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Question

A frictionless roller coaster is given an intial velocity of Vo at height h=30m. The radius of curvature of the track at point A is R=28m. The acceleration of gravity is 9.8m/s2.

Find the maximum value of Vo so that the roller coaster stays on the track at A solely because of gravity.

h at point A is 2/3h

Also, how do you determine the value of h' (at point B) that is neccesary if the roller coaster just makes it to point B

how do you find vo^2??

Answer 1

The centrifugal force on the coaster at the top of the track is m*V^2/R (R= 28m); this must be less than m*g (the weight of the coaster) for the coaster to stay on the track. Then

m*V^2/R < m*g

V^2/R < g

V < √[g*R]

The velocity at A is determined by the energy difference between the start and point A. The height difference is h - (2/3)h = (1/3)h or 10m. The energy difference is them m*g*10. Since the point A is lower than the start, energy is gained and velocity will increase. The relation is

Start kinetic energy + potential energy change = end kinetic energy

0.5*m*Vo^2 + m*g*10 = 0.5*m*V^2

V^2 = Vo^2 +2*g*10

V=√[Vo^2 +2*g*10]

and from the first result

√[Vo^2 +2*g*10] < √[g*R]

Vo^2 + 2*g*10 ≤ g*R

Vo < √[g*R - 2*g*10]

The coaster will make it to point B if the potential energy difference equals the initial kinetic energy. If ∆h is the difference in elevation between 30m and h', the potential energy difference is m*g*∆h, and this must equal 0.5*m*Vo^2:

m*g*∆h = 0.5*m*Vo^2

∆h = (0.5*Vo^2)/g

find ∆h and add to 30m to get h'

NOTE: You found Vo as the answer to the first part of the problem: Vo^2 = g*R - 2*g*10. Then ∆h = 0.5*(R - 20).