Roller Coaster Loop Problem

1. The problem statement, all variables and given/known data
The diagram below shows a roller coaster ride which contains a circular loop of radius r. A car of mass m begins at rest from point A and moves down the frictionless track from A to B where it then enters the vertical loop (also frictionless), travelling once around the circle from B to C to D to E and back to B, after which it travels along the flat portion of the track from B to F (which is not frictionless).

1. Find the centripetal acceleration of the car at points B and C.
2. Find the normal force at point B.

2. Relevant equations
Fnet = mv^2/r
ac = v^2/r


3. The attempt at a solution
1. I got Ac = Fn/m - g for point B. For point C i got Ac = -Fn/m + g
2. I got Fn = mv^2/r + g


I got -2 marks off for this problem.. where did I go wrong?

By the way, the loop goes counter clockwise (B, C, D, E)
 

LowlyPion

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If those are your answers, I'd say you didn't answer the question.

They asked for centripetal acceleration. Just curious why you involved g?
 
If those are your answers, I'd say you didn't answer the question.

They asked for centripetal acceleration. Just curious why you involved g?
Fn - mg = mAc
 

LowlyPion

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Well you are expected to come up with a value for that. Your answer is no answer at all.

For instance what is it at the point B?

I can't see your diagram but I presume it is at the base of the loop where the horizontal angle is 0°. So what is the Fn at that point?
 
Well you are expected to come up with a value for that. Your answer is no answer at all.

For instance what is it at the point B?

I can't see your diagram but I presume it is at the base of the loop where the horizontal angle is 0°. So what is the Fn at that point?
Fn at point B is Fg ??
 

LowlyPion

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it equals 0? =S
That might have made a better answer then. Now if your point B is just a hair past the 0° point its acceleration will be v2/r. In fact it may be open to debate whether right at B there is centripetal acceleration depending on how you want to interpret the drawing, (which I am disadvantaged in not being able to see).

What I can tell you is that answering Fn - mg as the centripetal acceleration is not showing how much you should know.
 
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That might have made a better answer then. Now if your point B is just a hair past the 0° point its acceleration will be v2/r. In fact it may be open to debate whether right at B there is centripetal acceleration depending on how you want to interpret the drawing, (which I am disadvantaged in not being able to see).

What I can tall you is that answering Fn - mg as the centripetal acceleration is not showing how much you should know.
Right understandable.

So at point C, what would be the appropriate answer if the tangent to the circle at this point is perpendicular to the x axis?
 

LowlyPion

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I will guess though that it is at 3 o'clock.

In which case the centripetal acceleration will be determined by its speed after losing 1 radius worth of potential energy.
 
I will guess though that it is at 3 o'clock.

In which case the centripetal acceleration will be determined by its speed after losing 1 radius worth of potential energy.
I don't understand. We didn't learn potential energy yet.. How can I calculate the acceleration without it?

And also for part B, There is no way to get value for Fn because there is no mass??
 

LowlyPion

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That's surprising, because these problems are typically addressed in terms of kinetic and potential energy.

Specifically

1/2*mVC2 = 1/2mVB2 - m*g*(HC-HB)

The m cancels out and you are left with

Vc2 = VB2 - 2*g*(HC-HB)

ac = 1/R*(VB2 - 2*g*(HC-HB))

And what is Hc - HB ?

That's simply R, so

ac = 1/R*(VB2 - 2*g*R)
 

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