# Determine if a car can pop a wheelie?

• bbbl67
In summary: This is how you should be thinking of it. This is how we all think of it. It is a very basic concept in mechanics.The vast majority of the mass is ahead of the rear axle, but there is still a non-negligible amount of mass behind the axle.I don't see how this can matter. The front wheels are off the ground. There is no force on the front wheels pushing them down. So the front wheels are irrelevant at this point. The rear wheels are still in contact with the ground. So why make it more complex than it needs to be?Yes, but obviously this is not a realistic car, it's
bbbl67

## Homework Statement

Okay you have a rear-wheel-drive car, we can model it as a uniform rectangular block, 12 ft long, 4 ft tall, and weighs 4000 lbs. The front and rear axle centerlines are positioned within 2 ft from the far edges of the block. What kind of forward acceleration (in g's) must it achieve to lift its front wheels off the ground at the start? If it maintained the same level of acceleration throughout the run, what kind of 0-60 mph times would it achieve?

## Homework Equations

-Center of mass equations
-torque equations
-Pythagorean theorem

## The Attempt at a Solution

We need to split the car into 2 sub-blocks. The front block is everything from the rear axle centerline to the front, while the rear block is everything from the rear axle centerline to the rear.

Front block:
center of mass:
x = (12 - 2) ft / 2 = 5 ft
y = 4 ft / 2 = 2 ft
d^2 = x^2 + y^2
d = (x^2 + y^2)^1/2
= (5^2 + 2^2)^0.5
= 5.385 ft

m = 4000 lb * 10 / 12
= 3333 lb

Rear block:
center of mass:
x = 2 ft / 2 = 1 ft
y = 4 ft / 2 = 2 ft
d = 2.236 ft

m = (4000 - 3333) lb
= 667 lb

So now we have to add up the balance of rotational forces:
t1 = force of gravity on front block
= 3333 lbf * 5 ft
= 16665 ft * lbf
f = engine linear force
= ?
t2 = engine torque on front block
= f * 5.385 ft
t3 = engine torque on rear block
= f * 2.236 ft
t4 = force of gravity on rear block
= 667 lbf * 1 ft
= 667 ft * lbf

The sum of torques on the overall body:
t1 = t2 + t3 + t4
16665 = 5.385f + 2.236f + 667
15998 = 7.621f
f = 2099 lbf

a = f / m
= 2099 lbf / 4000 lbm
= 9337 N / 1814 kg
= 5.146 N/kg
= 5.146 m/s^2
= 11.511 mph/s
(= 0.52475 g)

I converted to metric in the middle because it was easier to work with, before converting back to US units.

0-60 mph:
T = 60 mph / 11.511 mph/s
= 5.2 s

So the car will pop a wheelie, if it accelerates slightly faster than 0.52475 g or 5.2 s in the 0-60 mph trap. Is there any thing wrong with this methodology? Did I miss anything?

bbbl67 said:
We need to split the car into 2 sub-blocks.
That is unnecessary. Just work with the mass centre of the whole vehicle.
Btw, in the real world the mass centre would not be at half the height of the vehicle, but rather lower.
bbbl67 said:
d^2 = x^2 + y^2
Do you understand about lever arms and how you find the torque a force exerts about an axis?
bbbl67 said:
f = engine linear force
What do you mean by this exactly? Where does this force act and in what direction? Have you drawn a force diagram?

jbriggs444
bbbl67 said:
Is there any thing wrong with this methodology? Did I miss anything?
Two main points.

1. There is no need to split the car into two blocks. It is enough to consider it as one block with gravity acting downward on the center of gravity and the vertical force of the pavement acting on the rear wheel. You then have several options for how to factor the forward acceleration into the torque balance.

2. You have computed an "engine torque" based on an unknown force f multiplied by a diagonal distance from the rear axle to the center of mass of each block. Can you more clearly identify this force f. Is it at right angles to those diagonals?

Edit: Took too long thinking and composing, so Haruspex beat me to it.

jbriggs444 said:
Two main points.

