# After several hours of trying to understand how to do part a), I absolutely cannot figure out the reasoning behind the answer. Please help!?

### 1 Answer

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- Barry GLv 74 years agoFavorite Answer
a) Consider a small cube of fluid with dimensions dr x dr x dr at distance r from the axis. Suppose the pressure on the sides of the cube is P radially outward and P+dP radially inward. The net force on the cube radially is F=AdP where A=dr*dr. The mass of the fluid in this cube is m=ρdV=ρAdr. The centripetal force on the cube of fluid, keeping it moving in a circle, is mrω^2. This is provided by the difference in pressure. So

AdP=mrω^2=(ρAdr)rω^2

dP/dr=ρrω^2.

b) P=(1/2)ρr^2ω^2+Po.

c) The pressure P acts in all directions, vertically as well as horizontally. So P also equals the pressure at a depth h below the curved surface, which is Po+ρgh. Therefore

Po+ρgh=(1/2)ρr^2ω^2+Po

h=r^2ω^2/2g.

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