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

In summary, the conversation discusses the proper use of inequality symbols in a problem involving a coaster car and its weight and reaction forces. The main concern is whether the inequality should be included in the calculation or not. The solution is to introduce the governing equation and argue that because the reaction force must be greater than or equal to zero, the expression must also be greater than or equal to zero. This is deemed logically sound and the person is satisfied with the explanation.
  • #1
Ebby
41
14
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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  • #2
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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  • #3
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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