# Non-constant angular momentum in gyroscope?

I want to ask about a simple gyroscope experiment: the one in which a man is holding a gyroscope, and sits on a rotating chair, and when he rotates the gyroscope, he himself starts rotating in regard of the conservation of angular momentum.

For reference, I am putting up this link to Feynman. Initially, only the gyroscope is rotating, and the total angular momentum of the system is a vector pointing in the right direction, which basically is the angular momentum of the gyroscope. Now, the man rotates the gyroscope in such a way that now it is facing in the upwards direction, towards the sky. Here, the gyroscope has an angular momentum in the upwards direction, so to negate this, the man and chair start rotating at the same rate, but in the opposite direction. However, in this case, the angular momentum becomes $0$. It should instead be a vector again pointing in the right direction.

What is the problem that I am making here?

• Why do you think that the angular momentum becomes zero? Are you thinking that it cancels out exactly, i.e. that the magnitude of the man+chair's angular momentum is equal to that of the gyroscope? Why would it? – Emilio Pisanty Sep 21 '17 at 17:42
• @EmilioPisanty even if the angular momentum is not zero, its direction is either perpendicular up or perpendicular down. Its not to the right, which is again a fallacy. – codetalker Sep 22 '17 at 14:55
• The naive reading of your question has both the initial and final gyroscope rotation axes vertical. If you mean something different, you should specify clearly what you mean - without relying on external resources. – Emilio Pisanty Sep 22 '17 at 15:23

## 1 Answer

The horizontal components of the total angular momentum are not conserved in this situation, because the axle of the chair is able to exert a torque about those axes. The only axis with respect to which the man-plus-gyroscope can be considered as isolated is the vertical (since the axle doesn't exert a torque, other than friction, and neither does gravity), and Feynman's analysis is confined exclusively to that component.