# Roller coaster physics

1. Dec 30, 2015

### Rumplestiltskin

1. The problem statement, all variables and given/known data

b) Calculate the maximum velocity of the empty carriage at D. (3 marks)

2. Relevant equations
/

3. The attempt at a solution
750 x 9.8 = 7350 N

No clue. Can't use SUVAT on the vertical plane to find time because v = u = 0 from peak/trough to peak/trough. Horizontal plane from C to D:
s = ?
u = ?
v = ?
a = 0
t = ?

Maybe I could find u by finding v from B to C? Horizontal plane from B to C:
s = ?
u = 0
v = ?
a = 0
t = ?

Another dead end. Should I be using SUVAT at all?

2. Dec 30, 2015

### theodoros.mihos

Why problem ask for the maximum velocity on C? Can be less? Why? Which word we use 3 times on these questions?

3. Dec 30, 2015

### Rumplestiltskin

Maximum velocity at D. Carriage? It could be less when accounting for friction.

4. Dec 30, 2015

### theodoros.mihos

What friction do (take)?

5. Dec 30, 2015

### Staff: Mentor

No, you don't know the shape of the roller coaster.
You calculated the potential energy at point B already. How can you use this?

6. Dec 30, 2015

### Rumplestiltskin

Huh, so SUVAT only works on regular trajectories?
Most of it is converted to kinetic energy at point C, and back to GPE at point D.
GPE at point D = 750 x 9.8 x 20 = 147000. Ek would be another 147000 since the GPE is half at what it was at B.
147000 = 0.5 x 750 x v2 (this formula was only vaguely familiar, just touched upon it on the syllabus)
...
v = 19.8 m/s?

7. Dec 30, 2015

### Staff: Mentor

Only for constant acceleration.
Apart from missing units in the calculation, this is right.

Edit: Oops, copy&paste error from my side.

Last edited: Dec 30, 2015
8. Dec 30, 2015

### Rumplestiltskin

Huh? I did v2. Thanks!

9. Dec 30, 2015

### Rumplestiltskin

Woops, so there's 3 more sub-questions related to this.
c) The actual velocity of the carriage at D is less than the value calculated in part b. Explain the reason for this discrepancy by considering the energy changes that take place from B to D. (1 mark)
GPE is converted to kinetic energy but also other forms of energy such as sound energy. (Can someone confirm? Should I mention friction?)

d) Suggest how you would expect your answers to parts a and b to be affected if the carriage were to be fully occupied with people. (2 marks)
GPE would increase.
Max velocity would decrease.

e) The electric motor used to raise the carriage from A to B has a maximum useful power output of 11 kW and completes the lift in 1 minute. Determine the maximum number of people, of average mass 70 kg, that can be allowed in the carriage if the motor is to operate normally.
Really lost, this is kinda going outside of where we're at in the syllabus so far. Help?

10. Dec 30, 2015

### Staff: Mentor

Friction is much more relevant than sound (also, sound production is usually included in friction unless there is some active production with speakers or whatever).
Why?
Consider energy again. That is a general hint for all mechanics problems. If energy is conserved, it is usually the easiest approach.

11. Dec 30, 2015

### Rumplestiltskin

Sound was just the example, but true. How would I explain how energy is lost due to friction?

I could go into that, but the answer field was two bullet points and it's a two mark question, so I presume that would suffice.

Yeah, that just hit me.
11 kW = 11000 J/s. Not sure where we go from here. Energy isn't described in J/s, and it's given that the motor operates for a minute, so I guess I'd do 11000 / 60 = 183 J but I'm not sure what I'm describing. EDIT: Of course I do. Energy used, per second, to operate the motor.
Where P = number of people, 70P kg < ...something.
(70P kg x m2) / s2 < 183 J?

Last edited: Dec 30, 2015
12. Dec 30, 2015

### Staff: Mentor

I don't think it needs an explanation beyond that.
Well, one of the answers is wrong, so I asked how you got them.
You divide J/s by seconds, so the result has units J/s2, which is not a useful quantity (and it certainly does not have J as unit).

