Relative angular acceleration

Question

Two solid bodies rotate about stationary mutually perpendicular and intersecting axes with constant angular velocity $X$ and $Y$.
Find the angular acceleration of one body relative to another .

My approach: well, differentiating the velocity would get me acceleration but velocity is constant so i need to differentiate the direction ??? How do i do that? No clue , No idea.

Let one axis be along $x$ and another along $y$ axis. but what next ? pls help.

Answer 1

I think an intuitive solution is to consider the angular velocities as vectors in space. Technically they might be pseudovectors but the answer works out the same.

Remember that velocity is composed of speed and direction, so if the direction is changing then the velocity is not constant. Only the speed of rotation of the second body is constant in the first body’s frame.

A graphical solution is to take the vector representing the velocity, attach the “starting” end of it to a fixed point, and see how the other end moves as the velocity vector changes in magnitude, direction, or both. The speed and direction in which the end of the vector moves is the magnitude and direction of acceleration.

It is not really so different from using vectors to calculate the linear acceleration of a point mass moving at constant speed around a circle. Again the speed is constant and only the direction is changing. That means the acceleration is perpendicular to the velocity vector.

In the non-rotating frame both vectors are fixed; therefore in the frame of the object rotating around $X,$ the vector $Y$ is rotating with angular velocity $-X.$ This gives an angular acceleration of $-X\times Y,$ using essentially the same math that gives the angular acceleration (and hence the torque) for a gyroscope processing around an axis perpendicular to its own axis.

Answer 2

$$acceleration=\vec{{\alpha}}=d\omega/dt=d/dt(X\vec{{i}}-Y\vec{j})$$
and $$d/dt(\vec{i})=-Y\vec{k}$$
and $$d/dt(\vec{j})=0$$ Hence, $$\vec{\alpha}=-XY$$