- #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.

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?

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?