A cylindrical pot is slowly filled with water. The centre of mass of the empty pot is at a height of 10 cm, the mass of the pot is 1 kg, and its inner area is 0.4 m2. What is the height (in cm) of the water in it, if the centre of mass of the system is at the lowest position? (Take density of water 1000 kg/m3 )

Engineering

Physics

Center of Mass

Question

A cylindrical pot is slowly filled with water. The centre of mass of the empty pot is at a height of 10 cm, the mass of the pot is 1 kg, and its inner area is 0.4 m². What is the height (in cm) of the water in it, if the centre of mass of the system is at the lowest position?
(Take density of water 1000 kg/m³ )

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The position of centre of mass of the system is y_cm.

$y_{cm} = \frac{m_{1} \times H + m_{2} \times \frac{h}{2}}{m_{1} + m_{2}}$

Where m₁ = 1 kg, m₂ =(0.4 × h × 10³) kg
= (400 h) kg

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

for y_cm to be lowest (minimum)

$\frac{{dy}_{cm}}{dh} = 0$

200 h² + h –H = 0
h = 2 cm

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