Question

How high can a roller coaster be? A roller coaster car is to roll down a frictionless ramp into a loop of a radius r. If the riders can withstand an acceleration of 9g's and not black out - i.e. their apparent weight at the bottom of the loop is nine times their actual weight; what is the maximum height of the incline if the cars starts from the rest?

Answer 1

Answer:

h=4r

Explanation:

To solve the problem it is necessary to apply the energy conservation equations for the roller coaster.

The energy conservation equations warn that:

$\Delta KE = \Delta PE$

Where,

$\Delta KE = \frac{1}{2} mv^2 \rightarrow$ Kinetic Energy

$\Delta PE = mgh \rightarrow$ Potential Energy

Equating,

$\frac{1}{2}mv^2 = mgh$

Re-arrange for V,

$V^2 = 2gh$

For balance of forces, according to the announcement, those who are on a roller coaster can withstand up to a maximum of 9g.

Therefore, considering the centripede speed and the speed of the fall, we obtain that,

$F_w+F_a = F_t$

$mg+ma = 9mg$

The centripetal acceleration is given by the equation

$a = \frac{V^2}{r}$

Where

V = Tangencial velocity

r = Radius

Then replacing in the equation of Force,

$mg + m\frac{V^2}{r} = 9mg$

$mg + m \frac{(2gh)}{r} = 9mg$

$1+\frac{2h}{r} = 9$

$h= \frac{8r}{2}$

$h= 4r$

Therefore the maximum height of the incline if the cars starts from the rest is 4 times the raidus of the inclination