What is the acceleration of the edge of the rotor?

Question

An electrical motor spins at 1575.0 rpm, its radius is 7.204 cm.

Answer 1

The (inward) acceleration of the (outer) edge of the rotor is
~1959.7 m s^(-2), or ~ 1960 m s^(-2), since only 4 sig figs. are really justified as the radius r is itself only given to 4 sig figs.

Here's how I obtained this value:

1575.0 rpm (revs per minute) = 1575.0 / 60 rps (revs per second)

= 1575 2 pi / 60 radians per second = 'w' (Greek omega), the ANGULAR velocity.

Inward acceleration of (outer) edge of rotor = w^2 r cm s^(-2)

= [(1575 pi / 30)^2] 7.204 cm s^(-2)

= 0.01 [(1575 pi / 30)^2] 7.204 m s^(-2) = ~1959.7 m s^(-2).

(Rounding to the number of sig. figs justified by the input data gives
~ 1960 m s^(-2).)

Live long and prosper.

Answer 2

My guess aprox 5-1 ratio