Bicycle Translational Acceleration vs Angular Acceleration

[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?

 

[Chestermiller]

So I did that and this is what I got.
View attachment 87442
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?

When you ask about the force of the chain, are you referring to the force in the upper part of the chain, the force in the lower part of the chain, their difference, or their sum?

Chet

 

[UMath1]

The sum

 

[UMath1]

I mean the net force of the upper and lower tensions. Tu-Tl

 

[Chestermiller]

The sum

The vector sum of the tensions in the upper and lower parts of the chain are determined mainly be the preloading of the chain. Prior to riding, the tensions in both sections of the chain are equal, and pulling forward on the rear wheel assembly and backward on the front pedal assembly. When the bike is accelerating, the tension in the upper part of the chain increases, while the tension in the lower part of the chain decreases. However, the lower part of the chain does not go into compression. So, during acceleration, the tensions in the two sections are still pointing in the same direction. In my judgement, the resultant force they impose on the rear wheel does not change significantly, nor does the resultant force they impose on the front pedal assembly. However, the moments that they impose on the rear wheel assembly and on the front pedal assembly change substantially. This is because of the difference in tension that develops between the upper portion of the chain and the lower portion.

Chet

 

[UMath1]

Right. So, is there a way to calculate the difference in the two force based soley on the applied force on the pedal and the radii of the pedal and pedal sprocket? Similarly is a way to calculate the forcd of static friction soley based on this difference of the two tensions and the radii of the rear gear and rear wheel?

I am trying to find proof for why, as haruspex said, the torque of static friction lags behind the torque the chain puts on the wheel.

Earlier my calculation for the force of static friction was: (Tu-Tl)Rg/Rw. But this cannot be the case, because it were you'd have break the maximum force of static friction to get the bike to accelerate.

 

[jbriggs444]

The vector sum of the tensions in the upper and lower parts of the chain are determined mainly be the preloading of the chain. Prior to riding, the tensions in both sections of the chain are equal, and pulling forward on the rear wheel assembly and backward on the front pedal assembly. When the bike is accelerating, the tension in the upper part of the chain increases, while the tension in the lower part of the chain decreases.

This is not correct for real bicycles and real chains. For a typical multi-speed bicycle using derailleur gears the bottom portion of the chain is kept at a low and more-or-less fixed tension with a spring-loaded tensioner. You can see several relevant pictures here. https://en.wikipedia.org/wiki/Derailleur_gears

For a three-speed bicycle or a single-speed bicycle pedaled in the forward direction, the situation is similar. The lower chain is essentially slack. The only tension is that resulting from its sag under gravity.

[If the frame were perfectly rigid and the chain were perfectly unstretchable and there were no slack then the situation would be statically indeterminate with no way to determine tension in both chain segments and compression in the frame, even given perfect information about the bicycle. The slack in the bottom chain avoids this difficulty]

What changes when you pedal harder is the tension in the top chain and the compression in the parallel frame member. But the point of application of the force from the frame is exactly at the two axes of rotation -- the hub of the rear wheel and the hub of the crank. So frame compression is irrelevant when computing torques.

 

[Chestermiller]

Right. So, is there a way to calculate the difference in the two force based soley on the applied force on the pedal and the radii of the pedal and pedal sprocket?

You already have that if you are willing to make the reasonable assumption that the rotational inertia of the pedal assembly is negligible.

Similarly is a way to calculate the forcd of static friction soley based on this difference of the two tensions and the radii of the rear gear and rear wheel?

You already have to equations necessary to determine that.

I am trying to find proof for why, as haruspex said, the torque of static friction lags behind the torque the chain puts on the wheel.

The lag is probably not very important, but probably depends on how long it takes the tire and chain to deform. But why don't you just ask him?

Earlier my calculation for the force of static friction was: (Tu-Tl)Rg/Rw. But this cannot be the case, because it were you'd have break the maximum force of static friction to get the bike to accelerate.

On what basis do you say this? Did you divide the force of static friction that you calculate by the normal force (which is on the order of the weight of the rider plus bike) and compare the quotient to the coefficient of static friction?

Chet

 

[UMath1]

No, I thought the force the bike wheel applies on the ground could be obtained by a torque balance and hence the equation: (Tu-Tl)Rg/Rw. Then by Newtons 3rd Law the force static friction applies should be the same.

 

[jbriggs444]

No, I thought the force the bike wheel applies on the ground could be obtained by a torque balance and hence the equation: (Tu-Tl)Rg/Rw. Then by Newtons 3rd Law the force static friction applies should be the same.

Right. The force of static friction on the bike's rear wheel is given by this formula. Draw a free body diagram for the bicycle as a whole. You have the value for the force from static friction acting at the rear wheel. What other external forces is the bicycle subject to?

 

[UMath1]

Rolling resistance