Bicycle Translational Acceleration vs Angular Acceleration

[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

 

[Chestermiller]

JBriggs444: I just wanted to let you know that I've already gone through all this stuff with UMath1 in posts #47 and #55. Please check it out to save yourself some effort.

Chet

 

[haruspex]

Ok so did thatView attachment 87477

I'm not sure what your last line represents, since it is not an equation.
The middle two lines are equations, each involving the same two unknowns, F_SF and a. Decide which unknown you are interested in determining. If it is the frictional force, use one of your equations to replace a in the other equation.

 

[UMath1]

The last line is a simplification of the middle line. Jbriggs444 was trying to help me eliminate "a" by using alpa=a/r. But I am not sure how to progress.

 

[haruspex]

The last line is a simplification of the middle line. Jbriggs444 was trying to help me eliminate "a" by using alpa=a/r. But I am not sure how to progress.

You could do worse than what I suggested in post #107. (107!)

 

[UMath1]

Worse?

 

[Chestermiller]

You could do worse than what I suggested in post #107. (107!)

Hi haruspex,

I've been through all this already with the OP. Please see my post to jbriggs444 in post #106.

Chet

 

[haruspex]

Worse?

Perhaps English is not your first language?
Anyway, try what I suggested in post #107. I see Chet has already tried to help you through this. I don't understand why you are so reluctant to try it.

 

[UMath1]

So I did try that.

But now I have staticfriction on both sides of the equation.

 

[UMath1]

Am I doing something wrong?

 

[jbriggs444]

So I did try that.

It would be much friendlier if you could transcribe that equation.

$F_{sf} = \frac{(T_u - T_l)R_g - M_{w+g}R_w\frac{F_{sf}}{M_{total}}}{R_w}$

But now I have staticfriction on both sides of the equation.

That equation looks complicated. But almost everything there is a constant. If you roll all of the constants together, the equation takes the general form:

$F_{sf} = k F_{sf} + c$

Could you solve that equation for $F_{sf}$

 

[UMath1]

Would I need to integrate both sides?

 

[jbriggs444]

Would I need to integrate both sides?

Huh? Try subtracting $k F_{sf}$ from both sides to start.

 

[UMath1]

F_SF=c/1-k

 

[jbriggs444]

F_SF=c/1-k

That is correct -- or would be if we fix up the notation slightly. When written on one line in text form, the conventional interpretation of what you have written would be to divide c by 1 and then subtract k. i.e.

$F_{sf} = \frac{c}{1} - k$

What you want is

$F_{sf} = \frac{c}{1-k}$

Which you would write on one line in text form as F_sf = c/(1-k).

Now you just have to go back to your original equation...

$F_{sf} = \frac{(T_u - T_l)R_g - M_{w+g}R_w\frac{F_{sf}}{M_{total}}}{R_w}$

and perform the same manipulation you just did on the simplified version.

 

[UMath1]

So it would be

F_SF= [(Tu−Tl)Rg]/ [Rw(1+(M_w+g/RwM_total))]

Correct?