Wheelie of a car coming out of a ditch: what is the correct model?

[alex33]

Sorry, not sure what you are saying there.

I mean the car doesn't skid sideways (x-axis: horizontal, z-axis: vertical, y-axis: lateral)

As I wrote, it's not whether the tyres actually leave the ground; it's whether the compression in the suspension is greatly reduced, making the chassis effectively in free fall. That said, I will continue to describe matters as where the wheels are, as though there is no suspension.

OK !

It is a homework forum, so I rather assumed you had high school background in ballistics.

In fact I did a good scientific high school in Tuscany, the land of Leonardo da Vinci and Galileo. But that was from 1986 to 1991 !!!

Can you do that, or do you want to say, hey, it's not homework, just give me the answers?

Of course I'd rather get the right solutions myself, but you have to be very lenient with me, because my notions are very rusty. And keep in mind that I don't have to deliver any papers to any professors, or take any exams...

The simplest way to look at it is that the centre of mass follows a parabola while the chassis rotates about it at a steady rate.

May I ask what is the force (or what are the forces) that cause this second component of the motion?
As active forces in this scenario I can imagine the driving force of the vehicle, the weight force, the dynamic friction, the centripetal force. However, the latter ceases its action when the front wheels come out of the hole (to then act again on the rear wheels when they go up again).

Again, please accept the rigid body view, i.e. it does bounce in effect.

Ok... for the moment...

Just had to make it big enough that the rear does not continue down into the ground, but not so large that it increased the KE.

Of course total KE must be preserved

At this point, the front wheels were still 0,05 m off the ground.

Sorry, in doing the calculations did you use the right g that I posted? It's not the earthly one...

 

[haruspex]

May I ask what is the force (or what are the forces) that cause this second component of the motion?

Once it is in free fall, the only force is gravity. This means the horizontal velocity of the mass centre and the rotation about the mass centre are constant. The vertical velocity of the mass centre follows $v_y(t)=v_{y0}-gt$.

As active forces in this scenario I can imagine the driving force of the vehicle, the weight force, the dynamic friction, the centripetal force.

For a vehicle driving without skidding, the driving force is the frictional force. Without friction you'd get nowhere. Since we're discussing free fall here, neither applies.
Centripetal force is not an 'active' force. It is merely that component of the net force which is orthogonal to the velocity.

Of course total KE must be preserved

Not all. When it bounces (in your view, when the suspension goes through maximum compression) some is surely lost. But maybe not much in this case. Consider a horizontal uniform bar dropped so that one end strikes a fixed object. It is not hard to show that at most 1/4 of the KE is lost. The car landing in the dip is like that.
Indeed, if you don't see much bounce in the video then the loss must be close to the maximum.

you use the right g that I posted?

Yes.

 

[alex33]

Ok. I understand almost everything.

Once it is in free fall, the only force is gravity. This means the horizontal velocity of the mass centre and the rotation about the mass centre are constant.

With respect to rotation, you didn't answer explicitly... Do you mean that the rotation produced before the free fall by the circular guide of the dip is conserved around the centre of mass? Therefore it's different from a "uniform bar" which, when leaving the dip, follows a linear trajectory according to the tangent (as occurs in the sport of hammer throwing). Is that?

 

[alex33]

Ok. I understand almost everything.

With respect to rotation, you didn't answer explicitly... Do you mean that the rotation produced before the free fall by the circular guide of the dip is conserved around the centre of mass? Therefore it's different from a "uniform bar" which, when leaving the dip, follows a linear trajectory according to the tangent (as occurs in the sport of hammer throwing). Is that?

Oh no... it's the Angular Momentum !

 

[haruspex]

it's different from a "uniform bar" which, when leaving the dip, follows a linear trajectory according to the tangent (as occurs in the sport of hammer throwing).

Not sure if you are thinking of javelin, but in that case there's drag, which may cause some angular acceleration during flight. There is also the possibility that the thrower applied some downward rotation in the throw.

Oh no... it's the Angular Momentum !

Yes.

 

[alex33]

Rear wheel angular velocity retrieval
$mgh=\frac {1}{2}Jω^2$
I make the model for a parallelepiped of height h, width w, length l:
$J=\frac {1}{12}m(h^2+w^2)$
$ω=\sqrt{\frac {24gh}{h^2+w^2}}$

Here again the basic motion data:
Centre of the arc AB (the dip arc): (X=0; Z=0)
Length of the arc: AB = 2,082 m
Car wheelbase Length L: 2,286 m
Car Length L: 2,286 m (for the moment in order to simplify)
Radius of the arc: r = 2,875 m (reviewed)
Angle subtended by the arc AB: β = 2α = 41,5° (reviewed)
Angle between ground and arc tangent at A and B: α = 20,75° (reviewed)
Initial speed of the car at the entrance to the dip: V₀= 2,61 m/s (reviewed)
Car Centre of Mass = 1/2 L (for the moment in order to simplify)
g = 1,62 m/s²
X₀= r*SEN(α)
Z₀= r*COS(α)
V_x0= V₀*COS(α)
V_z0= V0*SEN(α)

