Answer to Question #239646 in Quantum Mechanics for Matthias

Question

What is the centripetal acceleration of a point on the trim of flywheel 1.20 m in diameter, turning at a rate of 1,200 r/min?

Answer 1

The centripetal acceleration of a point on a rotating object is given as follows:

$$a= \dfrac{v^2}{R}$$

where $v$ is the linear speed of the point and $R = 1.20m/2=0.6m$ is the distance from the center of rotation to the point.

Let the frequency be $f = 1200r/min = 20Hz$. Then the speed is:

$$v = 2\pi fR$$

Thus, obtain:

$$a = \dfrac{(2\pi fR)^2}{R} = 4\pi^2 f^2R\\ a = 4\pi^2\cdot (20Hz)^2\cdot 0.6m \approx 9.47\times 10^3m/s^2$$

Answer. $9.47\times 10^3m/s^2$.