Normal Force at a certain point in a loop

[mickjagger]

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 0^o from the center of the radius and point B is at the top of the loop.

Homework Equations

1/2mvf²+PE_i=1/2mv_f²+PE_f
Sum of all of the forces is = to ma_c 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.

 

[mickjagger]

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.

 

[kerimek]

I would approach this problem by first drawing a free body diagram of the mass at point A. This would help me visualize all the forces acting on the mass at that point. From the given information, I know that the mass has a weight (mg) acting downwards, a normal force (N) acting perpendicular to the track, and a centripetal force (Fc) acting towards the center of the loop.

Using Newton's second law, I know that the sum of all the forces acting on the mass at point A must be equal to its mass (m) times its centripetal acceleration (ac). Therefore, I can write the following equation:

ΣF = N - mg - Fc = mac

Since the mass is moving along a circular path, the centripetal force can be calculated using the equation Fc = mv^2/r, where v is the speed of the mass at point A and r is the radius of the loop.

To find the speed of the mass at point A, I can use the conservation of energy principle. At point C, the mass has only potential energy (PE = mgh), and at point A, it has both potential and kinetic energy (KE = 1/2mv^2). Therefore, I can write the following equation:

mgh = 1/2mv^2

Solving for v, I get v = √(2gh), where g is the acceleration due to gravity (9.8 m/s^2).

Substituting this value for v in the equation for the centripetal force, I get:

Fc = m(√(2gh))^2/r = 2mgh/r

Now, I can substitute this value for Fc in my original equation and solve for the normal force (N):

N = mac + mg + 2mgh/r

Since the mass is at the same level as the center of the loop, the radius (r) is equal to the height of the slope (1.2m). Therefore, the final equation for the normal force at point A is:

N = mac + mg + 2mgh/h = ma + mg + 2mg = m(a + 2g)

Substituting the given values for mass and acceleration, I get:

N = (5.4g)(9.8 m/s^2 + 2(9.8 m/s^2