[Physics] Effect of the tail of the cat in the falling cat problem

angular momentumconservation-lawsnewtonian-mechanics

To explain why a falling cat can turn by 180 degree without external torque and without violation of the conservation of angular momentum, one usually models the cat as two cylinders as in

http://en.wikipedia.org/wiki/Falling_cat_problem

This may explain the turn. However I often heard contrary to this that she can rotate her body simply because she rotates her tail very fast into the opposite direction (and essentially keeps the rest of the body rigid).

So, what effect does the tail have in reality? Is there any detailed model, which takes the tail rotation into account and calculates how large its effect is?

Best Answer

The two answers are physically equivalent. If you watch the Wikipedia animation

enter image description here

you see that the fastest part of her body on the top animation is the tail. It makes a substantial contribution to the angular momentum. At any rate, the angular momentum of the tail is included in the angular momentum of the "back cylinder". The trick the cat needs to achieve the task is to perform a fast relative motion of the back of her body (including, importantly, the tail) and the front of her body.

Related Solutions

[Physics] Conservation of angular momentum in helicopter

The air is of course the key to understand your questions. To start with the last question, as you figured out right yourself, in vacuum body of the helicopter would not turn at all. But in vacuum helicopter could not fly at all!

enter image description here

At least part of the principle of rotor is that by pushing air down makes helicopter airborne. However in order to do that, blades of the rotor are a bit inclined, so when blade hits air right with velocity $\vec{v}$, it pushes air mostly down and but also to right with force $\vec{F}_\text{B}$. Third Newton law tells us that in that case air pushes blade mostly up, but also to left with force $\vec{F}_\text{A}$.

The vertical push $\vec{F}_\text{V}$ makes helicopter airborne. However, the horizontal push $\vec{F}_\text{H}$ to blades creates torque which changes total angular momentum of the helicopter. In order to stop body of helicopter spinning you have to create oposite torque and you do that by tail rotor (or in your case by friction of the floor). If there is no tail rotor or friction of the floor, body of the helicopter would spin faster and faster until the drag of the air due to body of the helicopter spinning is so large, that creates apropriate torque.

Newtonian Mechanics – How to Simulate Rotational Instability

$\vec\omega = I^{-1} \vec L$, and $\vec L$ is constant in the absence of external forces. The bit that I think you're missing is that $I$ rotates with the rigid body, so it is not constant in general and neither is $\vec\omega$.

I played with your online example, and the angular velocity does seem to always remain constant when I'm not poking the block, which is consistent with an inertia tensor that's a scalar multiple of the identity. I've never used Unity, but it seems to support arbitrary tensors of inertia via Rigidbody.inertiaTensor and Rigidbody.inertiaTensorRotation, so maybe you just need to set those properly.

If you end up having to roll your own physics, here's some naive, untested, and potentially wrong pseudocode:

const double time_step = ...;
const Quaternion L = {0, Lx, Ly, Lz};  // angular momentum
const double K1 = 1/I1, K2 = 1/I2, K3 = 1/I3;  // reciprocals of principal-axis moments of inertia; no idea if there's a standard letter for this
Quaternion orientation = {1, 0, 0, 0};
while (1) {
    Quaternion transformed_L = conjugate(orientation) * L * orientation;
    Quaternion transformed_omega = {0, K1 * transformed_L.i, K2 * transformed_L.j, K3 * transformed_L.k};
    Quaternion omega = orientation * transformed_omega * conjugate(orientation);
    orientation = quaternion_exp(time_step * omega) * orientation;
    // ... or orientation = normalize((1 + time_step * omega) * orientation);
    // ... or orientation = normalize((1 + time_step * omega + 0.5 * (time_step * omega)**2) * orientation);
    output(orientation);
}

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Related Solutions

[Physics] Conservation of angular momentum in helicopter

Newtonian Mechanics – How to Simulate Rotational Instability

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