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A horizontal force is applied to the top of a cylinder, creating a torque on it trying to tip it over. Now I have wondered to what height should the cylinder be filled with water to render it most stable. I first tried to calculate at what filling level will the center of mass be lowest, since at that state the torque created by the total mass of the cylinder opposing the force tipping it is largest. (Will be furthest horizontally from the cylinder-ground pivot point when tipping begins:

h- height of center of mass of the filled cylinder

m_1- mass of empty cylinder

h_1- height of center of mass of the empty cylinder

m_2- mass of the body of water

h_2- height of center of mass of the body of water

p_2- density of water

v_2- volume of water

r- radius of cylinder base

h = (m_1h_1 + m_2h_2)/(m_1 + m_2)

h = (m_1h_1 + p_2v_2h_2)/(m_1 + p_2v_2)

h = (m_1h_1 + p_2(2h_2pier^2)h_2)/(m_1 + p_2(2h_2pier^2)) //height of the body of water is double the height of its center of mass

Now I will fill in the constants to make things clearer m_1 = 0.1 kg

h_1 = 0.3 m

p_2 = 997 kg/m^3

r = 0.05 m

h = (0.015 + 15.66*(h_2)^2)/(0.1 + 15.66*h_2)

Now I will differentiate this equation for center of mass and equal to 0.

0 = (0.1 + 15.66h_2)31.32h_2 - (0.015 + 15.66(h_2)^2)*15.66

Solved this part by calculator to spare the rows: h = 0.0252 m or 2.52 cm.

QUESTION:

First of all are my calculations correct and is the center of mass lowest with this height of water at this cylinder?

Second of all, I realize that adding more water will increase the cylinder momentum there by reducing the force impact upon it, despite heightening the center of mass.

So how could I get both aspects into account to find the actual value (are there any more contributing aspects- considering perfect cylinder, no air resistance and the like)?

Qmechanic
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Ahdriam
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  • It really depends on the force being applied. If the force remains at the same absolute elevation, the leverage it has on the fulcrum will be reduced. If it remains at the same elevation as the top corner of the cylinder, the math will be simpler. – Daniel Jun 11 '23 at 00:22
  • You may already understand this. But you haven't mentioned anything about figuring out what torque it takes to tip over the cylinder. Toppling of a cylinder on a block – mmesser314 Jun 11 '23 at 02:43
  • It is an instantaneous force. And does it matter what torque will tip it over? For two different cylinders the same torque will nudge the cylinder to a different extent. And I wander how to get the least nudge for any torque, not necessarily preventing it from tipping over – Ahdriam Jun 11 '23 at 09:00

0 Answers0