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A roller coaster car has a mass of 1360kg when fully loaded with passengers. As the car passes over a circular hill with radius 20m, its speed is not changing.
A.) What are the magnitude and direction of the force of the track on the car at the top of the hill if the cars speed is 11m/s?
B.) What are the magnitude and direction of the force of the track on the car at the top of the hill if the cars speed is 15m/s?

I have the correct answers according to the website I had to submit these answers to:
A.) 5100N upward
B.) 1972 downward

It was my understanding that at the top of the hill the acceleration due to circular motion and the acceleration due to gravity were in the same direction so an increase in speed would just increase the normal force which would be force of the track on the car. I don't get how there can be a force working in the downward direction.

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When the velocity is low, gravity is more than enough to keep the car on the track; instead of becoming weightless, the car's force on the rails is just reduced.

However, when the roller coaster is even faster, gravity alone is not enough to keep the car on the rails, and the rails have to pull down on the car. Because of the mechanism of the wheels on a roller coaster, this is possible to do. Perhaps this diagram helps - the direction of $F_r$ can be up or down, depending on what you need to provide the right $F_c$, and given that the magnitude and direction of $F_g$ are constant:

enter image description here

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This is kind of a special case, because of the how the wheels of roller coasters work: enter image description here

When a roller coaster passes over a a hill, inertia gravity alone is not enough to bend its path, and it would fly of if it weren't for the wheels pushing against the rails, creating a normal force downwards.

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Would it be correct to say Fnet = Fnor -Fg -Fcentrip ?
The centripetal force being equal to (mv^2)/r
Then Fg is the m
a = -mg
If this is the case, then Fg is constant and the only thing to change would be the Fcentrip because of speed changing.
Fnet needs to be equal to zero because the cart doesn't move, and Fcentrip only get bigger in the negative direction with a speed increase, and Fg is already in the negative direction, why wouldn't Fnor increase positively to cancel out the sum of two negative numbers that grow in magnitude in the negative direction?

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  • $\begingroup$ Did you take a look at the diagram I drew? I think you are confusing yourself with the signs in your equation. The blue arrow in my diagram is always the vector sum of green and red. If the velocity is large, then $F_c$ will be larger than $F_g$ and you need $F_r$ to point inwards. If it's small, then $F_r$ (the force from the rails on the car) can point outwards. $\endgroup$ – Floris Sep 29 '17 at 15:44

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