Normal Force at a certain point in a loop

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SUMMARY

The discussion centers on calculating the normal force acting on a 5.4g mass at point A of a loop-the-loop, where point A is at the same height as the center of the loop. The mass is released from a height of 1.2m, and the radius of the loop is 0.36m. The calculations involve energy conservation principles and centripetal force equations, leading to a derived velocity of 4.85 m/s at the top of the loop. The final normal force calculation yielded an incorrect value of 299.88 N, indicating a misunderstanding in the application of forces at point A.

PREREQUISITES
  • Understanding of energy conservation principles (PE = KE)
  • Knowledge of centripetal force calculations
  • Familiarity with Newton's laws of motion
  • Basic algebra for solving equations
NEXT STEPS
  • Review the concept of centripetal acceleration in circular motion
  • Learn how to apply the conservation of mechanical energy in dynamic systems
  • Study the relationship between gravitational force and normal force in circular motion
  • Practice similar problems involving forces on objects in loops
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of energy conservation and force calculations in circular motion.

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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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