, the rear sprocket pulls against that tension, taking up the extra slack and largely maintaining the tension. The tension in the upper chain will now be higher than before, while the tension in the lower chain is less than before, but they are both still positive. It is the increased difference in the tensions that resists the torque of the pedals and turns the wheel.

 

[UMath1]

Ok so I did that.

Haruspex, what is a derailleur spring?

 

[Chestermiller]

Good. Now $τ_{SF}=R_WF$, where R_W is the radius of the rear wheel and F is the forward frictional force exerted by the ground on the rear wheel. This is what causes the combination of bike and rider to accelerate forward. Call M the combined mass of the bike and rider. What is the acceleration?

Chet

 

[UMath1]

F/M

 

[Chestermiller]

F/M

Yes, but please express F in this equation in terms of the torque in the pedal and the sprocket radii.

 

[UMath1]

 

[Chestermiller]

Great. I think this answers your original question, correct?

Chet

 

[haruspex]

Haruspex, what is a derailleur spring?

A derailleur is the most common form of gear changer. When you change gears, the chain needs to be a different length, so you need a way to take up slack in the chain. Derailleurs have an extra couple of sprockets on a spring-loaded arm.
For the purposes of the question, we could instead assume it's fixed gear, so no derailleur or spring. But the rest stays the same: the lower part of the chain is suspended between the two sprockets and is therefore under tension, just as a rope would be.

 

[UMath1]

Not exactly, this tells me what the static friction output would be for a given input pedal force. My question was that since static friction is the net force on the bike, even a minute force from the pedal would be sufficient to get the bike accelerating translationally. Then the issue was that to the bike to angularly accelerate the torque from ststic friction would need to be less than the torque of the chain. For that to happen, a much greater force would need to be applied. So how come both translational and angular acceleration coincide?

 

[haruspex]

For that to happen, a much greater force would need to be applied.

Why much greater? Demonstrate it with algebra.
Remember, the static friction starts off at zero. Only when a force is applied to the pedals does a static friction force arise. If only a small force is applied then the static friction force will be small, and its torque will be small.

 

[Chestermiller]

Yes. I agree with Haruspex. Static friction only tells you how much torque you can apply before the bike starts to peel out.

Chet

 

[UMath1]

 

[Chestermiller]

There's no friction for the chain. The chain openings mesh with the gear teeth on the sprockets. If the torque is too high, the chain just breaks.

Chet

 

[haruspex]

View attachment 87078

Yes, the torque from the chain will exceed the torque from the actual frictional force, but as we keep pointing out, that will generally be much less than the maximum frictional force. By what reasoning do you say it will exceed the maximum friction?

 

[Chestermiller]

View attachment 87078

You seem intent on including the rotational inertia of the rear wheel in your analysis. So let's now include it:
$$(T_U-T_L)R_g-τ_{SF}=Iα$$
where I is the moment of inertia of the rear wheel and α is its angular velocity. Since the vast majority of the mass of the rear wheel is concentrated at its outer periphery, its moment of inertia is given by:
$$I=mR_W^2$$
where m is the mass of the rear wheel and R_W is the rear wheel radius.
The angular acceleration of the rear wheel is related to the acceleration of the bike kinematically by $α=a/R_W$. So if we combine these equations, we obtain:
$$(T_U-T_L)R_g-τ_{SF}=mR_Wa$$
Why don't you now combine this with the torque equation for the pedal assembly by again eliminating the tension difference and see what you obtain for the final results for the acceleration. Also note that, if we include the inertia of the rear wheel, we probably ought to be doing moments about the front wheel, and including the frictional force and inertia of the front wheel too. Of course, the frictional force on the front wheel is in the direction opposite to that of the rear wheel, and is of much smaller magnitude.

Chet

 

[UMath1]

Yes, the torque from the chain will exceed the torque from the actual frictional force, but as we keep pointing out, that will generally be much less than the maximum frictional force. By what reasoning do you say it will exceed the maximum friction?

So I think my issue is a conceptual one. It was my understanding that as you increase the torque of the chain, you increase the force the wheel applies to the ground thereby increasing the force of static friction. So the issue was that since the force of static friction continues to rise as the more torque is applied, it continues to match the torque of the chain. Based on this understanding, the only way to get the wheel to angularly accelerate would be to increase the torque of the chain more than the maximum frictional torque.

But that obviously seems to not be the case. Could you explain the concept?

 

[Chestermiller]

So I think my issue is a conceptual one. It was my understanding that as you increase the torque of the chain, you increase the force the wheel applies to the ground thereby increasing the force of static friction. So the issue was that since the force of static friction continues to rise as the more torque is applied, it continues to match the torque of the chain. Based on this understanding, the only way to get the wheel to angularly accelerate would be to increase the torque of the chain more than the maximum frictional torque.

