Normal Force at a certain point in a loop

AI Thread Summary
To find the normal force at point A in a loop-the-loop problem, the mass is released from a height of 1.2m, and the radius of the loop is 0.36m. The energy conservation principle is applied, equating potential energy at the top of the slope to kinetic energy at the top of the loop. The calculated velocity at point A is approximately 4.85 m/s. The normal force is then determined using the equation Fn = mg - m(v^2/r), but the resulting value of 299.88 N seems incorrect. The discussion seeks clarification on the calculation process and potential errors in deriving the normal force.
mickjagger
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Homework Statement


A 5.4g mass is released from rest at C which has a height of 1.2m above the base of the loop-the-loop and a radius of .36m
Finde the normal force pressing on the track at A, where A is at the same level as the center of the loop. Answer in units of N

So the picture is of a ball (at point C) at the top of a 1.2m slope that goes down into the beginning of the loop the loop where point A is at 0o from the center of the radius and point B is at the top of the loop.


Homework Equations


1/2mvf2+PEi=1/2mvf2+PEf
Sum of all of the forces is = to mac or mass times the centripetal acceleration.


The Attempt at a Solution


I need to make and equation that works for this problem but I don't know where to start.
 
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Top of loop elevation = 2 * 0.36 = 0.72 m
Top of track= 1.2 m

Figure change in energy from top of track to top of loop

PE = KE
MGH = 0.5MV^2
5.4 * 9.8 * 1.2 = 0.5*5.4 V^2

2.7 V^2 = 63.504
V^2 = 23.52
V = 4.85 m/s

Fn = Force of Gravity - Centripetal Force
Fn = MG - M* (V^2/r)
Fn = 5.4 * 9.,8 - 5.4 *(4.85^2 / 0.36)
Fn = 5.4 * 9.8 - 352.8 = 52.92 - 352.8 =299.88 N That is the force of the track down against the ball

It seems as if this isn't the answer my prof is looking for. can anyone tell me where I went wrong. I have an hour left to get this problem so I'm frantic right now.
 
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