Could pressing the brakes on a car in mid-air affect its pitch rotation? In a racing video game series named Trackmania, there is a game mechanic where when you hit a jump and the car is in mid-air, you can stop the car from pitching downward by tapping on the brakes.
I am curious if this mechanic has some basis in real physics. What would happen if a car were pitching nose down while in the air, and you hit the brakes thus stopping the rotating wheels. Would that affect the car's pitch, and if so, how?
 A: It will affect pitch, but not the way it works in the game. In the game, braking while in the air stops your pitch axis rotation. In real life, it does something completely different.
The wheels have angular momentum. When the wheels are slowed, this angular momentum must be conserved. This results in a transfer of angular momentum from the wheel to the vehicle.
Since the wheels are moving clockwise in relation to a vehicle travelling left-to-right (i.e. the top of the wheel is moving toward the front of the car, and the bottom of the wheel is moving toward the rear) the car will also start to rotate clockwise, pitching the nose down. For a car already starting to pitch down after a jump, this will cause it to pitch down even faster, the opposite of what it does in-game.
We can even calculate a rough magnitude of the effect!
For simplicity, let's state some assumptions:

*

*this is a rear-wheel drive vehicle with stationary front wheels

*both rear wheels are moving at the same angular velocity (no slip diff)

*both rear wheels are the same size and have the same mass

*the rear axle is a rigid balanced cylinder (i.e. its center of mass is the center of the axle)

*the brakes bring the wheels to a complete stop

*we ignore the motion of the driveshaft, flywheel, clutch, gearbox, and other parts of the drivetrain

*we ignore air resistance, lift, and all other aerodynamics

Each of the two rear wheels can be approximated as a cylindrical mass whose center of mass is the axle joint. The same can be said for the rear driveshaft - it's a long cylinder. As such, the momentum of each of these bodies can be described by spin angular momentum, which is angular momentum about the center of mass. This is in contrast to orbital angular momentum, which is angular momentum about an arbitrary point.
Angular momentum is expressed as $L=I\omega$, where $I$ is the angular moment of inertia and $\omega$ is the angular velocity in radians per second. You can think of the angular moment of inertia as a way to describe the mass distribution of an object about its axis (or axes) of rotation. A cube, a cylinder, and a sphere all have different moments of inertia, and those moments also change depending on where you put the axis of rotation (through the center, on an edge, etc.)
A cylinder with mass $m$ and radius $r$ rotating about its $z$ axis has an angular moment of inertia described by $I = \frac 1 2 mr^2$.

As such, the moment of inertia for each wheel can be approximated by $I = \frac 1 2 mr^2$, where $m$ is the mass of the wheel and $r$ is the radius of the wheel. The moment of inertia for the axle can be described similarly, since we can model that as a cylinder too.
Given that we have two wheels rotating about the same axis, we can think of them as a combined cylinder of the same radius but with twice the mass, which cancels out the $\frac 1 2$ term. We can then add the moment of inertia for the axle to get the total moment of inertia:
$$I_T = m_W {r_W}^2 + \frac 1 2 m_A {r_A}^2$$
(with $T$ meaning total, $W$ meaning wheels, and $A$ meaning axle)
This can then be plugged into the angular momentum equation, $L=I\omega$, where $\omega$ is the angular velocity in radians per second.
$$L = \omega \left(m_W {r_W}^2 + \frac 1 2 m_A {r_A}^2\right)$$
If we assume that the vehicle's wheels have remained at a constant angular velocity since leaving the ground, we can estimate $\omega$ from the vehicle's land speed at the time of take-off and the radius of the wheel including the tyre.
One revolution of the wheel moves the vehicle forward by the circumference of that wheel, and the circumference is $2\pi r$. If we take the car's velocity in meters per second (1mph ≈ 0.447m/s) and divide it by the wheel circumference, that tells us how many times the wheel was rotating per second. One rotation is 360°, or $2\pi$ radians. As such:
$$\omega \approx \frac {v_C} {2\pi {r_W}} \times 2\pi = \frac {v_C} {r_W}$$
Where $v_C$ is the car's velocity at the point of take-off, and $r_W$ is the wheel radius.
Substituting this into our previous equation, we get:
$$L = \frac {v_C} {r_W} \left(m_W {r_W}^2 + \frac 1 2 m_A {r_A}^2\right)$$
Where $L$ is the angular momentum, $v_C$ is the velocity of the car at take-off (for the purposes of angular velocity estimation), $r_W$ is the radius of the rear wheels including the tyre, $m_W$ is the mass of each of the two rear wheels including the tyre, $m_A$ is the mass of the rear axle, and $r_A$ is the radius of the rear axle.
For the sake of simplicity in this worked example, we'll assume that the front wheels aren't spinning, even though in practice it would make sense for the front wheels to be rotating at the same angular velocity as the rear wheels. While it is entirely possible to calculate the resultant angular velocity of the car as a result of both the front and rear wheels, including the case where the front wheels are not facing straight forward, the calculations are much easier to follow in a system with angular momentum being transferred between two bodies in a single axis.
Let's try a quick test-case:

*

*Each wheel weighs 25 kg including the tyre.

*The rear wheels have a radius of 25 cm (approximating a 16" diameter alloy with 2" thick tyres).

*The rear axle is 6 cm in diameter and weighs 50 kg.

*The car was travelling at 40 m/s (roughly 90 mph) when it left the ground.

