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 13.9 cm. When the cylinder is rotating at 1.50 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
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 13.9 cm. When the cylinder is rotating at 1.50 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
@Physics

Answer 1

\[a=v^2/r\]the circumference of the bowl is \[2 \pi r=2\pi*13.9\approx87.3cm\]the velocity at the edge is \[v=d/t=87.3*1.5\approx131cm/s\] plug those number into our first equation\[:a=v^2/r=131^2/13.9\approx1230cm/s^2\]

Answer 2

PS: d is the distance a point on the edge of the bowl travels in one second
so \[d=2\pi r*1.5\approx131cm\]in case you didn't see where that number came from.

Answer 3

why mutiply by time? thought v is distance/time taken