When you put water in a kitchen blender, it begins to travel in a 9-cm-radius circle at a speed of 1 m/s. How hard must the sides of the blender push inward on 0.006 kg of the spinning water?

The centripetal force on a rotating mass m is F = mv^2/R

Here m =0.006 kg, R=9cm = 0.09 m, v= 1m/s

F=0.006*(1)^2/0.09=0.067 N

Note: here I am assuming that all the water is concentrated in a relatively small volume, and the force is pressuring this volume.

There could also be a possibility that the water spread uniformly around the perimeter. In that case we can only give the force per unit length by dividing the the above result by the perimeter of the blender : 2*pi*R

F/perimeter = 0.067N/(2*pi*0.06m) =0.177 N/m

Apr 30th, 2015

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