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

`a_(cp)=7.77m/s^2`

Step-by-step explanation:

The equation for centripetal acceleration is
`a_(cp)=(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=(C)/(T)=2\pi r f`

Putting all together:

`a_(cp)=((2\pi r f)^2)/(r)=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2`