How Does Roller Coaster Design Affect Cart Speeds at Loop Apex?

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The discussion focuses on calculating the speed of a roller coaster's carts at the top of a loop, considering energy loss along the track. The roller coaster has a length of 10m, a mass of 4000kg, and experiences energy loss at a rate of 1300 J/m. Participants suggest treating the gravitational potential energy of each cart independently due to varying heights, while the kinetic energy can be considered uniform across the carts. It is emphasized that the distance between the lead and last cart is crucial for accurately accounting for energy loss. The consensus is to avoid complicating the calculations with the center of mass approach and instead treat the carts as point masses.
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Homework Statement


Consider a roller coaster with length = 10m and mass 4000kg consisting of 5 identical carts. Ignore the length of the connector between the carts. Suppose the rolercoaster is on a track where it loses energy at a rate of 1300 J/m. The rollercoaster approaches a loop with a speed of 40m/s with radius 20m. Find the speed of the first and the last cart of the rollercoaster when each of them is at the top of the loop.

Homework Equations


Kinetic and potential energy formulas

The Attempt at a Solution



we can solve the speed of the center of the mass of the rollercoaster using the following:

E_{K1} = E_{K2} + mg (2r) + 1300 (\pi r )

However, how do we accoutn for the actual length of the rollercoaster? can we treat each cart independantly? If that is the case, then would simply change the value of the last term in the above expression to account for the length of the rollercoaster - for example, if we consider the lead cart, 1300 would be multiplied by pi* r - 4. Is this correct? Do we have to include the friction of each cart in this calculation?

Thanks for your help
 
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stunner5000pt said:

Homework Statement


Consider a roller coaster with length = 10m and mass 4000kg consisting of 5 identical carts. Ignore the length of the connector between the carts. Suppose the rolercoaster is on a track where it loses energy at a rate of 1300 J/m. The rollercoaster approaches a loop with a speed of 40m/s with radius 20m. Find the speed of the first and the last cart of the rollercoaster when each of them is at the top of the loop.

Homework Equations


Kinetic and potential energy formulas

The Attempt at a Solution



we can solve the speed of the center of the mass of the rollercoaster using the following:
I advise against bringing the entire rollercoaster's center of mass into things. See more advice below.

E_{K1} = E_{K2} + mg (2r) + 1300 (\pi r )
That looks close to okay for the lead car, but not quite there yet.

A couple of things:
  • You should be able to substitute something into E_{K1} easily enough.
  • Treat the gravitational potential energy separately for each of the 5 cars (the cars are not all at the same height :wink:).

However, how do we accoutn for the actual length of the rollercoaster? can we treat each cart independantly?
Yes, I would treat the gravitational potential energy of each of the cars independently, since they are not all that the same height. A little bit of trigonometry is involved.

On the other hand, there is no need to do this with the rollercoaster's kinetic energy since they are all traveling at the same speed. Similarly, you don't need to break up the 1300 J/m either.

If that is the case, then would simply change the value of the last term in the above expression to account for the length of the rollercoaster - for example, if we consider the lead cart, 1300 would be multiplied by pi* r - 4. Is this correct?
I'm not sure where your "- 4" part comes from. But anyway, if you treat the gravitational potential energy of each cart separately, things should fall into place.

[Edit: But yes, as part of the process of finding the speed when the end car is at the top, you'll need to calculate the distance from the center of the lead car to the center of the end car.]

Do we have to include the friction of each cart in this calculation?
No, I don't think so.
 
collinsmark said:
I advise against bringing the entire rollercoaster's center of mass into things.
I second that. You could consider the KE of the mass centre, but then you'd have to add in the rotational KE around that centre. Much simpler to treat it as a number of point masses that all happen to have the same KE. (Of course, this ignores the rotational KE of each cart about its own centre, but that's a rather smaller error.)
as part of the process of finding the speed when the end car is at the top, you'll need to calculate the distance from the center of the lead car to the center of the end car.
Really? Is that necessary for finding the PE of the system?
 
haruspex said:
collinsmark said:
as part of the process of finding the speed when the end car is at the top, you'll need to calculate the distance from the center of the lead car to the center of the end car.

Really? Is that necessary for finding the PE of the system?

It's necessary for the "it loses energy at a rate of 1300 J/m" aspect of the problem. There's some energy lost from when the first cart is on top of the loop to when the last cart is on top.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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