Bicycle Translational Acceleration vs Angular Acceleration

[jbriggs444]

Rolling resistance

Any other forces?

 

[UMath1]

static friction of front wheel

I am considering the bike and the rider one system

 

[jbriggs444]

static friction of front wheel

I am considering the bike and the rider one system

Any other forces?

 

[UMath1]

Air resistance

 

[jbriggs444]

Air resistance

Any other forces?

 

[UMath1]

I can't think of any

 

[jbriggs444]

I can't think of any

Gravity? The normal force of the pavement on the front tire? The normal force of the pavement on the rear tire?

 

[UMath1]

Right. Yes, gravity and normal force, but they cancel out, right?

 

[jbriggs444]

Right. Yes, gravity and normal force, but they cancel out, right?

For some possible calculations, the force from gravity and the normal force on the two tires will cancel out, yes. But if we are trying to be careful, we should list them first and argue that they cancel for the particular calculation we are trying to perform afterward. For instance, we could do a torque balance on the bicycle as a whole to see how large static friction could be before the bicycle pulls a "wheelie". Gravity and the normal forces would be crucial for that.

But back to the problem at hand. You were saying that the bicycle as a whole cannot accelerate unless

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.

We have this free body diagram. It has forces from static friction at the rear wheel, static friction at the front wheel, rolling resistance, air resistance, gravity and the normal force on the two tires.

We can neglect gravity and the normal force on the two tires since they are vertical and we are considering horizontal movement. We can neglect air resistance because it is approximately zero for a near-motionless bicycle in still air. We can neglect rolling resistance as long as the bearings are well oiled and the tires are properly inflated.

Do you agree that we can neglect the force of static friction between pavement and front wheel? Or should we calculate that?

 

[UMath1]

I think it could be neglected for the sake of simplicity. Even if it were a unicycle, you still wouldn't have to break the maximum force of static friction.

 

[jbriggs444]

I think it could be neglected for the sake of simplicity. Even if it were a unicycle, you still wouldn't have to break the maximum force of static friction.

So you now agree that the force of static friction does not need to exceed the maximum force from static friction in order for the bicycle to accelerate?

 

[UMath1]

I agree because otherwise you'd run into the original issue of sliding before rolling. However I don't understand why. If the equation is not (Tu-Tl)Rg/Rw, what is it?

I originally thought the equation was (Tu-Tl)Rg/Rw because it would show that a bigger gear would give more traction.

 

[jbriggs444]

I agree because otherwise you'd run into the original issue of sliding before rolling. However I don't understand why. If the equation is not (Tu-Tl)Rg/Rw, what is it?

You still have not told us what makes you think that $(T_u - T_l)\frac{R_g}{R_w}$ has to be incorrect.

 

[UMath1]

It has to be incorrect because that would mean that static friction keeps growing as the torque from the chain grows. It would keep growing until it reaches the maximum value. But I have been told repeatedly that reaching the maximum value is unnecessary and that would be "burning rubber".

Also, it brings forth another problem, the equation cannot possibly be applicable after the maximum friction is reached because the friction cannot increase beyond that ppint.

 

[jbriggs444]

It has to be incorrect because that would mean that static friction keeps growing as the torque from the chain grows. It would keep growing until it reaches the maximum value. But I have been told repeatedly that reaching the maximum value is unnecessary and that would be "burning rubber".

And what makes you think that the torque from the chain grows and grows without bound?

 

[UMath1]

It has to in order give a torque greater than the torque of static friction according to the equation.

 

[Chestermiller]

I must be missing something. You apply a torque to the pedal, and it causes a corresponding torque to be applied by the ground to the rear wheel (to accelerate the bike). You hold the torque on the pedal constant while the bike is accelerating, and the corresponding torque applied by the ground to the rear wheel stays constant. So, what's the problem?

Chet

 

[UMath1]

In order for the wheel to rotationally accelerate the torque of the chain MUST be greater than the torque of static friction. Based on that equation the only way that can happen is if maximum static friction value is attained. Otherwise, static frictions torque would continue to match the torque of the chain, meaning no angular acceleration.

