Bicycle Translational Acceleration vs Angular Acceleration

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
  • #1
UMath1
361
9

Homework Statement


Given: Wheel radius is 20 CM, Gear radius is 5 CM, Coefficient of Static Friction is .2, Weight on rear wheel is 50 N.

What is the minimum force that must be applied to the pedal for the wheel to begin accelerating on a level surface?

Homework Equations


Net T = I * a
T = R X F
Net F= m*a
0c02725f116ef3cad99527413171d4bb.png


The Attempt at a Solution



I know that for the wheel to begin angularly accelerating a net torque must be present, meaning that the torque from the chain on the wheel must exceed the torque from static friction on the wheel. So I calculated the maximum possible torque from static friction. First, using
0c02725f116ef3cad99527413171d4bb.png
, I find that the maximum force of static friction is 10 N. Then, using T = R X F, I find that the maximum torque from static friction is 2.5 N*m. This means that the torque from the must be greater than 2.5 N*m. Then, solving for Force in the equation T = R X F, I find that the force from the pedal must be more than 50 N.

That all seems fine until you consider the case for translational acceleration. The only net force acting on the wheel is the force of static friction (the force of the chain is canceled out by the force of the axle). Then for the wheel to begin translationally accelerating even a force of 1 N on the pedal would be sufficient. Going through the process above in reverse, I would find that the 1 N force on the pedal would result in a force of 0.2 N from static friction. Since this would be the net force, the wheel should begin translationally accelerating prior to rotational acceleration.

But obviously this cannot be the case. I have never seen a bike slide before it rolls. How is it that rotational and translational acceleration coincide?
 
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  • #2
Usually these problems say something about rolling without slipping.

If the wheels roll without slipping, and there is no rolling friction (only static), then any downward force on a pedal in the right direction will provide enough torque for forward acceleration.
 
  • #3
I don't understand, the static force needed on the wheel is 10N, to gt the torque why you multiplied the force with the gear radius but not with the wheel radius?

I am not an expert in this, but I think you need to find out the torque for the wheel radius, then put the value on the gear to get the force needed to be applied on the gear.

Note. I am not sure if the two torques will be same or proportional,
 
  • #4
I am quite new in solving Newtonian mathematics. However, I had a class on this topic about two weeks ago. There I learned whatever force you apply on a static object if the force does not exceed the limiting value of frictional force then it would not move. So, the bicycle will not move just with a force of 1N. If it gives 1N force, it will also get same value of frictional force to minimize until it exceeds the limiting value of frictional force. When it will reach that value it will tend to move, and after exceeding it will move.

It can be noted that kinetic frictional force is slight less than the limiting value of static frictional force
 
  • #5
Dr. Courtney said:
If the wheels roll without slipping, and there is no rolling friction (only static), then any downward force on a pedal in the right direction will provide enough torque for forward acceleration.

How is that possible? What about the torque from static friction? That goes in the counterclockwise direction.
 
  • #6
UMath1 said:
How is that possible? What about the torque from static friction? That goes in the counterclockwise direction.

I guess I was picturing the static friction between the tire and the ground with the rest of the system frictionless.

A coefficient of friction of 0.2 is way too big for the inner workings of a bicycle.
 
  • #7
Dr. Courtney said:
I guess I was picturing the static friction between the tire and the ground with the rest of the system frictionless.

A coefficient of friction of 0.2 is way too big for the inner workings of a bicycle.
Right...so how come the angular and translational acceleration of the bike coincide, when any minute force is enough to get a net force from static friction but significantly more force is needed to overcome the ccw torque of static friction?
 
  • #8
UMath1 said:
Right...so how come the angular and translational acceleration of the bike coincide, when any minute force is enough to get a net force from static friction but significantly more force is needed to overcome the ccw torque of static friction?
Your premise is false. Any arbitrarily small force is adequate to overcome the torque of static friction and produce a clockwise acceleration of the wheel.
 
