Normal Force at a certain point in a loop

  • Thread starter mickjagger
  • Start date
  • #1

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.
 

Answers and Replies

  • #2
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 isnt 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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