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A cylindrical container is of mass M, radius R and height H. The cylinder is filled with a liquid of density D. The base of the cylinder has an outlet tap of radius r through which the liquid can flow out of the cylinder. Initially the center of mass of the cylinder and the liquid in the cylinder system is at a height of H/2 along the axis of the cylinder. At time t = 0, the outlet tap at the base of the cylinder is opened and the liquid starts flowing out. We want to study the locus of the center of mass of the system. As liquid flows out, the center of mass will start coming down till it reaches a minimum point ans then start rising up again until it returns back to the height of H/2 along the axis when the cylinder is completely empty.

Assumptions:

  1. The upper surface of the cylinder is closed.

  2. The experiment is done at an altitude A above sea level.

Questions:

  1. Where will the center of mass of the system be at time t = T?
  2. Find T at which the center of mass of the system will be lowest.
  3. How much liquid will be present in the cylinder when the center of mass of the system is lowest?
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  • $\begingroup$ Is the upper surface of cylinder open to atmosphere? $\endgroup$
    – evil999man
    Apr 17, 2014 at 5:41
  • $\begingroup$ @Awesome: Assume that the upper surface of the cylinder is closed. $\endgroup$ Apr 17, 2014 at 10:23
  • $\begingroup$ And you are not doing this experiment in vacuum? $\endgroup$
    – evil999man
    Apr 17, 2014 at 10:25
  • $\begingroup$ @Awesome: Assume that experiment is done at an altitude of A above sea level. This might complicate the calculation a bit I think this is a fair assumption because the experiment can be done anywhere. If A >> 1 we are in vacumn and if A = 0, we are at sea level. $\endgroup$ Apr 17, 2014 at 10:35
  • $\begingroup$ I will assume that $H<<A$ $\endgroup$
    – evil999man
    Apr 17, 2014 at 10:40

1 Answer 1

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Let the outside pressure be $P_0$. You can find its variation with altitude here.

Using Bernoulli Equation, speed of liquid coming out is : $$\rho gh=\frac{1}{2}\rho v^2+P_0$$ $$v(h)=\sqrt{2\rho gh-P_0/\rho}$$

$h$ is the height of water remaining. It is assumed that $r<<R$ and thus $V$ is very small compared to other terms and is neglected.

By Continuity Equation :

$$\pi R^2V=\pi r^2v$$

$V$ is speed with which water level comes down.

Equate it to $-\frac{dh}{dt}$

Then You have $h$ as function of time. Take a point as origin and use $$r_{cm}=\frac{m_1r_1+m_2r_2}{m_1+m_2}$$

You can use symmetry to find their respective centre of mass.

You can now set its derivative to zero to find maxima/minima(depending where you took origin)

I guess you can take from here.

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