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One can get that by solving this equation: $$\vec \nabla P = - \rho \cdot \vec a$$ substitue $a = \omega^2\cdot r \cdot \hat r$ and solve the equation. (Full answer cannot be provided because of homework-exercises tag) How is the above equation derived? $\mathbf{\vec \nabla P = - \rho \cdot \vec a}$ Take a small volume of liquid in the container $dV = ...


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You put considerable effort on merry go rounds (Even when seated) in order to subconsciously show there is no such thing/s. Or you are just wired up to face the rotating system you are in and your brain is taking charge. and $a=0$. You can't fool the Coriolis part is a lot more sneaky. Just look at that term with the angular velcoity vector crossed with ...


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The confusion is not realizing that real forces are real, and fictitious forces are not real. Therefore, they can not "cancel each other out" in real life. Using a rotating frame of reference and introducing fictitious forces is just adding and equal quantities (a force, and a mass $\times$ acceleration) to both sides of the math equations. It has nothing ...


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For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we (I suppose) have to write. It is a real force, after all, making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this? If you are in a frame ...


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Short answer: centrifugal force may not be a real force but centripetal force certainly is. Long answer: A weighing machine (which I am assuming is equivalent to a spring balance) is not measuring the force with which the Earth attracts an object. It is actually measuring the reaction force that the ground (or, to be exact, the spring) exerts in the object. ...


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Psuedo-forces aren't reality, they are the need of humanity! You must account for a centrifugal force in a non-inertial reference frame otherwise you can't use $F = ma$ to figure out the dynamics of the problem. Though as you may have noticed Newton's third law still isn't valid in such frames. And you must have observed that Earth is a non-inertial ...


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A quick edit: This calculation results in a negative oblation (of correct magnitude). I couldn't find my mistake (I even tried to integrate surface tension to it below), however I found this source from UTexas that finds a formula using potentials rather than forces (and lots of analytical mechanics). $$\gamma\frac{2}{R}S=\frac{GMm}{R^2}$$ $$-\frac{GMm}{(R+\...


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