Question

a cylindrical pot is filled with water. the center of mass of the empty pot is at a height 10cm the mass of the pot is 1 kg and it's inner area is 0.4m^2.what is the height of water in it,if the center of mass of the system is at the lowest position?​

Answer 1

Answer:

I believe you're missing the height of the pot.  Or...V = Pi*r^2 h

so, dV/dt = Pi*(9)hdh/dt = 8.1Pi*h = Constant.  But you have to get h; something like, "When h is..."  should be in the question.  Just multiply by that h when you find it and you got your answer

Explanation:

Answer 2

The height of water in the cylinder is 2cm at the center of mass of the system at the lowest position.

Given:

m1 = 1 Kg , H= 10cm , A = 0.4 m² and ρ =1000 Kg /m³

To Find:

The height of water in cylinder.

Solution:

y coordinate of the  center of mass of cylinder is

$y_{cm} =\frac{m1y1+m2y2}{m1+m2}$

y1 = H and y2= h/2

m2 = 0.4 *h *1000 = 400h kg

Substituting the values in the equation

$y_{cm} =\frac{1*H+400h*\frac{h}{2} }{1+400h}$

$y_{cm} =\frac{H+200h^{2} }{1+400h}$

The Center of mass of the system will be the lowest position when dy /dh = 0

Differentiating the equation with respect to h

$\frac{dy_{cm} }{dh} =\frac{(1+400h)(0+400h) -(H+200h^{2})(400) }{1+400h}$

$0 =\frac{400h+160000h^{2} -400H-80000h^{2} }{1+400h}\\\\0=200h^{2} +h -H$

Here a= 200 ,b = 1 and c= -H

$h =\frac{-1+\sqrt{1+4*200*0.1} }{2*200} \\\\h =\frac{8}{400}\\\\ h = 0.02m = 2cm$

∴ The height of water in the cylinder is 2cm at the center of mass of the system at the lowest position.

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