Question

What is the acceleration of the outer edge of a 14 inch diameter record that is spinning at 33 1/3 rpm? report answer in g's (1g=9.8m/s^2)

Answer 1

\[a_c = \frac{v^2}{r}=\omega^2r \]\[33\frac{1}{3} rpm = 33\frac{1}{3} * \frac{2\pi}{60} = 3.49 rad/s\]\[14 inches = 14*2.54 = 35.6 cm = 0.356 m\]\[a_c = \omega^2r=(3.49)^2*0.356 = 4.34 m/s^2 = 0.442 g\]

Answer 2

Thank you so much :)

Answer 3

But where did the 2.54 come from?

Answer 4

That's how many centimeters are in an inch

Answer 5

oh oops okay haha thank you again!

Answer 6

Sure :)