The Rotational Variables — Question 1

Question 1

An encoder wrap is not a backward turn

A shaft encoder reports angles only in $[0,360^\circ)$. It reads $350^\circ$ and then $20^\circ$ after $0.10\,\mathrm s$. The shaft is known to turn counterclockwise by less than one full revolution during that interval.

1. Find its unwrapped angular displacement and average angular velocity.

2. Without the direction and less-than-one-turn conditions, what family of displacements fits the same readings?

Question 1 - Solution

An encoder wrap is not a backward turn

(a) Derivation and calculation

The encoder reports an orientation modulo $2\pi$. For the specified forward crossing, $$\Delta\theta=(360^\circ-350^\circ)+20^\circ=30^\circ =30\frac{\pi}{180}=\boxed{\frac{\pi}{6}\ \mathrm{rad}}.$$ $$\bar\omega=\frac{\Delta\theta}{\Delta t}=\frac{\pi/6}{0.10} =\boxed{\frac{5\pi}{3}=5.236\ \mathrm{rad/s}}.$$

(b) Result and interpretation

Without the extra motion information, the unwrapped difference is $$\boxed{\Delta\theta_k=(20-350)^\circ+360^\circ k =-\frac{11\pi}{6}+2\pi k,\quad k\in\mathbb Z}.$$ The conditions $0<\Delta\theta<2\pi$ select $k=1$.

Physical check

The positive displacement crosses zero once and is less than one full turn.