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EpicFishFingers

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**1. So the loop is known to be 12m in diameter (6m radius). Assume the car weighs 1250kg - how fast must the car go to clear the loop, and what will be its max G-forces endured?**

This was the question, which I have broken into 3 sections:

a) Velocity at top of loop (done)

b) Velocity at start of loop (help?)

c) Max Gs

This was the question, which I have broken into 3 sections:

a) Velocity at top of loop (done)

b) Velocity at start of loop (help?)

c) Max Gs

**(I know the method, but it requires answer to part b)**

## Homework Equations

:[/B]**a) f = mv^2/r, where f = force (N) [used to calculate G-forces], m = mass of car (kg), v = velocity (m/s), r = loop radius (m)**

**b) v^2 = u^2 + 2as where u = start velocity, v = end velocity (both m/s), a = acceleration (m/s^2), s = distance traveled (m)**

**c) f = mv^2/r again**

## The Attempt at a Solution

I assumed I'd need to pull a minimum of 1.5Gs on the top of the loop, using the extra 50% as my safety factor. I rearranged the equation to make this:

v = (fr/m)^1/2

From which I got the velocity value of 9.4m/s, where f = 1.5 x 9.81 x 1250 = 18400N.

I undertook part b by simply calculating the velocity the car would gain by being dropped from a height of 12 metres, and adding that to the velocity attained from part a, and that would be a conservative stab at how fast the car would need to enter the loop.

Therefore v = 9.4 + 15.34 = 24.74m/s

Assuming the car does not accelerate or brake on the loop, I calculated part c using this value, and got f = 127,514.1 N, which converts to 10.4Gs, which is grossly excessive.

Nevertheless I emailed this to my teacher and told them it seemed wrong, most likely because of the 1.5Gs I stated as the minimum at the top of the loop.

Here's the part where I'm confused: they said I was over-complicating things regarding the entry velocity of the car (part b), and that I should think about conservation of momentum.

Now I'm sure this would involve kinetic and gravitational potential energy equations, I just can't think how to apply them to this scenario.

I can work out the car's kinetic energy at the top of the loop from the equation KE = 1/2 m v^2, but then what? If I try to work out the kinetic energy at the start of the loop, I don't know the entry velocity, or the kinetic energy.

What do I do??

EDIT: I found this: http://www.real-world-physics-problems.com/roller-coaster-physics.html but it's what I did anyway.

It factors both PE and KE together to give the equation v = (2gh)^1/2.

So if we let the height = 12m, g = 9.81m/s^2, we can calculate v... this seems too easy.

So v at the start of the loop = (2 x 9.81 x 12)^1/2 = 15.34m/s, which is what I got in the first place for part b before I added 9.4m/s, so this is just the same thing again. Still stuck...

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