Proper use of inequality symbols in equations to do with circular motion

AI Thread Summary
The discussion focuses on the proper use of inequality symbols in equations related to circular motion, specifically for a roller coaster car at the top of a loop. The key equation involves the reaction force (A) and the weight (W = mg), establishing that A must be greater than or equal to zero to prevent the car from falling off the tracks. The user initially struggles with how to incorporate the inequality logically into their calculations, particularly when determining the minimum velocity (v) required for the car to stay on the track. After some deliberation, they conclude that their approach to introducing the governing equation and the inequalities is sound. The conversation emphasizes the importance of correctly applying inequalities in physics equations to ensure logical consistency.
Ebby
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Homework Statement
What is the minimum initial speed of the rollercoaster such that it will complete the vertical loop without falling away?
Relevant Equations
F_net_centripetal = mv^2/R
G.P.E._initial + K.E._initial = G.P.E._final + K.E._final (assuming no friction losses)
The problem itself is easy. My question is regarding the proper use of inequality symbols.

loop1.jpg

loop2.jpg

loop3.jpg


I only need to do the first part to show where I am having the issue.

The forces I need to consider are the coaster car's weight ##W = mg## and the reaction ##A## of the tracks acting on it. With the car at the top of the loop, the reaction ##A## must be greater than or equal to zero or the car will have fallen away from the tracks: $$A >= 0$$
Ignoring the inequality for the moment (yes, this isn't very mathematical), I'll just take the case when: $$A = 0$$
The governing equation is: $$A + mg = \frac {mv^2} {R}$$
When ##A = 0##, I can solve for ##v##: $$v = \sqrt {gR}$$
And now I consult my intuition. I visualise the coaster performing the loop and I "see" that I must reinsert the inequality like this:$$v >= \sqrt {gR}$$
But this isn't very satisfactory. Surely I can do this inequality stuff so that one step in the calculation properly follows another - without any tricks. Perhaps I should introduce the governing equation like so: $$A = \frac {mv^2} {R} - mg$$
And then I could argue that because actually ##A >= 0## it must also be the case that: $$\frac {mv^2} {R} - mg >= 0$$
But I'm not sure this is logically sound... Can someone explain how to do this so it makes sense to me?
 
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Ebby said:
Perhaps I should introduce the governing equation like so: $$A = \frac {mv^2} {R} - mg$$
And then I could argue that because actually ##A >= 0## it must also be the case that: $$\frac {mv^2} {R} - mg >= 0$$
But I'm not sure this is logically sound... Can someone explain how to do this so it makes sense to me?
Looks perfectly logical to me. What worries you?
 
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haruspex said:
Looks perfectly logical to me. What worries you?
I think I'm ok with it now. I thought that there was some circular logic in there, but on further consideration I think it's all right.

Thanks :)
 
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