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miss Akunina [59]
3 years ago
7

A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen

terline, and the water level drops at the center while it rises at the edges.
Determine the angular velocity at which the bottom of the tank will first be exposed.


Also determine the maximum water height at the moment.
Engineering
1 answer:
arlik [135]3 years ago
7 0

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2}   }{4g} (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s

if you consider the equation 1 for the free surface at the edge is equal to

z_{s(R)} =h_{0} +\frac{w^{2}R^{2}   }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft

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T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

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