1. There is no need to split the car into two blocks. It is enough to consider it as one block with gravity acting downward on the center of gravity and the vertical force of the pavement acting on the rear wheel. You then have several options for how to factor the forward acceleration into the torque balance.
The reason I split the body into two blocks was because I was trying take into account the seesaw effect of the bit of the body behind the rear axle. The vast majority of the mass is ahead of the rear axle, but there is still a non-negligible amount of mass behind the axle. The mass behind the axle would add a bit of additional torque that would help the engine torque to raise the front of the car up.

jbriggs444 said:
2. You have computed an "engine torque" based on an unknown force f multiplied by a diagonal distance from the rear axle to the center of mass of each block. Can you more clearly identify this force f. Is it at right angles to those diagonals?

Edit: Took too long thinking and composing, so Haruspex beat me to it.
Yes, it's the tangential force from the torque of the engine. But of course, it's not really the engine at this point, the torque is coming from the rear axle and wheel at this point.

haruspex said:
That is unnecessary. Just work with the mass centre of the whole vehicle.
Btw, in the real world the mass centre would not be at half the height of the vehicle, but rather lower.
Yes, but obviously this is not a realistic car, it's just a simplefied uniform rectangular box.

bbbl67 said:
it's the tangential force from the torque of the engine
You need to be more specific, in your own mind at least. Where does this force act on the car? In what direction? You also need to answer my question:
haruspex said:
Do you understand about lever arms and how you find the torque a force exerts about an axis?
Does the expression "perpendicular distance" ring any bells?

bbbl67 said:
The reason I split the body into two blocks was because I was trying take into account the seesaw effect of the bit of the body behind the rear axle.
As @jbriggs444 and I have indicated, there is no need to split it into two parts. Just treat it as a point mass located at the mass centre of the whole vehicle. The cancellation between the portions in front of and behind the rear wheels will be taken care of by the algebra.

jbriggs444
haruspex said:
As @jbriggs444 and I have indicated, there is no need to split it into two parts. Just treat it as a point mass located at the mass centre of the whole vehicle. The cancellation between the portions in front of and behind the rear wheels will be taken care of by the algebra.
Alright if we do as you say, and treat the whole car as a single body, then the center of mass will occur 4 ft horizontally in front of the rear axle, and gravity will work from that location. Similarly the engine torque/axle torque will act against a center of mass that is 4 ft horizontally in front and 2 ft vertically up from the rear axle.

Grav torque:
t1 = 4000 lbf * 4 ft
= 16000 ft-lbf

Axle torque:
t2 = f * (4^2 + 2^2)^0.5 ft
= 4.472f ft-lbs

t2 = t1
4.472f = 16000
f = 3578 lbf

So "f" being the same tangential force of the torque being applied to the axle/wheel and ground.

a = f / m
= 3578 lbf / 4000 lbm
= 8.772 m/s^2
= 0.8945 g

From previous work, where I split the body into two parts, I had gotten an acceleration of only 0.52475 g. So the two methods don't agree with each other. Which method is wrong, and why?

bbbl67 said:
Axle torque:
t2 = f * (4^2 + 2^2)^0.5 ft
You are still getting this part completely wrong. Draw a diagram and answer this question: what external force acts on the car to accelerate it? It is not a rocket. Where does this force act and in what direction?

Good to see that you have now correctly calculated the weight torque by using the perpendicular distance (4 ft) to the weight force. You need to do the same for the driving force.

haruspex said:
You are still getting this part completely wrong. Draw a diagram and answer this question: what external force acts on the car to accelerate it? It is not a rocket. Where does this force act and in what direction?