  • #9
jbriggs444 said:
Your premise is false. Any arbitrarily small force is adequate to overcome the torque of static friction and produce a clockwise acceleration of the wheel.
Yes, but I feel that needs a bit more explanation.
Until a force is applied to the pedals there is no static frictional force. That force only arises in consequence of the force applied to the pedals, and only as necessary to prevent the tyre slipping on the ground. Because the bicycle is free to move forwards at the slightest horizontal force, the static frictional force does not increase enough to apply the same magnitude torque at the wheel's centre as is exerted by the pedals.
 
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  • #10
haruspex said:
Because the bicycle is free to move forwards at the slightest horizontal force, the static frictional force does not increase enough to apply the same magnitude torque at the wheel's centre as is exerted by the pedals.
Can you explain this in terms of the example? Was my calculation of the required 50 N to have rotational acceleration incorrect? If so, how do you calculate the force of static friction based on the force of the pedal?
 
  • #11
UMath1 said:

Homework Statement


Given: Wheel radius is 20 CM, Gear radius is 5 CM, Coefficient of Static Friction is .2, Weight on rear wheel is 50 N.

What is the minimum force that must be applied to the pedal for the wheel to begin accelerating on a level surface?
...
Is this the complete statement of the problem as it was given to you?

If not, please state the complete problem.
 
  • #12
I actually created this problem myself. I was just trying to understand this concept.
 
  • #13
UMath1 said:
Can you explain this in terms of the example? Was my calculation of the required 50 N to have rotational acceleration incorrect? If so, how do you calculate the force of static friction based on the force of the pedal?
Your calculation seems confused. At a guess, you were [almost] computing was the pedal force required to make the rear wheels slip.

Your calculation again:
So I calculated the maximum possible torque from static friction. First, using [PLAIN]https://upload.wikimedia.org/math/0/c/0/0c02725f116ef3cad99527413171d4bb.png, I find that the maximum force of static friction is 10 N.
So far, so good. The hardest the ground could possibly be pushing on the rear wheel is the contact force (50 N) times the coefficient of static friction (0.2) which yields 10N.

Then, using T = R X F, I find that the maximum torque from static friction is 2.5 N*m.
The relevant radius here is 20 cm = 0.2 meters.

Then, solving for Force in the equation T = R X F, I find that the force from the pedal must be more than 50 N.
The relevant R this time is the radius of the sprocket on the rear wheel. Let's say that's 5 cm or 0.05 meters. What you are solving for is the tension in the chain. In order to get the pedal force, you'd need to know the radius of the teeth on the front chain ring and the radius of the pedal from the center of the front chain ring.

But that would just tell you how hard you would have to pedal to start "burning rubber". Few cyclists accomplish that feat.
 
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  • #14
jbriggs444 said:
Your calculation seems confused. At a guess, you were [almost] computing was the pedal force required to make the rear wheels slip.

Your calculation again:

So far, so good. The hardest the ground could possibly be pushing on the rear wheel is the contact force (50 N) times the coefficient of static friction (0.2) which yields 10N.The relevant radius here is 20 cm = 0.2 meters.The relevant R this time is the radius of the sprocket on the rear wheel. Let's say that's 5 cm or 0.05 meters. What you are solving for is the tension in the chain. In order to get the pedal force, you'd need to know the radius of the teeth on the front chain ring and the radius of the pedal from the center of the front chain ring.

But that would just tell you how hard you would have to pedal to start "burning rubber". Few cyclists accomplish that feat.
Yes, I agree. To do this problem, you need to include the pedal gear/sprocket in the analysis. This transmits torque from the pedal to the chain, which then transmits torque to the rear sprocket. The top part of the chain is going to have a greater tension than the bottom part of the chain. The difference between these two tensions puts torque on the pedal sprocket and on the rear sprocket.

You need two free body diagrams to analyze this problem. Split the chain vertically (conceptually) half way between the front and rear sprockets. One free body diagram includes the rear half of the split chain and the rear sprocket. The other free body diagram includes the front half of the split chain and the front sprocket. All that is needed are moment balances about the rear wheel axis and about the pedal axis.