The equations of motion of the rear wheels:
$Z(t) = Z_0 + V_z0 t + \frac {1}{2} gt^2 - \frac {1}{2}L * SEN(ωt)$
$X(t) = X_0 + V_x0 t + (\frac {1}{2}L - \frac {1}{2}L * COS(ωt))$

If they're right, I'll do the spreadsheet tomorrow.
THANK YOU

 

[haruspex]

Rear wheel angular velocity retrieval
$mgh=\frac {1}{2}Jω^2$

How do you get that? The right hand side looks like energy, but I don't see why it should equal the term on the left.

As I wrote, you need to start by finding the velocity components at the front wheels just as they reach the top of the dip. If the rear wheels have velocity v (all horizontal), what is the horizontal velocity of the front wheels? Given that the front wheels are still going up the last bit of the ramp, what is their vertical velocity?
What, then, are:
- the vertical and horizontal velocities of the mass centre
- the angular velocity of the chassis?
This much is just kinematics; no need to consider forces, torques, momentum or energy.

 

[alex33]

How do you get that?

I thought that a key to solving the problem related to the falling motion of the rear wheels could be the conservation of energy. The potential energy mg x height of the fall h should be conserved with the kinetic energy as a function of the moment of inertia and the angular velocity... so isn't ?

just as they reach the top of the dip. If the rear wheels have velocity v (all horizontal), what is the horizontal velocity of the front wheels?

...in that moment the car is launched with the same speed value V and direction according to the tangent of the arc just crossed, so I think:
V_x0 = V₀*COS(α)
...when the rear wheels are in free fall also, a rotational motion will be added with angular speed ω which on the X axis will be:
$ω_x = \frac{ω}{\sin(ωt)}$

What, then, are:
- the vertical and horizontal velocities of the mass centre
- the angular velocity of the chassis?

During the free fall the centre of mass have:
V_x0 = V₀*COS(α)
V_z0 = V₀*SEN(α)

Angular velocity of the chassis:
$ω = \frac{2V}{L}$

 

[haruspex]

I thought that a key to solving the problem related to the falling motion of the rear wheels could be the conservation of energy

We need to find the angular velocity when the rear wheels reach the dip. They have not descended yet. Besides, most the potential energy will be turned into vertical KE. That will then turn into elastic potential energy in the suspension. Suspensions are not designed to be highly elastic; after going over a bump you don't want to be bouncing up and down for the next 3 km. They are damped systems, converting the energy to heat.
Depending on the exact characteristics, the bounce might help the vehicle get up the other side of the dip, but it depends on the timing.

...in that moment the car is launched with the same speed value V and direction according to the tangent of the arc just crossed, so I think:
V_x0 = V₀*COS(α)

No. The car has a fixed length, and the rear is moving horizontally, directly towards the front wheels. If the front wheels have a different speed in that direction the car must be getting shorter or longer.
So the horizontal component of the front wheels' velocity is also v.
However, they are going up a ramp at angle $\alpha$, so what is their vertical component?

 

[alex33]

What you are saying is gospel to me!

However, they are going up a ramp at angle α, so what is their vertical component?

$V_z = V_x * \tan (α)$

 

[haruspex]

What you are saying is gospel to me!$V_z = V_x * \tan (α)$

Right.
So, can you figure out the angular velocity of the car?

 

[alex33]

However, this implies that the total speed of the front wheels is higher than the rear ones. It's correct?

 

[haruspex]

However, this implies that the total speed of the front wheels is higher than the rear ones. It's correct?

Yes, well spotted. It's $v\sec(\alpha)$.

 

[alex33]

... but for the Centre of Mass, what I'm guessing is that the vertical speed is lower than that of the front wheels? It's possible?

 

[haruspex]

... but for the Centre of Mass, what I'm guessing is that the vertical speed is lower than that of the front wheels? It's possible?

Yes. What should it be there, half way along the car?

 

[alex33]

more or less the centre of the dip

 

[haruspex]

more or less the centre of the dip

No, I mean what is the vertical velocity of the mass centre, in terms of $v, \alpha$?

 

[alex33]

this is what i don't understand... the tilt angle is the same (we are supporting the car on the rear wheels), since the relationship between Vx and Vz is the same i would be tempted to say that the vertical velocity of the center of mass is the same as the front wheels, but it is clear that it is not so... I don't understand what is missing...

 

[haruspex]

this is what i don't understand... the tilt angle is the same (we are supporting the car on the rear wheels), since the relationship between Vx and Vz is the same i would be tempted to say that the vertical velocity of the center of mass is the same as the front wheels, but it is clear that it is not so... I don't understand what is missing...

