Question

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on track. The biggest loop is 40.0m high. Suppose the speed at the top is 14.4m/s and the corresponding centripetal acceleration is 2g.

(a) What is the radius of the arc of the teardrop at the top?

(b)If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top?

(c) Suppose the roller coaster had a circular loop of radius 21.4 m. If the cars have the same speed, 14.4 m.s at the top, what is the centipetal acceleration at the top?

Answer 1

a)radius of the arc of the teardrop at the top is 10.58m

b)T = mg

c) the centipetal acceleration at the top is 14.5m/s

Step-by-step explanation:

Part A

Given that roller coaster is of tear drop shape

So the speed at the top is given as

v = 14.4 m/s

acceleration at the top is given as

a = 2 g

`a = 2(9.8) = 19.6 m/s^2`

now we know the formula of centripetal acceleration as

`a = (v^2)/(R)\\`

`19.6 = (14.4^2)/(R)\\R = 10.58 m`

Part B

now

the total mass of the car and the ride is M

Let the force exerted by the track be n

By Newton law

`n +Mg =(Mv^2)/(r) \\\\n=(Mv^2)/(r) -Mg\\\\=M((v^2)/(r)-g )\\\\=M(2g-g)\\\\T=Mg`

Part C

If the radius of the loop is 21.4 m

speed is given by same v = 14.4 m/s

now the acceleration is given as

`a = (v^2)/(R)`

`a = (14.4^2)/(21.4) \\\\= 9.69 m/s^2`

Now for normal force at the top is given by force equation

`F_n + mg = ma\\F_n = m(a-g)`

The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²

So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e
`a \prec g`

`(v^2)/(r) \prec √(g) \\\\v = √(rg) \\\\v = √(21.4 * 9.8) \\\\v = 14.5m/s`