A tank of water and a mass that is its perfect counterbalance on a pivot. The pu

Question

A tank of water and a mass that is its perfect counterbalance on a pivot. The puncture is in the shape of an annulus centered about the tank's center. An annalus is a ring shaped object bounded by two concentric circles.

The annulus has an inner diameter of Di and an outer diameter of Do. Water leaks through the shaded area bounded by the two diameters.

What is the change in mass necessary to keep the tank balanced after the puncture is made (compared with the mass before the puncture is made)? Additional given variables are: p,h.

Answer:

m=-(p*pi*h/2)(Do^2-Di^2)

Explanation / Answer

As initially tank and mass are in perfect counterbalance, it can be concluded that there initial masses are equal.

Now, after puncturing the tank, there will be leakage and some mass will be lost. If that mass is exactly reduced from the counterbalancing mass also, then balance can be maintained. Hence, the mass lost due to leakage is:

= Density * Volume

Taking, density = p

Volume = Area * Height

Mass lost due to leakage = p * Area * h

Now, area of the leakge = pi * ( Do^2 - Di^2 ) / 4 ....................(Area of ab annulus)

Hence, mass lost due to leakge = p * pi * h * ( Do^2 - Di^2 ) / 4

This same quantity of mass has to be reduced, hence, change in mass = - ( p * pi * h / 4 ) ( Do^2 - Di^2 )

It is -ve, as it reduction in mass.

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