Is angular velocity (the horizontal rotation of a massive disc falling through the air) sensitive to time dilation or invariant to it? Will its angular velocity (speed of rotation) increase, decrease or stay invariant when reaching lower altitudes?
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The angular velocity of the falling disk will slow down as measured by a stationary (non inertial) observer standing on the surface of the Earth, for two reasons:
As it falls, its vertical velocity increases and it will experience time dilation proportional to the velocity time dilation factor $\sqrt{1-v^2/c^2}$.
As it falls it moves from a higher gravitational potential to a lower gravitational potential and is subject to gravitational time dilation proportional to $\sqrt{1-\frac{2GM}{Rc^2}}$ where M is the mass of the Earth or whatever the main gravitational body is.
Total slow down of the angular velocity measured by a ground observer is proportional to: $$\sqrt{\left(1-\frac{v^2}{c^2}\right) \left(1-\frac{2GM}{Rc^2}\right)}$$
The clock of a free falling observer comoving with the disk, will also slow down by the same factor, so a co-falling inertial observer will see no change in the angular velocity of the disk (but the ground based non inertial observer will). Basically, the disk is a form of clock and is subject to velocity and gravitational time dilation like any other ideal clock. (A pendulum is not an ideal clock as it requires gravity to operate).
As a side note, what will happen to the pendulum as it falls? If the pendulum is at the top of its swing when it is released, it will appear to stop instantly the moment it is in free fall, according to a co-falling inertial observer. If it is not at the top of its swing, it continues in circles around its pivot with the tangential velocity it had at the instant it was released, rather than oscillate to and fro.
EDIT I have answered the main question. My answer to the subsidiary question of a pendulum, has been migrated to this related question.
KDP
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