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 13.6 cm. When the cylinder is rotating at 2.73 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

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 13.6 cm. When the cylinder is rotating at 2.73 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

Answer 1

To find the magnitude of the centripetal acceleration at the outer wall of the rotating container, we can use the formula for centripetal acceleration:

a = (v^2) / r

where:
a is the centripetal acceleration
v is the linear velocity
r is the radius

First, let's calculate the linear velocity of the outer wall of the container. The linear velocity can be found by multiplying the circumference of the outer wall by the number of revolutions per second:

v = 2πr * revolutions

In this case, the radius is given as 13.6 cm (or 0.136 m), and the container is rotating at 2.73 revolutions per second. Plugging in these values, we get:

v = 2π * 0.136 * 2.73 = 2.229 m/s

Now, we can plug the calculated velocity and the radius into the formula for centripetal acceleration:

a = (v^2) / r
= (2.229^2) / 0.136
= 36.47 m/s^2

So, the magnitude of the centripetal acceleration at the outer wall of the rotating container is approximately 36.47 m/s^2.