Chet
 
  • #15
I am kind of confused. My understanding was that the pedal sprocket applies a force to the chain and that same force is transmitted to the rear gear. Doing a moment balance would mean that the force transmitted to the read gear would be less the force applied on the chain. Why should the torques equal?

I did do a FBD like you said though. Is this correct?
Screenshot_2015-08-05-22-39-15.png
 
  • #16
UMath1 said:
My understanding was that the pedal sprocket applies a force to the chain and that same force is transmitted to the rear gear.
Why should the torques equal?
Chet didn't say the torques would be equal. Your understanding is correct. But the force exerted on the pedal will be different from the tension in the chain.
What is the relationship between the force on the pedal and the torque on the rear wheel?
UMath1 said:
how do you calculate the force of static friction based on the force of the pedal?
You need to introduce an unknown for the acceleration and solve the equations. You need to know the whole mass, not just the weight on the rear wheel.
 
  • #17
UMath1 said:
I am kind of confused. My understanding was that the pedal sprocket applies a force to the chain and that same force is transmitted to the rear gear. Doing a moment balance would mean that the force transmitted to the read gear would be less the force applied on the chain. Why should the torques equal?

The torques are not equal.

I did do a FBD like you said though. Is this correct?View attachment 86916
Your FBD is correct except for one thing. You omitted the torque applied to the rear wheel by the ground.

Now that you've drawn this free body diagram, you need to split it in half into two separate free body diagrams at the location where you have your vertical split. You then need to do a moment balance about the pedal axle and about the rear wheel axle.

I'm going to do it for the pedal axle, and leave it up to you to do it for the rear wheel axle.

Let τP be the torque applied by the rider to the pedal
Let TU be the tension in the upper part of the chain
Let TL be the tension in the lower part of the chain
Let RP be the radius of the pedal sprocket

Then

$$τ_P-(T_U-T_L)R_P=0$$

In doing the moment balance about the rear axle, start out by neglecting the rotational inertia of the rear wheel. This will make the problem much simpler and, in my judgement, is a pretty good approximation. Once you have used the two moment balances to establish the relationship between the applied torque on the pedal and the torque exerted by the ground on the rear wheel, you can then apply the force balance in the horizontal direction to get the acceleration of the bike.

Also note that another relationship you can derive from your free body diagram is the kinematic relationship between the angular velocity of the pedal and the angular velocity of the rear wheel.

Chet
 
  • #18
I am not sure I understand. Why does there have to be a moment balance? The bicyclist applies a downward force on the pedal. This force is a torque because the radius of the pedal sprocket is a moment arm. The pedal sprocket then applies a force on the chain, but that force has no moment arm right?
 
  • #19
The bicycle is not spinning. Chet has wisely suggested that we neglect the angular momentum of the spinning wheel. So total angular momentum must be constant. So the external moments must balance. In fact, no matter what piece of the bicycle we consider, the external moments on that piece must balance.
 
  • #20
UMath1 said:
I am not sure I understand. Why does there have to be a moment balance?
When you analyze a system involving rigid body rotations, you need to employ moment balances. In this problem, the rotating rigid bodies involved are the
  • front pedal / front sprocket assembly
  • rear wheel / rear sprocket assembly
The bicyclist applies a downward force on the pedal. This force is a torque because the radius of the pedal sprocket is a moment arm. The pedal sprocket then applies a force on the chain, but that force has no moment arm right?
Does the upper part of the chain exert a tangential (tensile) force on the front sprocket? Does the lower part of the chain exert a tangential (tensile) force on the front sprocket? Do these forces exert moments on the front sprocket? What is the moment arm for these forces? You would have seen all this if you had done the split FBDs that I recommended doing.

Chet
 
  • #21
Is this correct? My equation for net torque pertains to the rear gear.

Screenshot_2015-08-06-18-01-04.png


I am not sure if my force diagram is correct for the pedal sprocket. It seems as if Tu, Tl, and the force of the pedal create torques in the same direction.
 
