Question

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 10.3 cm. When the cylinder is rotating at 2.30 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

Answer 1

.a = 849.05 m / s²

Step-by-step explanation

The centripetal acceleration is

a = v² / r

Linear and angular velocity are related

v = w r

Angular velocity and frequency are related by

w = 2π f

Let's replace

a = w² r

a = 4π² f² r

Let's reduce to the SI system

f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

.r = 10.3 cm = 0.103 m

Let's calculate

a = 4π² 14.45² 0.103

.a = 849.05 m / s²