Exploring Weightlessness on a Frictionless Roller Coaster

[DrunkApple]

Homework Statement

Consider a frictionless roller coaster such as
depicted below.
The acceleration of gravity is 9.8 m/s2.
Passenger cars start at point A with zero
initial speed, accelerate as they go down to
point B, swing around the circular vertical
loop B ! C ! B of radius 30 m, then go
on towards further adventures (not shown).
When a car goes through the top of the loop
(point C), the passengers feel weightless (for
just a moment).
What is the height hA of the starting
point A above the loop’s bottom B?
Answer in units of m

Homework Equations

KE = .5mv^2
PE = mgh

The Attempt at a Solution

I know that the system is conservative energy since there is no friction and whatnot, so mgh(a) is the total energy. But what I don't get is, how does weightlessness come into this?
If it's weightless, does that mean normal force doesn't exist?

 

[lightgrav]

did you forget ΣF = ma ? what about circular paths?

 

[DrunkApple]

then does gravity becomes the acceleration of centripetal force?

 

[lightgrav]

yes, to get the speed at the top ... Energy conservation then gets height at begining

 

[rwisz]

And if so, how would you calculate the height of point A above the loop's bottom B?

Weightlessness in this context means that the passengers in the car feel as though they have no weight or are floating. This is because at the top of the loop, the normal force acting on them is equal to zero. This is due to the fact that the centripetal force (provided by the car's acceleration) is equal to the force of gravity. As a result, the passengers are not experiencing any acceleration due to gravity and therefore feel weightless.

To calculate the height of point A above the loop's bottom B, we can use the conservation of energy principle. At the top of the loop, the car has maximum potential energy and zero kinetic energy. As it reaches point A, it will have maximum kinetic energy and zero potential energy. Therefore, we can equate the two using the equations KE = .5mv^2 and PE = mgh.

.5mv^2 = mgh

Since we know the radius of the loop (30 m) and the acceleration due to gravity (9.8 m/s^2), we can solve for h.

h = .5v^2/g

To find the velocity at point A, we can use the conservation of energy again. At point A, the car starts with zero initial speed and reaches a maximum speed at the top of the loop. Therefore, we can equate the potential energy at point A to the kinetic energy at the top of the loop.

mghA = .5mv^2

Solving for v, we get:

v = √(2ghA)

Substituting this into our previous equation, we get:

h = .5(2ghA)/g

Therefore, the height of point A above the loop's bottom B is:

hA = h = (2ghA)/g = 2(30 m)(9.8 m/s^2)/9.8 m/s^2 = 60 m

So, the starting point A must be 60 meters above the bottom of the loop in order for the passengers to experience weightlessness at the top of the loop.