 

[jbriggs444]

In order for the wheel to rotationally accelerate the torque of the chain MUST be greater than the torque of static friction.

If you recall, we assumed that the wheel has negligible moment of inertia in order to simplify the calculations. Accordingly, it takes negligible net torque for it to accelerate.

If you want to re-do the calculations with a non-negligible moment of inertia in the rear wheel then we can do so.

 

[Chestermiller]

In order for the wheel to rotationally accelerate the torque of the chain MUST be greater than the torque of static friction. Based on that equation the only way that can happen is if maximum static friction value is attained. Otherwise, static frictions torque would continue to match the torque of the chain, meaning no angular acceleration.

If the rotational inertia of the rear wheel is neglected, that equation tells you that the chain torque is equal to the static friction torque. If the rotational inertial of the rear wheel is not neglected, then the chain torque has to be a little higher than the static friction torque, but part of the chain torque is consumed in accelerating the rear wheel, so the static friction torque is the same.

Chet

 

[UMath1]

Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached

 

[Chestermiller]

If you recall, we assumed that the wheel has negligible moment of inertia in order to simplify the calculations. Accordingly, it takes negligible net torque for it to accelerate.

If you want to re-do the calculations with a non-negligible moment of inertia in the rear wheel then we can do so.

We've already done that in previous posts.

Chet

 

[Chestermiller]

Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached

Huh?

 

[jbriggs444]

Even if it takes a neglible net torque, it still requires one torque to be greater than the other. With the equation (Tu-Tl)Rg/Rw, the torques will be equal until the maximum static friction is reached

I want to echo Chet's puzzlement. The torques will be negligibly different until maximum static friction is reached. [At which time the angular acceleration of the wheel becomes significant in spite of the wheel's neglibible moment of inertia]

 

[UMath1]

The equation is based on neglecting moment of inertia and taking the torques to be equal, but they are not. They are negligibly different as you say...but they are still different.

So what equation indicates that the torque of the chain is little higher than the torque of static friction.

 

[Chestermiller]

The equation is based on neglecting moment of inertia and taking the torques to be equal, but they are not. They are negligibly different as you say...but they are still different.

So what equation indicates that the torque of the chain is little higher than the torque of static friction.

Your own first equation in post #52.

Chget

 

[UMath1]

Right but that doesn't express static friction solely in terms of the force of the chain and the two radii. It only arbitrarily shows that there is net torque but it doesn't show how a bigger gear would cause more static friction.

 

[jbriggs444]

Right but that doesn't express static friction solely in terms of the force of the chain and the two radii. It only arbitrarily shows that there is net torque but it doesn't show how a bigger gear would cause more static friction.

It leads to a set of simultaneous equations involving the mass of rider and bicycle and the moment of inertia of the front and rear wheels which can be solved to obtain a formula for the force of static friction in terms of the force of the chain, the two radii, the mass of the rider and bicycle and the moment of inertia of the front and rear wheels.

 

[UMath1]

How can I eliminate acceleration?

 

[jbriggs444]

How can I eliminate acceleration?

Can you write an equation for the angular velocity of the wheel in terms of the linear velocity of the bicycle?

 

[UMath1]

W=v/r

 

[jbriggs444]

W=v/r

With some type setting, that's $\omega = \frac{v}{r_{wheel}}$ where $\omega$ is the angular velocity of the wheel, $v$ is the velocity of the bicycle and $r_{wheel}$ is the radius of the rear wheel.

Now can you use that to write down a formula for angular acceleration of the wheel in terms of the linear acceleration of the bicycle?

 

[UMath1]

Right..but I don't want angular acceleration or acceleration in the equation. It would alpha=a/r.

 

[jbriggs444]

Right..but I don't want angular acceleration or acceleration in the equation. It would alpha=a/r.

I am trying to lead you there. We'll get rid of the angular acceleration and the linear acceleration in due course. First I want you to write down the equation.

 

[UMath1]

Ok so did that