Question

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per minute) if the wheel's diameter is 35 cm?

Answer 1

Answer:

$a_{cp}=7.77m/s^2$

Explanation:

The equation for centripetal acceleration is $a_{cp}=\frac{v^2}{r}$.

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference $C=2\pi r$ and the period T, then we have:

$v=\frac{C}{T}=2\pi r f$

Putting all together:

$a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2$