But that obviously seems to not be the case. Could you explain the concept?

That effect is contained in the moment balance equations that I wrote down in post # 47. So why don't you follow my advice in post #47 and complete the solution?

Chet

 

[Chestermiller]

So I think my issue is a conceptual one. It was my understanding that as you increase the torque of the chain, you increase the force the wheel applies to the ground thereby increasing the force of static friction. So the issue was that since the force of static friction continues to rise as the more torque is applied, it continues to match the torque of the chain. Based on this understanding, the only way to get the wheel to angularly accelerate would be to increase the torque of the chain more than the maximum frictional torque.

But that obviously seems to not be the case. Could you explain the concept?

You do understand that, in the limit of very low mass for the wheel (i.e., low moment of intertia of the wheel), the torque applied to the wheel by the chain only needs to differ from the torque applied by the ground to the wheel by a tiny amount in order to give the wheel any desired angular acceleration, correct?

Chet

 

[haruspex]

So I think my issue is a conceptual one. It was my understanding that as you increase the torque of the chain, you increase the force the wheel applies to the ground thereby increasing the force of static friction. So the issue was that since the force of static friction continues to rise as the more torque is applied, it continues to match the torque of the chain.

Yes, the static friction increases, but not so much as to match the torque from the chain. It lags behind. The difference of the two torques provides the angular acceleration of the wheel.
As Chet remarked, a cyclist with cleats could in theory apply so much torque to the pedals that the rear wheel skids, but my guess is it would require superhuman strength.

 

[UMath1]

Yes I do understand that.

I am still having some issues with the equation though. I can't seem to figure out how to eliminate the chains tension.

 

[Chestermiller]

Yes I do understand that.

I am still having some issues with the equation though. I can't seem to figure out how to eliminate the chains tension.View attachment 87196

Solve your second equation for (Tu - Tl), and then substitute it into your first equation.

By the way, do you really think that the rotational inertia of the pedal assembly is significant?

Chet

 

[UMath1]

So I solved for acceleration and this is what I obtained.

I am still not sure I understand the concept though. And the inertia of the pedal assembly is insignificant?

 

[Chestermiller]

So I solved for acceleration and this is what I obtained.
View attachment 87317
I am still not sure I understand the concept though.

That's because you didn't replace the torque that the ground exerts on the rear wheel τ_s in your equations with its value expressed in terms of the mass and acceleration of the rider and bike, and of the radius of the rear wheel:$$τ_S=MaR_W$$
where M is the total mass of bike plus rider.
If this is substituted into your equation, you get:
$$τ_p\frac{R_g}{R_s}-MaR_W=a(m_WR_W+m_pR_p)$$
where R_s is the radius of the pedal sprocket, and m_p and R_p are the mass and the radius of gyration of the pedal assembly, respectively. So, solving this equation for the acceleration gives:
$$a=\frac{τ_p}{R_W(M+m_W+m_pR_p/R_W)}\frac{R_g}{R_s}$$
I hope this makes more sense to you. It shows how the acceleration of the rider and bike is quantitatively related to the torque the rider puts on the pedal.

And the inertia of the pedal assembly is insignificant?

To get an idea of how significant the inertia of the pedal assembly is, ask yourself how much effort it would take to turn the pedal assembly if the chain were not attached.

Chet

 

[UMath1]

I still don't understand though. This proves that a minute amount of force from the is sufficient to get the bicycle to accelerate forward.

But what about the requirement for angular accleration. What is the relationship between the force applied on the pedal and the force the chain applies on the rear gear? What is the relationship between the force applied by the chain on the rear gear and force applied by static friction?

 

[Chestermiller]

I still don't understand though. This proves that a minute amount of force from the is sufficient to get the bicycle to accelerate forward.

But what about the requirement for angular accleration. What is the relationship between the force applied on the pedal and the force the chain applies on the rear gear? What is the relationship between the force applied by the chain on the rear gear and force applied by static friction?

This has already been included in the equations he have already derived. If you want the net force that the chain applies, you already have the relationships you need to determine this.

Chet

 

[UMath1]

Is there any way to express the force the chain applies only in terms of the pedal force? Using the equations we have, I can only express in terms of acceleration,
moment of inertia, and the pedal force.

 

[Chestermiller]

Is there any way to express the force the chain applies only in terms of the pedal force? Using the equations we have, I can only express in terms of acceleration,
moment of inertia, and the pedal force.

To get the acceleration out of there, combine the 2nd equation in post #52 with the third equation in post #55.

Chet

 

[UMath1]

So I did that and this is what I got.

Can I simplify this more? Is there a way I can find the force of the chain solely in terms of the force of the pedal and their respective radii? And likewise, is there a way to find the force of static fricition solely in terms of the force of the chain on the rear gear and the radii of the wheel and rear gear?