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 rooted 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 container and turn the crank to spin the off the water. The radius of the container is 13.7cm. When they 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 of the cylinder, we will use the formula for centripetal acceleration:
a = ω^2 * r
where a is the centripetal acceleration, ω is the angular velocity, and r is the radius.
Given:
ω = 1.71 revolutions per second
r = 13.7 cm = 0.137 m
First, let's convert ω from revolutions per second to radians per second:
1 revolution = 2π radians
So, ω = 1.71 revolutions/second * 2π radians/revolution.
ω = 1.71 * 2π ≈ 10.74 radians/second
Now, let's substitute the values into the formula:
a = (10.74 radians/second)^2 * 0.137 m
a ≈ 146.89 m/s^2
Therefore, the magnitude of the centripetal acceleration at the outer wall is approximately 146.89 m/s^2.