There is a clever kitchen gadget 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 13.7cm. When the cylinder is rotating at 1.71 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
To find the magnitude of the centripetal acceleration at the outer wall, we can use the formula for centripetal acceleration:
a = ω^2 * r
Where:
a = centripetal acceleration
ω = angular velocity
r = radius
Given:
ω = 1.71 revolutions per second
r = 13.7 cm = 0.137 m
First, let's convert the angular velocity from revolutions per second to radians per second:
1 revolution = 2π radians,
1.71 revolutions/s = 1.71 * 2π radians/s = 10.748π radians/s
Now we can substitute the values into the formula to find the magnitude of the centripetal acceleration:
a = (10.748π)^2 * 0.137
≈ 463.74 m/s^2
Therefore, the magnitude of the centripetal acceleration at the outer wall is approximately 463.74 m/s^2.