If I give you 5 apples per second, and do that over 60 seconds, how many apples do you get?

Don't forget the cart. No, you cannot just change the units. Think what you did for (a), the potential energy at the top of the hill. You can do the same thing with persons in the cart, and compare it to the energy available (which you still have to calculate in the right way).

13. Dec 30, 2015

### haruspex

Whether or not you can provide your reasoning as part of your official answer, you need to have a reason.
How will friction mean that more mass leads to a lower velocity?
The other significant loss of energy will be through drag. How will mass affect velocity when that is considered?
If you include the units all through that calculation you might spot a problem with that.

14. Dec 31, 2015

### Rumplestiltskin

I'm curious? :P

Thing is, this was a question on a paper we completed a while back, and I'm going back working out what went wrong.
"GPE would increase" was marked correct.
"Max velocity would increase" was marked incorrect. I figured that an increased mass would pull up the whole equation. And it would, but I forgot that the equation was for Ek and v is only a variable in that. So it would stay the same? Either that or my teacher is wrong.

300 apples! :D Yeah, that was clumsy. Thanks. So 11000 x 60 = 660000 J.

Right, 750 + 70P kg < something.
So 660000 J is needed to operate the motor in total. Not sure how to proceed. I know GPE = mgh is going to come into this but I'd be fumbling in the dark. The energy available?
EDIT: Okay, conservation of energy. That 660000 goes somewhere, into GPE.
660000 = m x 9.8 x 40.
...
m = 1684 kg. There's my something.
750 + 70P < 1684
70P < 934
P < 13.3

13 people?

I'd guess that increased mass would decrease velocity, but that was marked incorrect. With respect to maximum velocity, anyway.
I'd also guess that Newton's third law comes into this, but doesn't more mass mean more force, and that added force would at least equal the added friction?

Last edited: Dec 31, 2015
15. Dec 31, 2015

### Staff: Mentor

Well, there is friction between the cart and the track, and drag from the air.

It would stay the same - at least approximately if we can neglect air drag.
Right.
To a good approximation, everything scales linearly with mass: double the mass and you double the potential energy, kinetic energy, and energy lost to friction everywhere. The speed is related to the ratio "energy per mass", which stays the same.

16. Dec 31, 2015

### Rumplestiltskin

The question said you needed to consider energy changes. How would this relate to energy? That's why I opted for mentioning GPE >> unused energy.

Right. More mass wouldn't affect velocity on it's own because while your force increases, your friction increases in proportion to cancel that out.
A larger surface area is going to increase the drag, so technically you'd be right to say that the velocity would decrease slightly, but the maximum velocity is a theoretical value and so wouldn't be affected.

17. Dec 31, 2015

### Staff: Mentor

An energy change would change the energy?
Friction always reduces the mechanical energy.

18. Dec 31, 2015

### haruspex

Yes, part d only asks about the effect of increased mass on the answers to a and b. So in answering it you should ignore both drag and friction.
Had it asked about the effect on c then, as you say above, friction still would not change the velocity. But you are wrong on drag. Consider that while energy, momentum and friction all increase in proportion to mass, drag is related to the cross-sectional area. Since the carrriage appears closed, why would that change? Even if it were open, I doubt it would increase as fast as the mass.

19. Dec 31, 2015

### Rumplestiltskin

Why? And always? What if you could redirect the friction force and make it work in your favour?

Part D asks specifically how the answers to parts A and B would change if the carriage were to be "fully occupied with people"; increased mass was only an inferrence from this. That would increase the surface area and hence the drag. I was thinking to mention earlier that drag is related to the area, which is a separate property of the people to their mass and so is unaccounted for by the proportionality of mass to friction. So there'd be additional friction force from drag and consequently a slightly lower velocity.

20. Dec 31, 2015

### Staff: Mentor

Then it would not be called friction.
Right, but drag does not influence the answers to (a) and (b).
Drag would not increase linearly with mass. The energy loss per mass would go down, even with a higher total energy loss.