Roller coaster help

Question

roller coaster has a mass of 840 kg. It is launched horizontally from a giant spring with spring constant 31,000 N/m into a frictionless vertical loop-the-loop track of radius 6.3 m. What is the minimum amount that the spring must be compressed if the car is to stay on the track?

Answer 1

Let A be initial point

Let B be top of loop

Given

K = 31,000 N/m

m = 840 kg

r = 6.3 m

Initial Compression

Xa = ?

Final (spring is in natural state then)

Xb = 0

Now, please note to find minimum compression we need to find the velocity at the top of the hoop when the roller coaster is close to separation, meaning normal force is equal to zero

Sum force C = Mac

mg = mVb^2/R

g = Vb^2/R

gR = Vb^2

Now, we apply work energy method

-(1/2) * k* (Xb^2-Xa^2) - mg(2r) = (1/2) * m * (Vb^2-Va^2)

Va = 0

Xb = 0

Please note for work of weight the overall change of height is 2r, the diameter of the loop

-(1/2) * k * (-Xa^2) - mg(2r) = (1/2) * m * Vb^2

(1/2) * k * (Xa^2) - mg(2r) = (1/2) * m * Vb^2

Isolate the deformation Xa of the spring using algebra

(1/2) * k * (Xa^2) = (1/2) * m * Vb^2 + mg(2r)

Xa^2 = (2/k) * [(1/2) * m * Vb^2 + mg(2r) ]

Xa^2 = m/k * Vb^2 + m/k * g*4r

Sub in Vb^2

Xa^2 = m/k * (gR) + m/k * g*4r

Xa^2 = m/k * (gR + g4R) = m/k * (5gR)

Xa^2 = 840/31,000 * (5 * 9.81 * 6.3)

Xa^2 = 8.373

Xa = 2.89 m

You might ask why I solve everything in variables. In engineering applications we tend to do that to optimize our variables when we are designing something. Hope this makes sense :)