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 13.2 cm. when the cylinder is rotating at 1.70 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

Question

Elizabeth Oneill

Physics

18 March, 18:42

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 13.2 cm. when the cylinder is rotating at 1.70 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

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Cottonball · Answer 1

The formula for centripetal acceleration is v^2/r, where v is the tangential velocity, and r is the radius.

Alternatively, w^2r where w is the angular velocity and r is the radius.

The centripetal acceleration is ar = (((1.70 rev/s) * (2*pi rad/rev)) ^2) * (0.132m) = 15.06 m/s^2