- #1

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- Homework Statement
- A cone (with height H and radius R) has a total volume V, contains a fluid of density ρ and volume mV, where m is a fraction (0<m<1). Such cone is rotated about its axis of symmetry and accelerates very slowly. The magnitude of gravity is g. Find the volumetric flow rate at which the liquid overflows from the container.

- Relevant Equations
- p1-p2=pw^2(r1^2-r2^2)/2-pg(z1-z2)

I solved the case where m=0.99999. Then the height at which it overflows can be obtained with the equation, when points on the liquid surface are chosen. Then the cross-sectional area is given by the circumference of the circle times the height that the parabola reaches, that cross-sectional area multiplied by the linear velocity, which is obtained from v=wR, gives the volumetric flow.

I don't know if my approach is correct, but I want to solve for all cases, not just one. And I'm not quite sure how to do that, If m is small, it is troublesome to find the new hight, the one that starts at H so the liquid overflows. Also the fact that it accelerates very slowly, makes me think a minimum velocity is reached, then How big has to be the velocity for the cone to overflow? I can obtain the angular velocity before it overflows, but not while it overflows. I could find it if I knew the acceleration.

I don't know if my approach is correct, but I want to solve for all cases, not just one. And I'm not quite sure how to do that, If m is small, it is troublesome to find the new hight, the one that starts at H so the liquid overflows. Also the fact that it accelerates very slowly, makes me think a minimum velocity is reached, then How big has to be the velocity for the cone to overflow? I can obtain the angular velocity before it overflows, but not while it overflows. I could find it if I knew the acceleration.