Question

At the Six Flags Great America amusement park in Gurnee, Illinois, there is a roller coaster that incorporates some of the latest design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop (shown in the figure). The cars ride on the inside of the loop at the top, and the speeds are high enough to ensure that the cars remain on the track. The biggest loop is 35 m high, with a maximum speed of 31 m/s (nearly 69.44 mph) at the bottom. Suppose the speed at the top is 11 m/s and the corresponding centripetal acceleration is 2.9 g (where g is the free-fall acceleration). The acceleration of gravity is 9.8 m/s 2 .
What is the radius of the arc of the teardrop
at the top?
Answer in units of m.

If the total mass of the cars plus people is
2821 kg, what force does the rail exert on it
at the top?
Answer in units of N.

Suppose the roller coaster had a loop of radius
14 m.
If the cars have the same speed (11 m/s) at
the top, what is the centripetal acceleration
at the top?
Answer in units of m/s

What is the normal force at the top in this
situation?
Answer in units of N.

Answer 1

Given that roller coaster is of tear drop shape

So the speed at the top is given as

v = 11 m/s

acceleration at the top is given as

a = 2.9 g

$a = 2.9(9.8) = 28.42 m/s^2$

now we know the formula of centripetal acceleration as

$a = \frac{v^2}{R}$

$28.42 = \frac{11^2}{R}$

$R = 4.26 m$

PART B)

now if the total mass is given as

m = 2821 kg

now the Net force will be given as

$F = ma$

$F = 2821 \times 28.42 = 80172.8 N$

Now by force equation we will have

$F_n + mg = ma$

$F_n + 2821(9.8) = 80172.8$

$F_n = 52527 N$

PART C)

If the radius of the loop is 14 m

speed is given by same v = 11 m/s

now the acceleration is given as

$a = \frac{v^2}{R}$

$a = \frac{11^2}{14} = 8.6 m/s^2$

PART D)

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

$F_n + mg = ma$

$F_n + 2821(9.8) = 2821(8.6)$

so here normal force is coming with negative sign so it will not be possible to complete the circular path in this case

So normal force will be ZERO