# Homework Help: Max acceleration of a racecar

1. Jul 3, 2017

### decerto

1. The problem statement, all variables and given/known data
http://imgur.com/a/vA0H2

2. Relevant equations

$$F=ma$$
$$r \times F = I \alpha$$

3. The attempt at a solution

Denoting the forces at the back wheel with a 1 and front wheel with a 2 and calculating the torques around the CoM I have

$$F-f_1-f_2 = mA$$
$$N_1 + N_2 -mg = 0$$
$$hf_1 + h_f2 - hF +bN_1-aN_2=0$$

with $$f_i = \mu_i N_i$$

So just using this I can't seem to solve for A in terms of mu and g. I have some questions though, is mu the same for both wheels? The also mentions "rolling without slipping", am I meant I incorporate this somehow?, it doesn't give any explicit info and moments of inertias of the wheels so I don't see how.

Any help would be appreciated.

Last edited by a moderator: Jul 3, 2017
2. Jul 3, 2017

### jack action

The torque is applied to the rear wheels only and the resultant force of that torque is $F$. By definition, the force $F$ is also the friction force reacting to that torque. There is nothing on the front wheels other than the normal reaction to the weight $Mg$.

3. Jul 3, 2017

### Staff: Mentor

The coefficient of friction for the front wheel does not matter.

If the back wheel tries to accelerate too fast, the front wheel will go up (see motorbikes doing that, or drag races going wrong). What is the acceleration where this happens? Does the back wheel have enough friction to support this acceleration?

4. Jul 3, 2017

### decerto

Just so I am clear, Are you saying I just have

$$F=MA=\mu N_1$$?

5. Jul 3, 2017

### jack action

Yes. You are asked to find the acceleration. The only unknown is $N_1$. Solve the free body diagram to find it as a function of $\mu$ and $A$.

6. Jul 3, 2017

### decerto

So I just use this and the other two equations

$$N_1 + N_2 = Mg$$
$$bN_1-aN_2 -h F =0$$

?

Just so I have this clear in my head, if it says "roll without slipping" this means there is no frictional force that resists the motion. Instead when a torque is applied to the wheel (like the back wheel) or a force is applied through the CoM (like the front wheel) the wheel exerts a force on the road due to friction which results in an equal and opposite force which accelerates the car forward?

7. Jul 3, 2017

### jack action

No, it only means that you use the static friction coefficient. If it was slipping, there would be a relative motion between the tire and the ground, thus you would use the kinetic friction coefficient.

For you moment equation, note that you have introduce $N_2$ which is an unknown. To eliminate it, base your equation on the moments about the contact point at the front wheels and the ground instead. Then you should be able to isolate $N_1$. Also don't forget that $F = MA$.

8. Jul 5, 2017

### decerto

So calculating the moments about the point of contact of the front wheel I got

$$(a+b)N - aMg=0$$

which gives

$$A_{max} = \mu g \frac{a}{a+b}$$

Is this correct?

The next part of the question asks

but my answer and all my formulas are independent of h and if I go back to using the moments about the CoM to obtain a formula for h, I get h=0

9. Jul 5, 2017

### Staff: Mentor

Why did you take the front wheel? Do you expect the car to rotate around the front wheel?
Where is the acceleration in this equation?

10. Jul 5, 2017

### jack action

Nope. You are missing the $MA$ component.

11. Jul 5, 2017

### jack action

Because I told him to.

12. Jul 5, 2017

### decerto

I have

$$F= \mu N=MA$$

$$N=\frac{MA}{\mu}$$

which is how I got

$$A_{max} = \mu g \frac{a}{a+b}$$

in my original post.

Or are you saying I missing an "MA component" from my torque equation?

13. Jul 5, 2017

### jack action

The $MA$ vector creates a moment about the front wheel contact point. Take your time studying the free body diagram.

14. Jul 5, 2017

### decerto

The MA vector is one which acts through the CoM?, you are not talking about the F acting at the point of contact of the backwheel?

15. Jul 5, 2017

### jack action

Yes.
No.

