Roller Coaster Physics: How to Calculate Maximum Velocity on a Horizontal Plane

[Rumplestiltskin]

Homework Statement

[/B]

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

Homework Equations

/[/B]

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?

 

[theodoros.mihos]

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

 

[Rumplestiltskin]

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

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

 

[theodoros.mihos]

What friction do (take)?

 

[mfb]

Should I be using SUVAT at all?

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?

 

[Rumplestiltskin]

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?

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?

 

[mfb]

Huh, so SUVAT only works on regular trajectories?

Only for constant acceleration.

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?

Apart from missing units in the calculation, this is right.

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

 

[Rumplestiltskin]

Only for constant acceleration.
Apart from missing units in the calculation, this is right.
v² or v^2 for exponents.

Huh? I did v². Thanks!

 

[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?

 

[mfb]

GPE is converted to kinetic energy but also other forms of energy such as sound energy. (Can someone confirm? Should I mention friction?)

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).

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.

Why?

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?

Consider energy again. That is a general hint for all mechanics problems. If energy is conserved, it is usually the easiest approach.

 

[Rumplestiltskin]

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).

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

Why?

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.

Consider energy again. That is a general hint for all mechanics problems. If energy is conserved, it is usually the easiest approach.

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?

 

[mfb]

How would I explain how energy is lost due to friction?

I don't think it needs an explanation beyond that.

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.

Well, one of the answers is wrong, so I asked how you got them.

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.

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?

Where P = number of people, 70P kg < ...something.

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).

 

[haruspex]

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.

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?

I guess I'd do 11000 / 60 = 183 J

If you include the units all through that calculation you might spot a problem with that.

 

[Rumplestiltskin]

I don't think it needs an explanation beyond that.

I'm curious? :P

Well, one of the answers is wrong, so I asked how you got them.

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.

If you include the units all through that calculation you might spot a problem with that.

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?

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

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).

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?

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?

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?

 

[mfb]

I'm curious? :P

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

"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.

It would stay the same - at least approximately if we can neglect air drag.

13 people?

Right.

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?

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.

 

[Rumplestiltskin]

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

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.

It would stay the same - at least approximately if we can neglect air drag.

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.

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.

 

[mfb]

The question said you needed to consider energy changes. How would this relate to energy?

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

 

[haruspex]

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.

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.

 

[Rumplestiltskin]

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

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

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.

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.

 

[mfb]

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

Then it would not be called friction.

That would increase the surface area and hence the drag.

Right, but drag does not influence the answers to (a) and (b).

So there'd be additional friction force from drag and consequently a slightly lower velocity.

Drag would not increase linearly with mass. The energy loss per mass would go down, even with a higher total energy loss.