Plugging these numbers in, we get:
$$L = \frac {40~\mathrm{m~s}^{-1}} {0.25~\mathrm{m}} \left(25~\mathrm{kg} \times (0.25~\mathrm{m})^2 + \frac 1 2 50~\mathrm{kg} \times (0.06~\mathrm{m})^2\right) = 265.4~\mathrm{kg}⋅\mathrm{m}^2⋅\mathrm{s}^{-1}$$
Note that kg⋅m2⋅s−1 are the units for momentum.
This is all well and good, but what does this mean in terms of the movement of the car?
Since angular momentum must be conserved, the change in momentum in the wheels is passed on to the body of the car. The equations we used above can be used in reverse - we can start with angular momentum and a moment of inertia and use it to find the resulting angular velocity!
However, there's a bit of a hitch: the angular momentum isn't being applied at the center of mass of the car, but instead at the location of the rear axle. This means that the car's movement is described by orbital angular momentum, not spin angular momentum. The car also isn't a cylinder, so we need a different equation.
To keep things simple, let's imagine the car is a cuboid of uniform mass with the real axle running along one of the bottom edges:

The moment of inertia for such a cuboid is described by:
$$I = \frac {m(a^2 + b^2)} {12}$$
where $m$ is the mass, $a$ is the side of length a in meters, and $b$ is the side of length b in meters.
We can now derive the equation for estimating the angular momentum of the car, using $L=I\omega$:
$$L_C \approx \omega \times \frac {m_C(l^2 + h^2)} {12}$$
where $L_C$ is the angular momentum of the car, $\omega$ is the angular velocity of the car, $m_C$ is the mass of the car, and $l$ and $h$ are the length and height of the car respectively in meters.
The equation above is written in terms of $L_C$, so to find the resulting angular velocity of the car we need to rearrange it in terms of $\omega$:
$$\omega \approx \frac {L_C} {\left(\frac {m_C(l^2 + h^2)} {12}\right)} = \frac {12L_C} {m_C(l^2 + h^2)}$$
Let's continue with our test case by defining the last few parameters:

*

*The car is approximately 1.25 m tall and 4.75 m long.

*The car weighs 1600 kg. After subtracting the mass of the wheels and rear axle, that's 1500 kg. (edit: thanks to nitsua60 for pointing out that since the wheels and axle move as part of the car, their mass counts as part of the overall moment of inertia and should not be subtracted)
Since we know that the angular momentum being transferred from the wheels and axle to the car is 265.4 kg⋅m2⋅s-1, we can now plug everything in:
$$\omega \approx \frac {12 \times 265.4~\mathrm{kg}⋅\mathrm{m}^2⋅\mathrm{s}^{-1}} {1600~\mathrm{kg} \times (4.75^2 + 1.25^2)~\mathrm{m}^2} = 0.0825~\mathrm{rad/s}$$
This is equivalent to 4.73°/s of nose-down rotation - small, but fairly noticeable!
The approximations here are crude, but they give you a good idea of how the conservation of angular momentum results in the downward pitch when the brakes slow down the spinning wheels.
It is possible to calculate the system's behaviour more accurately by considering the three-dimensional moment of inertia around the rear axle, angular momentum transfer of the flywheel and drivetrain (the car will tend to tilt slightly to one side as the drivetrain slows), non-uniform mass distribution of the car, air resistance, lift, and other aerodynamic effects, but the calculations are significantly more complicated and beyond the scope of this answer.

As a final wrap-up point, if you express the conservation of angular momentum between two objects ($a$ and $b$) as a single equation, you can gain some intuition for the behaviour of the objects as a function of their mass, size, and velocity.
$$I_a\omega_a = I_b\omega_b$$
If we rearrange for $\omega_b$ we can see how a change in angular velocity on object $a$ affects the angular velocity of object $b$:
$$\omega_b = \frac {I_a} {I_b} \omega_a$$
When angular momentum is transferred, the change in angular velocity in object $b$ is a function of the ratio between the two angular moments of inertia. Since the moment of inertia of an object is proportional to its mass and size, a smaller lighter object imparts less angular velocity to a larger heavier object.
A: It’s a real effect, but it goes the other way. In an isolated system, the total amount of spinning (including the three-dimensional direction) is a constant. If a car is in the air and its  wheels are spinning forward, applying the brakes will better attach the car to the wheels, so the car will pitch nose-downwards.
You could in principle pitch the car nose-up by applying the gas. This is basically how you start a “wheelie” on a motorcycle, why you can’t do a wheelie at all in a front-wheel drive vehicle, and why the front brakes contribute more to your stopping power than the rear brakes.
A: Here's a video of a motocross high jump competition. The effects of braking are more pronounced since motorcycles have much less mass than cars. The first jump is at 1:33. You can see that the bike is pointing almost straight up and slowly rotating backwards after the launch off the ramp, but the nose is rotated down once the bar is cleared. In the instant replays starting at 2:10, you can see that the rear wheel is stopped on the way down. The brake is applied to get the front wheel down using conservation of angular moment. The rear wheel was spinning forward, but the brakes applied backward torque to the wheel. By Newton's third law, the wheel puts a forward torque on the bike, nosing it down.
A: Maybe I am missing something, but it seems to me that braking in air decreases angular momentum of the wheels, the total angular momentum must be conserved, so the body of the vehicle should start rotating in the same sense as the wheels, so I would expect nose pitching down to increase, not decrease.
If the above is correct, then we need acceleration, not braking, to prevent the nose from pitching down.
