Roller Coaster Physics and Energy

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SUMMARY

The discussion focuses on calculating potential and kinetic energy for a roller coaster with a mass of 750 kg at various heights: 35 meters, 10 meters, and 20 meters. Using the equations for potential energy (PE = mgh) and kinetic energy (KE = 1/2 mv²), participants confirmed that the potential energy at the highest point (35m) is 257,250 joules, while at the first dip (10m) it is 73,500 joules, resulting in a kinetic energy of 183,750 joules. The total mechanical energy remains constant throughout the ride, allowing for the calculation of energy at different points.

PREREQUISITES
  • Understanding of basic physics concepts: potential energy and kinetic energy
  • Familiarity with the conservation of mechanical energy principle
  • Ability to perform calculations involving mass, height, and gravitational acceleration
  • Knowledge of the equations KE = (1/2)mv² and PE = mgh
NEXT STEPS
  • Learn how to apply the conservation of energy in different physical systems
  • Explore the effects of friction on energy calculations in roller coasters
  • Investigate the relationship between speed and kinetic energy in roller coaster dynamics
  • Study advanced topics in roller coaster design, including G-forces and safety measures
USEFUL FOR

Students studying physics, educators teaching energy concepts, and engineers involved in amusement park ride design will benefit from this discussion.

sromick
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Homework Statement


1. Assume the mass of the roller coaster is 750 kg. Answer the following.

a) If the height of the first hill is 35 meters (measured from the ground), calculate the potential and kinetic energies at this highest position on the track.

b) When the riders are at the first "dip" in the track, the height is 10 meters (measured from the ground), calculate the potential and kinetic energies at position on the track (assume no loss of energy due to friction).

c) At a hill right before a loop, the riders are 20 meters above the ground. Calculate the potential and kinetic energies at position on the track (assume no loss of energy due to friction).

Homework Equations



KE= (1/2)*m*(v^2)
PE= m*g*h
PEi+KEi=PEf+KEf

The Attempt at a Solution



For A, I used PE=mgh which is (750kg)*(9.8)*(35) = 257250 joules. Also for A, KE is 0 because it is at the top of the hill. Is this correct? And how do I do b and c? I have no clue!

Thanks!
 
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sromick said:

The Attempt at a Solution



For A, I used PE=mgh which is (750kg)*(9.8)*(35) = 257250 joules. Also for A, KE is 0 because it is at the top of the hill. Is this correct? And how do I do b and c? I have no clue!

Thanks!

Good, you've found this correctly. Now you know the total energy at A = Total energy at B= Total energy at C right? (conservation of mechanical energy)

Since at point B, the roller coaster is at a height of 10m, what is the potential energy? Now your third equation posted

(KE+PE)1=(KE+PE)2
(1 = point 1, 2 = point 2)

can you find the KE at B using it?
 
No friction=> PE+KE is constant.
For b, you can calculate PE the same way you did for a. Since you know, that in total PE+KE are constant, and you know this value (for you've already calculate it in a). c - the very same way.
 
rock.freak667 said:
Good, you've found this correctly. Now you know the total energy at A = Total energy at B= Total energy at C right? (conservation of mechanical energy)

Since at point B, the roller coaster is at a height of 10m, what is the potential energy? Now your third equation posted

(KE+PE)1=(KE+PE)2
(1 = point 1, 2 = point 2)

can you find the KE at B using it?


Point B PE: m*g*h
:750*9.8*10=73500 joules
 
sromick said:
Point B PE: m*g*h
:750*9.8*10=73500 joules

right at point B, PE= 73500 J.

KE+PE (at B) = KE+PE (At A)

So what is the KE at B?
 
rock.freak667 said:
right at point B, PE= 73500 J.

KE+PE (at B) = KE+PE (At A)

So what is the KE at B?

257250 joules + 0 joules = 73500 joules + KE (b)
so..
257250 joules = KE (b) + 73500 joules
so..
KE (b) = 183750 joules correct?
 
sromick said:
257250 joules + 0 joules = 73500 joules + KE (b)
so..
257250 joules = KE (b) + 73500 joules
so..
KE (b) = 183750 joules correct?

that should be correct, it is a similar exercise for part c)
 
rock.freak667 said:
that should be correct, it is a similar exercise for part c)

So:

PE for c = m*g*h
=750*9.8*20= 14700 joules

I use PEi+KEi=PEf+KEf right?

PEi would be 14700 right?
KEi would be what I calculated in B right?
Whats PEf and Kef?
 
sromick said:
So:

PE for c = m*g*h
=750*9.8*20= 14700 joules

I use PEi+KEi=PEf+KEf right?

PEi would be 14700 right?
KEi would be what I calculated in B right?
Whats PEf and Kef?

No. Remember KE+PE = constant. This means that (KE+PE)A=(KE+PE)B=(KE+PE)C

So if you wanted to find at C, using what was found at B, you add the PE and KE at B and put that equal to the KE and PE at C
 
  • #10
rock.freak667 said:
No. Remember KE+PE = constant. This means that (KE+PE)A=(KE+PE)B=(KE+PE)C

So if you wanted to find at C, using what was found at B, you add the PE and KE at B and put that equal to the KE and PE at C

So
73500+183750=Pe+Ke

Pe at C= m*g*h = 750*9.8*20=147000

so
73500+183750=14700(pe) + ke
solve for ke = 242550 joules
right?
 
  • #11
Yes that should be correct.
 
  • #12
rock.freak667 said:
Yes that should be correct.

Thank you rock freak! My savior!
 
  • #13
rock.freak667 said:
Yes that should be correct.

One more question rockfreak, sorry about all this!

What would be the speed of the car at the bottom of the first hill?
 
  • #14
sromick said:
One more question rockfreak, sorry about all this!

What would be the speed of the car at the bottom of the first hill?

Same process as before. Except that at the bottom of the hill, what is the height relative to the ground?
 
  • #15
I would solve for velocity here right? The height is 10 m.

so v= sq root(2ke[at b]/m)
so v= sq root(2(183750)/(750))

so velocity = 22.13594362 m/s?
 
  • #16
sromick said:
I would solve for velocity here right? The height is 10 m.

so v= sq root(2ke[at b]/m)
so v= sq root(2(183750)/(750))

so velocity = 22.13594362 m/s?

Yes and no, remember the bottom of the hill is where you are measuring your heights from, if you are at the bottom of the hill, you have no height!.

So your PE =?

Remember the total energy is what you found at A. So you don't always need to pick the previous point, you can just pick a convenient point. (Else you will always have to add numbers together and might make an error)
 
  • #17
rock.freak667 said:
Yes and no, remember the bottom of the hill is where you are measuring your heights from, if you are at the bottom of the hill, you have no height!.

So your PE =?

Remember the total energy is what you found at A. So you don't always need to pick the previous point, you can just pick a convenient point. (Else you will always have to add numbers together and might make an error)

I do not understand, so what do I have to solve for or what should I do?
 
  • #18
sromick said:
I do not understand, so what do I have to solve for or what should I do?

At the bottom of the hill, the height above the ground is zero. So the PE is 0.

Hence your KE at C = KE+PE (at B)
 
  • #19
rock.freak667 said:
At the bottom of the hill, the height above the ground is zero. So the PE is 0.

Hence your KE at C = KE+PE (at B)

But I am solving for speed at b. That is velocity right?
 
  • #20
sromick said:
But I am solving for speed at b. That is velocity right?

ah sorry, I did not read your entire question. I thought the bottom of the first hill was at the bottom, I forgot the first hill ended at B. So your answer was correct. Sorry about that.
 

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