- #36
UMath1
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F/M
Bicycle Translational Acceleration vs Angular Acceleration
- Thread starter UMath1
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In summary, to calculate the minimum force needed for a bicycle wheel to begin accelerating on a level surface, you need to consider the net torque present. The torque from the chain on the wheel must exceed the torque from static friction on the wheel. Using T = R X F, it can be determined that the force from the pedal must be more than 50 N. However, this calculation does not take into account the translational acceleration of the bike. As the bike is free to move forwards with the slightest horizontal force, the static frictional force does not increase enough to apply the same magnitude torque at the wheel's center as is exerted by the pedals. Therefore, any arbitrarily small force applied to the pedals is enough to produce a clockwise
- #37
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Yes, but please express F in this equation in terms of the torque in the pedal and the sprocket radii.UMath1 said:
- #38
UMath1
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- #39
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Great. I think this answers your original question, correct?
Chet
Chet
- #40
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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.UMath1 said:
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.
- #41
UMath1
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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?
- #42
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Why much greater? Demonstrate it with algebra.UMath1 said:
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.
- #43
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Yes. I agree with Haruspex. Static friction only tells you how much torque you can apply before the bike starts to peel out.
Chet
Chet
- #44
UMath1
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- #45
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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
Chet
- #46
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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?UMath1 said:
- #47
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You seem intent on including the rotational inertia of the rear wheel in your analysis. So let's now include it:UMath1 said:
$$(T_U-T_L)R_g-τ_{SF}=Iα$$
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 RW 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
- #48
UMath1
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haruspex said:
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?
- #49
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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?UMath1 said:
Chet
- #50
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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?UMath1 said:
Chet
- #51
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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.UMath1 said:
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.
- #52
UMath1
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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.
I am still having some issues with the equation though. I can't seem to figure out how to eliminate the chains tension.
- #53
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Solve your second equation for (Tu - Tl), and then substitute it into your first equation.UMath1 said:
By the way, do you really think that the rotational inertia of the pedal assembly is significant?
Chet
- #54
UMath1
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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?
I am still not sure I understand the concept though. And the inertia of the pedal assembly is insignificant?
- #55
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That's because you didn't replace the torque that the ground exerts on the rear wheel τs 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:$$τ_S=MaR_W$$UMath1 said:
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 Rs is the radius of the pedal sprocket, and mp and Rp 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.
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
- #56
UMath1
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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?
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?
- #57
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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.UMath1 said:
Chet
- #58
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To get the acceleration out of there, combine the 2nd equation in post #52 with the third equation in post #55.UMath1 said:
Chet
- #60
UMath1
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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?
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?
- #61
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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?UMath1 said:
Chet
- #62
UMath1
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The sum
- #63
UMath1
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I mean the net force of the upper and lower tensions. Tu-Tl
- #64
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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.UMath1 said:
Chet
- #65
UMath1
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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.
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.
- #66
jbriggs444
Science Advisor
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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_gearsChestermiller said:
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.
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You already have that if you are willing to make the reasonable assumption that the rotational inertia of the pedal assembly is negligible.UMath1 said:
You already have to equations necessary to determine that.
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?
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
- #68
UMath1
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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.
- #69
jbriggs444
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UMath1 said:
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?
- #70
UMath1
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Rolling resistance