Circular motion of a rollercoaster car on a loop-the-loop

In summary, the circular motion of a rollercoaster car on a loop-the-loop involves the car traveling along a curved track that forms a vertical loop. As it ascends, gravitational potential energy increases, while kinetic energy decreases. At the top of the loop, the car experiences centripetal force required to maintain its circular path, relying on both its speed and the gravitational pull. The design of the loop ensures that the car remains on the track, with sufficient speed at the top to avoid falling. The interplay of forces, energy transformations, and the principles of physics are essential for the safe and thrilling experience of the ride.
  • #1
Jolene
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1
Homework Statement
You are designing a rollercoaster and want to add an upside loop to it (a vertical circle). If the top speed that the rollercoaster cars can generate is 175 km/h, what is the maximum radius that you can make the loop so that the roller coaster does not fall off the track during the loop?
Relevant Equations
Fy= N + Fg = mac
Can someone please check if I got the correct answer. Thank you!
I got:
Fy= N + Fg = mac
N + mg = mv^2/r
g = v^2/r
r = v^2/g
r = (48.61)^2/9.8
r = 241.1 m
 
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  • #2
Jolene said:
Homework Statement: You are designing a rollercoaster and want to add an upside loop to it (a vertical circle). If the top speed that the rollercoaster cars can generate is 175 km/h, what is the maximum radius that you can make the loop so that the roller coaster does not fall off the track during the loop?
Relevant Equations: Fy= N + Fg = mac

Can someone please check if I got the correct answer. Thank you!
I got:
Fy= N + Fg = mac
N + mg = mv^2/r
g = v^2/r
r = v^2/g
r = (48.61)^2/9.8
r = 241.1 m
:welcome:

Just typing in some formulas without saying what you are doing or what your assumptions are is not a good solution. Note that a radius of approx 240m is huge. The height of a tall skyscraper and 480m long. Does that answer sound realistic?
 
  • #3
PeroK said:
The height of a tall skyscraper and 480m long. Does that answer sound realistic?
Show me a rollercoaster that reaches 175 kph … without any friction losses that in itself would require a height drop of over 100 m … The unrealism is already present in the problem statement.
 
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  • #4
Orodruin said:
Show me a rollercoaster that reaches 175 kph … without any friction losses that in itself would require a height drop of over 100 m … The unrealism is already present in the problem statement.
The fastest rollercoaster gets to 240 kph. And the highest loop is about 50m. The problem is not that unrealistic, despite not taking mechanical energy loss into account.

A 480m high loop, however, is unrealistic.
 
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  • #5
@Jolene Can you figure out what you're missing here?
 
  • #6
PeroK said:
A 480m high loop, however, is unrealistic.
The point is that - given only the constraint of no negative g-force - that’s what you get if it travels at 175 kph at the top of the loop.
 
  • #7
If it's 175 km/h at the top, it won't be the maximum speed it "can generate". Unless it is a powered car, with an engine.
 
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  • #8
nasu said:
If it's 175 km/h at the top, it won't be the maximum speed it "can generate". Unless it is a powered car, with an engine.
Well, you already stated the obvious exception. We do not know what is actually assumed by the problem.
 
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  • #9
PeroK said:
:welcome:

Just typing in some formulas without saying what you are doing or what your assumptions are is not a good solution. Note that a radius of approx 240m is huge. The height of a tall skyscraper and 480m long. Does that answer sound realistic?
Thank you for the quick respond and sorry for not being specific. Here's what I did:

I considered the total force of the rollercoaster at the top of the loop: Fy = N + Fg = mac

Then, I re-write the equation as: N + mg = mv^2/r

As N approaches 0 (at the top of the loop), v goes to v max (when n=0), , the equation becomes: mg = mv^2/rI Rearranged the equation to solve for r r = v^2/g

Lastly, I plugged in the numbers to find the radius when the speed is at its top speed. r=241.1m
 
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  • #10
Jolene said:
Thank you for the quick respond and sorry for not being specific. Here's what I did:

I considered the total force of the rollercoaster at the top of the loop: Fy = N + Fg = mac

Then, I re-write the equation as: N + mg = mv^2/r

As N approaches 0 (at the top of the loop), v goes to v max (when n=0), , the equation becomes: mg = mv^2/rI Rearranged the equation to solve for r r = v^2/g

Lastly, I plugged in the numbers to find the radius when the speed is at its top speed. r=241.1m
If you read the other posts, you might see why that's an unrealistic analysis. Do you think the rollercoaster might slow down as it ascends the loop?
 
