# Critical Velocity in a roller coaster cart

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1. May 31, 2017

### Cactus

1. The problem statement, all variables, and given/known data
I was just wondering if a roller coaster can still pass through a loop with less than critical velocity/energy (Also if I'm assuming critical energy correctly). The loop can be of any size yet it must not exceed 5.7g at the entry point. The problem occurs when I work backwards from the critical velocity of the loop, and find that loop sizes 10m, 20m, 30m, 40m, 50m and 60m always work out to be 6.2g at the entry point. Whereas a friend worked from 5.7g at entry point to find the velocity at the entry point and then found if it had enough energy to complete the loop from there (finding kinetic and potential and subtracting energy loss etc), however, it had positive kinetic energy at the top, but not the critical kinetic energy that I got from the critical velocity (ex it had 115,000 J of Ek at top but critical velocity dictated it should have 315,000 J of kinetic energy (by using critical velocity in Ek equation)

2. Relevant equations
Critical velocity = Square root (rg)
G force at bottom as = ac + g/g
G force at top as = ac - g/g
Ac = centripetal acceleration
Kinetic energy = 1/2mv^2
Potential energy = mgh
Mass = 7000
G = 10
Radius = x (Any variable but for attempted solution = 10)
Energy loss = 5000 J per m displaced from level ground
3. The attempt at a solution (I.e in 20m diameter loop)
Working from critical velocity
v = Square root (rg)
v = square root 10x10

v = 10

Kinetic energy at top required
Ek = 1/2mv^2
Ek = 1/2 x 7000 x 10^2
Ek = 350,000 J

Potential Energy at top
Ep = mgh
Ep = 7000 x 10 x 20
Ep = 1,400,000

Total energy at top
Et = ep + ek
Et = 700,000 + 350,000
Et = 1,750,000 J

Total energy at bottom of loop
Et = Et from top + energy lost (adding energy lost as working backwards)
Et = 1,750,000 J + 20 x 5000
Et = 1,750,000 J + 100,000 J
Et = 1,850,000 J

At bottom et = ek (no height)
Therefore velocity at bottom
Ek = 1/2mv^2
1,850,000 J = 1/2 x 7000 x v^2
v = 22.99m/s
v = 23m/s

Centripetal acceleration
ac = v^2/r
ac = 23^2/10
ac = 52.9m/s^2

G force = ac + g/g
G force = 52.9 + 10/10
G force = 6.29g's
Which is over the 5.7g limit at entry yet is it critical energy for the loop (Same results for loop size 10m, 30m, 40m, 50m, 60m)

2. Jun 1, 2017

### Staff: Mentor

Where does the "energy loss" come from?
Please post the full problem statement.

Without "energy loss" the result is exactly 6 g, independent of the size of the loop, assuming a perfect circle. Friction or other losses will increase the force at the beginning of the loop.

Real rollercoasters don't have circular loops to avoid this problem.