The front wheels have a vertical velocity $v\tan(\alpha)$, while the rear wheels have no vertical velocity. The mass centre, being directly above the midpoint of those, therefore has a vertical velocity $\frac 12v\tan(\alpha)$.
So now you have enough to write the expressions for the x and y coordinates of the mass centre at time t after 'lift off'.

What about the angular velocity?

 

[alex33]

What about the angular velocity?

To answer this question I would use the relation $ω = V * R$
Since the radius, in our case is $L$ (the whole chassis rotates)
and that the speed we've to use is the vertical one $V * \tan(α)$
...if I'm lucky, the angular speed for the front wheels is...
$ω = V * tan(α) * L$
on the centre of mass we'll have:
$ω = \frac{1}{2}V * tan(α) * L$
...the rear wheels are our pivot so there can't be an angular speed at the moment. We will therefore have a load on the suspension.

 

[haruspex]

To answer this question I would use the relation $ω = V * R$

No such relation; it would be dimensionally wrong for a start.
Try $ω = V / R$.
You may be thinking of angular momentum, mvr.

the angular speed for the front wheels is

Angular speed is a property of the whole rigid object.
E.g. considering how the front axle rotates about the rear axle, the angular speed is $v\tan(\alpha)/L$, or how the mass centre rotates about the rear axle, the angular speed is $v\frac 12\tan(\alpha)/(L/2)$, which is the same.

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

 

[alex33]

Try ω=V/R.

what kind of bungler... I reversed the elements !!

Angular speed is a property of the whole rigid object.
E.g. considering how the front axle rotates about the rear axle, the angular speed is $v\tan(\alpha)/L$, or how the mass centre rotates about the rear axle, the angular speed is $v\frac 12\tan(\alpha)/(L/2)$, which is the same.

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

OK

 

[alex33]

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

Wait! we said we weren't in free fall yet: the rear wheels are still on the ground just before the dip. However we say that Vx remains constant because the driving force compensates the friction. Okay?

Unfortunately with α(t) i got stuck again. Please be patient... I would have:
$\frac{α}{t} = \frac{V* \tan(α)}{L}$ and so $\frac{α}{\tan(α)} = \frac{V * t}{L}$ but I have no idea how to get α if we have to divided it by its tangent.

 

[haruspex]

we said we weren't in free fall yet: the rear wheels are still on the ground just before the dip.

We have the velocity and rotation rate just as the rear wheels reach the dip. They are not going to change suddenly.

Unfortunately with α(t) i got stuck again. Please be patient... I would have:
$\frac{α}{t} = \frac{V* \tan(α)}{L}$ and so $\frac{α}{\tan(α)} = \frac{V * t}{L}$ but I have no idea how to get α if we have to divided it by its tangent.

$\alpha$ is in use as the half angle of the dip arc. Let's use $\omega$ for the rotation rate (taking the front rising as positive) and $\theta(t)$ for the angle reached (above horizontal).
$\theta=\omega t=\frac{V* \tan(α)}{L}t$.

 

[alex33]

Ok. I understand. θ(t) is the function for the rotation when the free fall start. α is the angle reached at the top of the dip and then fixed by sperimental data.

To stay close to the real event, I ask you to take into account the fact that from when the front wheels lift completely out of the hole, for 2/24 seconds the same front wheels travel without detaching from the ground (suspensions?) and at the same time the wheels rear they complete the last flat stretch before going down into the hole using the same 2/24 + other 2/24 seconds. This is proper event scanning.

Now I try to build the spreadsheet.

 

[haruspex]

when the front wheels lift completely out of the hole, for 2/24 seconds the same front wheels travel without detaching from the ground

Are you saying that the front wheels remain in contact with the ground for that time after exiting the dip but then lift off? Yes, that's just the suspension relaxing, but has negligible effect on what the chassis does. Once we have found the chassis path, it won't be hard to hang the wheels off it.

the same 2/24 + other 2/24 seconds

So the rear wheels reach the start of the dip 2/24 s after the front wheels leave it?

Now I try to build the spreadsheet.

Before you do that, please post the equations you have for:

• the x, y positions of the mass centre
• the attitude of the chassis, $\theta$
• the x, y positions of the rear wheels

as functions of time.

 

[alex33]

Are you saying that the front wheels remain in contact with the ground for that time after exiting the dip but then lift off?

Yes

So the rear wheels reach the start of the dip 2/24 s after the front wheels leave it?

No 4/24 after the front wheels leave the dip, 2/24 after the front wheels lift off from the ground

 

[alex33]

• the attitude of the chassis, $\theta$

...sorry the "attitude"?

 

[haruspex]

...sorry the "attitude"?

Angle to the horizontal.

 

[haruspex]

No 4/24 after the front wheels leave the dip, 2/24 after the front wheels lift off from the ground

That makes the car 0.43m longer than the dip. That will make a huge difference. The front wheels should be back on the ground before the rear wheels reach the dip.
No, it's ok. I made a mistake in how I adjusted the spreadsheet.