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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 :

cjmejiab

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:

[tex]\Delta KE = \Delta PE[/tex]

Where,

[tex]\Delta KE = \frac{1}{2} mv^2 \rightarrow[/tex] Kinetic Energy

[tex]\Delta PE = mgh \rightarrow[/tex] Potential Energy

Equating,

[tex]\frac{1}{2}mv^2 = mgh[/tex]

Re-arrange for V,

[tex]V^2 = 2gh[/tex]

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,

[tex]F_w+F_a = F_t[/tex]

[tex]mg+ma = 9mg[/tex]

The centripetal acceleration is given by the equation

[tex]a = \frac{V^2}{r}[/tex]

Where

V = Tangencial velocity

r = Radius

Then replacing in the equation of Force,

[tex]mg + m\frac{V^2}{r} = 9mg[/tex]

[tex]mg + m \frac{(2gh)}{r} = 9mg[/tex]

[tex]1+\frac{2h}{r} = 9[/tex]

[tex]h= \frac{8r}{2}[/tex]

[tex]h= 4r[/tex]

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

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