Once again:

16. Jul 5, 2017

### decerto

I understand what you are getting at but I don't understand why I need to include MA as a moment, what physical reason is there for it when the force F is clearly marked as acting at the base of the wheel.

edit * ok I can see the logic if we consider MA is a fictitious force due to it being a non-inertial frame, but I don't see how make sense of it in an inertial frame.

Last edited: Jul 5, 2017
17. Jul 5, 2017

### jack action

You know that $Mg$ is based on an acceleration? And you included it in your moment analysis? So why not include $MA$ as well?

The following is the free body diagram, Do $\sum F_x$, $\sum F_y$ and $\sum M$ about point $Z$:

18. Jul 5, 2017

### jbriggs444

Depending on the axis of rotation you choose to use, you can make one or more forces irrelevant. @jack action appears to be assuming an axis of rotation at the point of contact of the front wheel on the ground. [Not the choice I would have made, but it's viable]. With that axis of rotation in mind one has yet another choice...

1. Assume an inertial frame and note that the angular momentum of the race car is increasing in the counter-clockwise direction as the car accelerates leftward. [Note: key point about angular momentum -- it's not always about rotation!]

2. Assume an accelerating frame and note that the angular momentum of the race car is now constant but that there is a fictitious rightward inertial force acting on the center-of-mass of the race car.

@jack action has chosen option 2. Option 1 is equally viable.

Either way, there is still a net torque from the upward supporting force on the rear wheel and the downward force from gravity. Either way, the choice of rotation axis has made the frictional force of road on tire irrelevant to the angular momentum balance.

Or... choose the axis of rotation at the contact point on the rear wheel. Now you can ignore the supporting force from road on tire.

Or... choose an axis at the center of mass. Now the angular momentum is constant regardless of reference frame choice. But now the frictional force of road on tire becomes relevant.

Any choice will work and the analysis can proceed. But one needs to make a choice and see how the forces on the free body diagram work with that choice.

Last edited: Jul 5, 2017
19. Jul 5, 2017

### jack action

But that is the force we are looking for, why would we want to ignore it? The one we do not care about is the supporting force from road on front tire $N_f$. With point $Z$, both $N_f$ and $F$ are ignored, leaving us with $MA$, $Mg$ and $N_r$. You actually find $N_r$ with one equation (which should take less than 2 days to solve!).

20. Jul 6, 2017

### haruspex

That's not entirely true. You need to use that this cannot be negative.

21. Jul 6, 2017

### jbriggs444

We are (or should be) looking for the condition where the front wheel is about to lift off the ground. In that condition, the entire weight of the car is supported by the normal force of road on rear wheels. One can determine that force, $N_r$ easily without writing down an equation for angular momentum.

At the condition where the front wheels are about to lift off the road, $N_f$ is already known to be zero. Accordingly, it does not contribute to the angular momentum balance. Choosing an axis of rotation at point Z under the front wheels causes one term to drop out of the angular momentum balance. But it is the term associated with N_f which is already known to be zero. There is no point in eliminating a term that is already zero.

22. Jul 6, 2017

### decerto

This clears it up for me I think, dL/dt != 0 in the inertial frame and there is no MA 'force'?

23. Jul 6, 2017

### jbriggs444

Yes.

If you write down the equation for angular momentum balance in the inertial frame and then do so again for the accelerating frame, they should come out almost identical. You will have a term that moves from one side of the $Σ\tau = \frac{dL}{dt}$ equation to the other. In the inertial frame, that term is the $\frac{dL}{dt}$. In the non-inertial frame, that term is the fictitious torque from $Mah$.

By no coincidence, those two terms are equal and opposite.

24. Jul 6, 2017

### jack action

I must admit that I understood the problem differently, especially when only part (a) was available. I thought you needed to determine the maximum acceleration given $\mu$ and $g$, but also given $a$, $b$ and $h$ (which is a much more interesting and realistic problem). This will give you a relationship to determine the maximum acceleration for any position of the c.m.

By studying this relationship and knowing the fact that the front wheels must remain on the ground at all times (which determines the maximum friction force you can ultimately get), you can find the c.m. locations that will produce the ultimate maximal acceleration for a given $\mu$ only. In any other location, either the front wheels will be off the ground or the maximum acceleration will be lower (but there is still a valid maximum acceleration).