Cleonis said:
f a trajectory is circular then the centripetal force is not doing work. When the trajectory is a spiral then the work that is being done is causing rotational acceleration.
No.
Peculiarly, quite a few authors write statements such as: "The person on the chair starts rotating faster because his moment of inertia decreases." I believe that line of reasoning violates causality.
Quite a few authors say that because they are correct.
Can you lift yourself by your bootstraps? No. The reason is because pulling on your bootstraps is an internal force. An external force is required to change your momentum. Similarly, an external torque is required to change your angular momentum. In the ideal case, a person on a frictionless spinning chair, flailing ones arms about (or pulling them in) does not result in an external torque.
Look at the rotational equation of motion:
\frac{D\vec L}{dt} = \frac{d}{dt}\left(\mathbf I \vec {\omega}\right) = \vec{\tau}
No torque is required to change the angular velocity. In fact, if there is zero torque (as it is in the ideal case), angular velocity *must* increase to keep the angular momentum constant when the moment of inertia is decreasing.
The causal factor is the centripetal force.
A causal factor is indeed needed to change the moment of inertia. Work is being done here! However, it is mistaken to think of this as necessitating a torque. Torque is change in angular momentum, not angular velocity. The kinetic energy of the system is
T = \frac 1 2 \left(\vec L \cdot \vec {\omega}\right)
Differentiating wrt time,
\frac {dT}{dt} = \frac 1 2<br />
\left(\frac{d\vec L}{dt}\cdot \vec{\omega}<br />
+ \vec L \cdot \frac{d\vec{\omega}}{dt}\right)
With no external torque, this reduces to
\frac {dT}{dt} = \frac 1 2 \vec L \cdot \frac{d\vec{\omega}}{dt}
In the simple freshman physics case of rotation about an eigenaxis of the system, angular momentum is parallel to angular velocity and angular acceleration. The above vector equations become scalar equations.
\aligned<br />
\dot{\omega} &= -\,\frac{\dot I}{I} \omega \\<br />
\dot T &= \frac 1 2 \dot I \omega^2<br />
\endaligned
Work must be being done to accomplish that increase in kinetic energy. Just because work is being done does not mean that a torque must exist.