Question

here 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 10.7 cm. When the cylinder is rotating at 1.65 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall

Answer 1

11.5 m/s²

Step-by-step explanation:

The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s

So, a = rω²

a = r(2πN)²

a = 4π²rN²

substituting the values of the variables into the equation, we have

a = 4π²rN²

a = 4π²(0.107 m)(1.65 rev/s)²

a = 4π²(0.107 m)(2.7225 rev²/s)²

a = 4π² × 0.2913075 mrev²/s)²

a = 11.5 m/s²