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  • #11
PeroK said:
If you read the other posts, you might see why that's an unrealistic analysis. Do you think the rollercoaster might slow down as it ascends the loop?
I'm so confused, isn't that why I used the rollercoaster's top speed for my calculation? So, is my starting equation Fy=N+Fg=mac incorrect?
 
  • #12
Your equation is correct. However, "top speed" does not mean "speed at the top of the track." It means "the fastest the roller-coaster can ever move."
 
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  • #13
As an aside and for those interested, I found here an interesting little article about "real" roller coaster design.
 
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  • #14
Jolene said:
I'm so confused, isn't that why I used the rollercoaster's top speed for my calculation? So, is my starting equation Fy=N+Fg=mac incorrect?
Your equations are not wrong in themselves. But, they do not realistically model the motion of a rollercoaster.

I looked up on the Internet the fastest rollercoaster (which is faster than your one) and the highest loop, which is about 50m. That tells you that something has gone wrong somewhere in your calculations. Your loop is higher than the Empire State Building!

If a rollercoaster is doing 175 kph at the highest point, then won't it be going even faster when it comes back down?
 
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  • #15
PeroK said:
If a rollercoaster is doing 174 kph at the highest point, then won't it be going even faster when it comes back down?
That would depend on if it slams the breaks or not! 😉
 
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FAQ: Circular motion of a rollercoaster car on a loop-the-loop

What forces act on a rollercoaster car during a loop-the-loop?

During a loop-the-loop, the main forces acting on a rollercoaster car are gravity, the normal force from the track, and the centripetal force required to keep the car moving in a circular path. At the top of the loop, gravity and the normal force both act downward, while at the bottom of the loop, gravity acts downward and the normal force acts upward.

How does the speed of the rollercoaster car affect its motion through the loop?

The speed of the rollercoaster car is crucial for successfully navigating the loop. If the car is too slow, it may not have enough centripetal force to stay on the track at the top of the loop, potentially causing it to fall. Conversely, if the car is too fast, the forces experienced by riders can become uncomfortably high. The car must have sufficient speed to ensure that the centripetal force at the top of the loop is greater than or equal to the gravitational force.

What is centripetal force and how is it related to circular motion in a loop-the-loop?

Centripetal force is the inward force required to keep an object moving in a circular path. For a rollercoaster car on a loop-the-loop, this force is provided by the normal force from the track and, at the top of the loop, by gravity. The centripetal force ensures that the car follows the curved path of the loop rather than moving in a straight line.

Why do riders feel heavier at the bottom of the loop and lighter at the top?

Riders feel heavier at the bottom of the loop because the normal force from the track must counteract both the gravitational force and provide the necessary centripetal force to keep the car moving in a circle. This results in a higher normal force, which riders perceive as increased weight. At the top of the loop, the normal force and gravity both act downward, reducing the normal force needed to provide the centripetal force, making riders feel lighter.

What is the minimum speed required for a rollercoaster car to complete a loop-the-loop?

The minimum speed required for a rollercoaster car to complete a loop-the-loop can be calculated using the concept of centripetal force. At the top of the loop, the centripetal force must at least equal the gravitational force. This can be expressed as \( v = \sqrt{rg} \), where \( r \) is the radius of the loop and \( g \) is the acceleration due to gravity. This ensures that the car has enough speed to maintain contact with the track and not fall off at the top of the loop.

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