Answer to Question #166437 in Classical Mechanics for Ahmed Alzaabi

Question

An electrical motor spins at a constant 2857.0 rev/min. If the armature radius is 2.685 cm, what is the acceleration of the outer edge of the armature?

Answer 1

We can find the acceleration of the outer edge of the armature from the formula:

$$a=\dfrac{v^2}{r}=\dfrac{(r\omega)^2}{r}=\omega^2 r,$$ $$a=(2857\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot2\pi\ rad)^2\cdot0.02685\ m=2401\ \dfrac{m}{s^2}.$$