How Does a Rotating Pail Affect Water Surface Shape?

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The discussion focuses on determining the shape of the water surface in a rotating pail, specifically how the surface curves from the center to the outer edge due to angular velocity. The slope of the surface is defined as dy/dx = tan(theta), and resolving forces is necessary to express tan(theta) as a function of the radial position x. Given a pail with a diameter of 0.45 m rotating at one revolution per second, the problem requires calculating the height of the water surface at the outer edge. The approach includes using the ideal fluid assumption and analyzing forces acting on a point on the water surface. The discussion emphasizes the importance of understanding fluid dynamics in this context.
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Homework Statement


A rotating pail of water with an angular velocity w has a surface curving up from the centre to the outer edge. Find the shape of the surface. The slope of the surface is dy/dx = tan(theta) at position x is y (= function of x) is the shape of the surface. You must resolve forces to find tan(theta) as a function of x, and then solve for the height, y, as a function of x.

If the pail rotates once per second, and has a diameter of 0.45 m, how high above the centre is the outer edge?



Homework Equations





The Attempt at a Solution


All I see is that w = 2pi rad/sec and r = 0.225 m.
 
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Method I:
Ideal Fluid, by definition, can not tolerate any shear force. Use this fact.
Assume the centre of the water surface to be origin. Let P(x,y) be a point on the water surface. Consider forces on it. Impose the condition that tangential force will be zero.
I hope you can proceed now.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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