# Bicycle Translational Acceleration vs Angular Acceleration

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1. Jul 31, 2015

### UMath1

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Net T = I * a
T = R X F
Net F= m*a

3. 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 , 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?

2. Jul 31, 2015

### Dr. Courtney

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. Jul 31, 2015

### fireflies

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. Jul 31, 2015

### fireflies

I am quite new in solving Newtonian mathematics. However, I had a class on this topic about two weeks ago. There I learnt 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. Jul 31, 2015

### UMath1

How is that possible? What about the torque from static friction? That goes in the counterclockwise direction.

6. Jul 31, 2015

### Dr. Courtney

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. Jul 31, 2015

### UMath1

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. Jul 31, 2015

### jbriggs444

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. Jul 31, 2015

### haruspex

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.

10. Jul 31, 2015

### UMath1

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. Jul 31, 2015

### SammyS

Staff Emeritus
Is this the complete statement of the problem as it was given to you?

If not, please state the complete problem.

12. Jul 31, 2015

### UMath1

I actually created this problem myself. I was just trying to understand this concept.

13. Jul 31, 2015

### jbriggs444

Your calculation seems confused. At a guess, you were [almost] computing was the pedal force required to make the rear wheels slip.

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.

Last edited by a moderator: May 7, 2017
14. Jul 31, 2015

### Staff: Mentor

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. Aug 5, 2015

### UMath1

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?

16. Aug 5, 2015

### haruspex

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?
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. Aug 6, 2015

### Staff: Mentor

The torques are not equal.

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. Aug 6, 2015

### UMath1

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. Aug 6, 2015

### jbriggs444

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. Aug 6, 2015

### Staff: Mentor

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
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