# Is a falling cat's angular momentum conserved?

I found this question in my physics textbook:

From a certain height a cat is dropped back-side down. The cat rotates his body while falling and lands on his four legs. Does the cat's angular momentum change during the fall?

The answer is no, but I said yes, because I thought the gravitational force will change the angular momentum? Am I missing something?

• Your confusion may be that you have to define relative to which point you measure the angular momentum. E.g., relative to the floor immediately beneath my chair, the angular momentum will be different from zero when I spin my chair. But relative to the seat of my office chair, my angular momentum is always zero (the chair spins with me!). Likewise, in the frame of the cat, or, more generally, in any frame along the (straight) trajectory of the cat, angular momentum will be constant. But relative to an observer that isn't trying to catch the cat, its angular momentum will increase as it falls. – tobi_s Feb 20 at 9:42

To change angular momentum, a torque must be applied. Since gravity pulls every part of the cat with a force proportional to its mass (that is, with the same acceleration), there is no net torque on the free falling cat, and thus no change in angular momentum.

This is true for any free falling object, but not necessarily if it is supported at any point. The support together with the gravitational force can apply a torque and therefore change angular momentum.

As to how the cat manages to turn around even with no net torque, this is known as the Falling cat problem, and is visualized nicely in this very disturbing animation from Wikipedia The rotation is based on the fact that the cat is not a rigid body, and can thus bend in a way that results in its reorientation.

• Wow that is so interesting I didnt know there was a thing like this it seems like there is some physic tricks going behind this. I need more time to observe it to understand better but at least I know which way I have to think now, many thanks! – E.Berk Feb 19 at 17:29
• Slow Motion Flipping Cat Physics ... good video and channel :) – Steve Feb 20 at 2:26

An external force like the gravitational force acts on an object like if it acts on its center of mass. Since the cat's center of mass is on its rotational axis, this would mean that the gravitational force doesn't give any angular momentum to the cat.

In general, a force does not always give angular momentum to an object. It will if the force is applied at a certain distance from the rotation axis.

The gravitational force can indeed give angular momentum to a system. Think of a pendulum that you drop after raising it from it's rest position. In that scenario, the rotation axis is the pendulum holding point and the center of mass would be close to the end of the pendulum. Thus, the gravitational force acts away from the rotational axis and the pendulum will start to rotate around its holding point, thus gaining angular momentum.

• Assuming the cat's center of mass is very close to its rotational axis Why assume something that is (trivially) true? A free body will always rotate exactly around its center of mass. – Sanchises Feb 19 at 17:38
• No the reaction from the attachment point is perpendicular to the motion of the pendulum and thus the torque is 0. here it's truly gravity that affects angular momentum. – fgoudra Feb 21 at 10:15
• Yep you're right, I was mixing up some things. Second comment deleted! – Sanchises Feb 21 at 10:37

The problem is flawed because it does not specify which axis we are supposed to be measuring the angular momentum around.

Angular momentum is always specified relative to an axis of rotation. However, you can always split the total angular momentum of an object into two parts, $$L_{total} = L_{external}+L_{CM}$$.

The first part is the angular momentum of the whole object around some external axis of rotation. Specifically, $$L_{external} = \vec{R}_{CM}\times \vec{P}_{CM}$$ where $$\vec{R}_{CM}$$ is the position vector of the center of mass of the object, and $$\vec{P}_{CM}$$ is the momentum of the center of mass.

The second part is the internal angular momentum, or the angular momentum of the object measured around its own center of mass.

Gravity can change the external angular momentum, depending on the axis of rotation you choose. The torque from gravity around an external center of mass is just the weight of the object times the horizontal distance of the center of mass from the axis.

What the question is probably trying to get, however, at is the fact that gravity cannot change the internal angular momentum of an object. Since gravity "acts" at the center of mass, the torque of gravity around the center of mass is always zero!