  • #22
UMath1 said:
I am not sure if my force diagram is correct for the pedal sprocket. It seems as if Tu, Tl, and the force of the pedal create torques in the same direction.
In what direction does TL pull on the rear sprocket?
 
  • #23
UMath1 said:
Is this correct? My equation for net torque pertains to the rear gear.

View attachment 86976

I am not sure if my force diagram is correct for the pedal sprocket. It seems as if Tu, Tl, and the force of the pedal create torques in the same direction.
You've done the rear wheel correctly, once you correct the error that jbriggs444 pointed out, and you set the right hand side to zero (since we are neglecting the rotational inertia of the rear wheel). The moment balance I wrote down for the pedal was correct. So the next step is to combine these two moment balance equations to eliminate (TU - TL), which appears in both equations.

Chet
 
  • #24
This seems counterintuitive to me. Wouldn't the entireity of the chain pull in the same direction? Wouldn't it be pulling against itself otherwise?

And for the pedal sprocket are the directions in my FBD switched? As per your equation, the resultant torque from the chain must be counterclockwise because the torque from the pedal is clockwise.
 
  • #25
UMath1 said:
This seems counterintuitive to me. Wouldn't the entireity of the chain pull in the same direction? Wouldn't it be pulling against itself otherwise?
A chain can only pull, not push. Both upper and lower parts must be in tension, though the tension in the lower part can be quite small. All that will matter is the difference in the two tensions. The exact tensions cannot be deduced without further information, such as the derailleur spring settings.
UMath1 said:
And for the pedal sprocket are the directions in my FBD switched?
Only in the lower part of the chain.
 
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  • #26
UMath1 said:
This seems counterintuitive to me. Wouldn't the entireity of the chain pull in the same direction? Wouldn't it be pulling against itself otherwise?
Counterintuitive is one of those "in the eye of the beholder" kind of things. None of what we have said is counterintuitive to jbriggs444, haruspex, and me. As haruspex has indicated, both parts of the chain are under tension, so they are actually pulling against each other. The top half of the chain is just pulling harder than the bottom half.
And for the pedal sprocket are the directions in my FBD switched?
No. The only thing that is wrong about your FBD is that force exerted by the upper part of the chain on the pedal sprocket is drawn in the wrong direction.
As per your equation, the resultant torque from the chain must be counterclockwise because the torque from the pedal is clockwise.
Yes. The sum of the two torques must be zero. There is no net torque about the pedal axis. That's what we call a balance of moments.

Chet
 
  • #27
haruspex said:
Only in the lower part of the chain.
I think you meant that, in post #21, the force exerted by the upper part of the chain on the pedal sprocket is drawn in the wrong direction.

Chet
 
  • #28
Chestermiller said:
I think you meant that, in post #21, the force exerted by the upper part of the chain on the pedal sprocket is drawn in the wrong direction.

Chet
Sorry, yes - for some reason I was thinking in terms of the force on the chain, not the force on the sprocket.
 
  • #29
Is this right?
Screenshot_2015-08-07-14-59-15.png


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.

Also, are we assuming that the bike is traveling at a constant velocity? Otherwise the moments wouldn't balance, correct?
 
  • #30
UMath1 said:
Is this right?
View attachment 87011

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.
It doesn't matter what direction you draw the chain forces as long as they are consistent between the front sprocket and the rear sprocket. In your most recent diagram, they are consistent.
Also, are we assuming that the bike is traveling at a constant velocity? Otherwise the moments wouldn't balance, correct?
If we neglect the rotational inertia of the rear wheel and of the pedal assembly, then the moments must balance even if the bike is accelerating. (If I = 0, then Iα = 0).

Chet
 
  • #31
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?
 
  • #32
UMath1 said:
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
 
  • #33
UMath1 said:
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.
 
  • #34
Ok so I did that.
Screenshot_2015-08-08-09-17-28.png


Haruspex, what is a derailleur spring?
 
  • #35
Good. Now ##τ_{SF}=R_WF##, where RW 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
 

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