Good to see that you have now correctly calculated the weight torque by using the perpendicular distance (4 ft) to the weight force. You need to do the same for the driving force.
If it's a tangential force, then it's acting perpendicularly to the wheel where it touches the road. It's also acting on the diagonal center of mass from the axle, to tilt the car upwards. It's the same force either case.

bbbl67 said:
If it's a tangential force, then it's acting perpendicularly to the wheel where it touches the road.
Not sure what you mean by perpendicularly to the wheel. Do you mean horizontally or vertically?
What is the force usually called? (How would the car go trying to do this on ice?)

bbbl67 said:
It's also acting on the diagonal center of mass from the axle, to tilt the car upwards.
A force acting from where the tyre touches the road diagonally up through the centre of mass would not tilt the car up. It would tend to lift the car off the ground, but without tilting.

haruspex said:
Not sure what you mean by perpendicularly to the wheel. Do you mean horizontally or vertically?
What is the force usually called? (How would the car go trying to do this on ice?)
Okay, now you're just being patronizing, do you actually have something to add here anymore?
haruspex said:
A force acting from where the tyre touches the road diagonally up through the centre of mass would not tilt the car up. It would tend to lift the car off the ground, but without tilting.
Pedantic too?

haruspex said:
A force acting from where the tyre touches the road diagonally up through the centre of mass would not tilt the car up. It would tend to lift the car off the ground, but without tilting.
In particular, a diagonal force such as this would produce zero torque about the center of gravity. So it seems, @bbbl67, that you need to sharpen your thinking here.

jbriggs444 said:
In particular, a diagonal force such as this would produce zero torque about the center of gravity. So it seems, @bbbl67, that you need to sharpen your thinking here.
Well obviously! It's not a torque around the center of mass we're looking for, but a torque around the rear axle's center point.

bbbl67 said:
Well obviously! It's not a torque around the center of mass we're looking for, but a torque around the rear axle's center point.
No linear force originating at the rear axle can produce any torque about the rear axle.

Edit: In many problems involving angular momentum or torque, there are any number of choices for the reference axis to use when analyzing the problem. This problem is no different. If you want to use point of contact of the rear tires on the road as the reference axis, we can do that. It is perhaps not the best choice, but it will work.

Note that given this choice of reference axis, the linear acceleration of the car has a non-zero contribution to the rate of change of angular momentum over time.

bbbl67 said:
Okay, now you're just being patronizing, do you actually have something to add here anymore?
They were perfectly serious questions.
In post #8 you wrote, correctly, that the propulsive force acts tangentially to the wheel where it touches the road. (I was trying to get you to call it by its name, static friction.) But then you added that it acts diagonally up through the mass centre, which a tangential force at that point cannot do. So that was a bit confusing. Perhaps you were adding in the normal force from the road when you wrote that.

bbbl67 said:
Well obviously! It's not a torque around the center of mass we're looking for, but a torque around the rear axle's center point.
Actually, that is not a good choice. There are in general two valid choices: the mass centre of the body or any point fixed in an inertial frame. Anything else can lead you to the wrong answer.
In the present case, if the mass centre of the car is not at the same height as the chosen axis then you would have to allow for the car's linear acceleration constituting an angular acceleration about the axis. That makes the mass centre of the car an excellent choice. Another good one would be the point that is (instantaneously) in the same vertical line as the rear axle and the same height as the mass centre.

Last edited:

## 1. Can any car pop a wheelie?

No, not all cars have the capability to pop a wheelie. It depends on the car's power, weight distribution, and suspension.

## 2. How much horsepower does a car need to pop a wheelie?

The amount of horsepower needed varies depending on the weight and design of the car. Generally, a car needs at least 300-400 horsepower to pop a wheelie.

## 3. Is popping a wheelie dangerous?

Yes, popping a wheelie can be dangerous if not done properly. It can cause the car to lose control and potentially result in accidents.

## 4. Can a front-wheel drive car pop a wheelie?

It is possible for a front-wheel drive car to pop a wheelie, but it is much more difficult compared to rear-wheel drive cars. The weight distribution and power of the car play a significant role.

## 5. What modifications can be made to a car to help it pop a wheelie?

Modifications such as adding more horsepower, adjusting the suspension, and changing the weight distribution of the car can increase the chances of popping a wheelie. However, it is important to note that these modifications can also be dangerous and should be done with caution.

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