# Bicycle Translational Acceleration vs Angular Acceleration

• UMath1
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
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

jbriggs444
UMath1 said:
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, FSF 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.

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.

UMath1 said:
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!)

Worse?

haruspex said:
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

UMath1 said:
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.

So I did try that.

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

Am I doing something wrong?

UMath1 said:
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}##

Would I need to integrate both sides?

UMath1 said:
Would I need to integrate both sides?
Huh? Try subtracting ##k F_{sf}## from both sides to start.

FSF=c/1-k

UMath1 said:
FSF=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 Fsf = 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.

So it would be

FSF= [(Tu−Tl)Rg]/ [Rw(1+(Mw+g/RwMtotal))]

Correct?

Last edited:
This thread is in an endless loop, and is going nowhere. I am temporarily closing it.

Jbriggs444, haruspex, and I have agreed to collaborate on developing a complete analysis of this problem, which we will post once we have completed it. Immediately after that, the thread will be permanently closed. Hopefully, this will answer all UMath1's questions, and provide all the answers he/she desires. If not, all we can say is that we've done the best we could. There will be no further responses allowed in this thread.

Chet

This Bicycle Mechanics Analysis was developed by collaboration between jbriggs444, haruspex, and chestermiller. We all concur on its contents.

Moment Balance on Rear Wheel Assembly:

$$(T_U-T_L)R_{RS}-F_{SR}R_W=I_Rα\tag{1}$$
where

##T_U## = Tension in upper part of the chain
##T_L## = Tension in lower part of the chain
##R_{RS}## = Radius of rear sprocket
##F_{SR}## = Static friction force exerted by ground on rear wheel in forward direction
##I_R## =Moment of inertia of rear wheel assembly
##α## = Angular acceleration of wheels

Moment of Inertia of Rear Wheel
$$I_R=m_RR_W^2\tag{2}$$
where

##m_R## = mass of rear wheel assembly

Eqn. 2 makes the reasonable approximation that all the mass of the wheel assembly is concentrated at the rim.

Kinematic Equation Between Angular Acceleration of Wheels and Forward Acceleration of Bike
$$α=a/R_W\tag{3}$$
where a is the acceleration of the bike and rider.

Combining Eqns. 1-3 for Rear Wheel Assembly Moment Balance
$$(T_U-T_L)R_{RS}-F_{SR}R_W=m_RR_Wa\tag{4}$$

Moment Balance on Pedal Assembly (neglecting rotational inertia of pedal sprocket assembly)
$$τ_P=(T_U-T_L)R_{PS}\tag{5}$$
where

##τ_P=## = Torque applied by rider to pedal sprocket
##R_{PS}## = Radius of pedal sprocket

Moment Balance on Front Wheel Assembly
$$F_{SF}R_W=m_FR_Wa\tag{6}$$
where

##F_{SF}## = Static friction force exerted by ground on front wheel in backward direction
##m_F## = mass of front wheel assembly

Overall Force Balance on Bike and Rider
$$F_{SR}-F_{SF}=Ma\tag{7}$$
where

##M## = combined mass of bike and rider

Solution of Equations and Results
Equations 4-7 constitute a complete mathematical statement of our bicycle mechanics model. Our objective now will to be to use these equations to derive relationships for the:
• acceleration of the bike expressed as a function of the pedal torque applied by the rider
• static friction force on the rear wheel expressed as a function of the pedal torque applied by the rider
• static friction force on the rear wheel expressed as a function of the tension difference between the upper part of the chain and the lower part of the chain
If we combine Eqns. 4 and 5 to eliminate the tension difference between upper and lower portions of the chain, we obtain:
$$τ_P\frac{R_{RS}}{R_{PS}}-F_{SR}R_W=m_RR_Wa\tag{8}$$
Combining this with Eqns. 6 and 7 yields:
$$(M+m_R+m_F)R_Wa=τ_P\frac{R_{RS}}{R_{PS}}\tag{9}$$
Note from Eqn. 9 that including the rotational inertia of the wheel assemblies in the analysis is equivalent to increasing the total mass of the bike and rider (which already includes the mass of the wheel assemblies) by the mass of the wheel assemblies a second time. If we solve equation for the acceleration a, we obtain:
$$a=\frac{τ_P}{R_W(M+m_R+m_F)}\frac{R_{RS}}{R_{PS}}\tag{10}$$
If we substitute Eqn. 10 into Eqn. 8, we obtain the following relationship for the static frictional force on the rear wheel as a function of the torque applied by the rider on the pedal:
$$F_{SR}=\frac{τ_P}{R_W}\frac{R_{RS}}{R_{PS}}\frac{(M+m_F)}{(M+m_R+m_F)}\tag{11}$$
We can also express the static frictional force in terms of the tension difference between the upper part of the chain and the lower part of the chain (for whatever that's worth) by solving Eqn. 5 for ##τ_P## and substituting the result into Eqn. 11 to obtain:
$$F_{SR}=\frac{(T_U-T_L)R_{RS}}{R_W}\frac{(M+m_F)}{(M+m_R+m_F)}\tag{12}$$

The mass ratio term in Eqn. 12 captures the effect haruspex was referring to in his responses when he indicated that the frictional torque "lags" behind the propulsive torque.

This completes our analysis.

jbriggs444, haruspex, chestermiller

artyb

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