Garbage can suspended in the air

[Saitama]

Homework Statement

An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up the ground with speed $v_0$, at a constant rate $dm/dt$. The problem is to find the maximum height at which garbage can rides. What assumption must be fulfilled for the maximum height to be reached.

Ans. clue. If $v_0=20\, m/s$, W=10kg, dm/dt=0.5 kg/s, then $h_{max} \approx 17 \, m$

Homework Equations

The Attempt at a Solution

I don't know how to begin with this problem. I think that I have to find the force due to water but I can't see how to form the equations here. I need a few hints to begin with.

Any help is appreciated. Thanks!

 

[ehild]

Homework Statement

An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up the ground with speed $v_0$, at a constant rate $dm/dt$. The problem is to find the maximum height at which garbage can rides. What assumption must be fulfilled for the maximum height to be reached.

Ans. clue. If $v_0=20\, m/s$, W=10kg, dm/dt=0.5 kg/s, then $h_{max} \approx 17 \, m$

Homework Equations

The Attempt at a Solution

I don't know how to begin with this problem. I think that I have to find the force due to water but I can't see how to form the equations here. I need a few hints to begin with.

Any help is appreciated. Thanks!

Does the momentum of the water change when it interacts with the can? Can you solve the problem if there were balls shot up instead of water?

ehild

 

[rude man]

Think F = dp/dt with p = p(height).

BTW I got a very different answer. (I assume W = 10*9.81 = 98.1 N).
@ehild?

 

[NascentOxygen]

The geyser alone reaches 20.4m

 

[ehild]

Think F = dp/dt with p = p(height).

BTW I got a very different answer. (I assume W = 10*9.81 = 98.1 N).
@ehild?

I also think that the example solution is wrong.

ehild

 

[Saitama]

Does the momentum of the water change when it interacts with the can? Can you solve the problem if there were balls shot up instead of water?

ehild

Yes, the momentum of water changes. If the base of garbage can is at a height h, the velocity with which water strikes is $\sqrt{v_0^2-2gh}$. The problem is how do I find the change in momentum of mass of water striking the can. The question doesn't state the type of collision between water and can.

 

[Enigman]

1. The problem is to find the maximum height at which garbage can rides.

So under what conditions does it?

 

[ehild]

Yes, the momentum of water changes. If the base of garbage can is at a height h, the velocity with which water strikes is $\sqrt{v_0^2-2gh}$. The problem is how do I find the change in momentum of mass of water striking the can. The question doesn't state the type of collision between water and can.

Yes, when is it maximum? At elastic or inelastic collision?

ehild

 

[Saitama]

Yes, when is it maximum? At elastic or inelastic collision?

ehild

Elastic.
Does it mean the water bounces back with $\sqrt{v_0^2-2gh}$? I guess that would be wrong because the garbage can is also moving. Correct?

 

[Enigman]

Does it mean the water bounces back with $\sqrt{v_0^2-2gh}$? ?

Nope...

I guess that would be wrong because the garbage can is also moving. Correct?

Yep, so what are you missing? What does the water do?

 

[ehild]

What does maximum height mean? Does the can raise higher from that? ehild

 

[D H]

I also think that the example solution is wrong.

Same here. I think the correct answer for a 10 kg garbage can (and calling this "weight" in a physics text is not good) is zero. 20 m/s * 0.5 kg/s = 10 Newtons. Double that for a pure elastic collision and you get 20 Newtons, which is not enough to lift a 10 kg object off the ground.

 

[voko]

The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.

 

[ehild]

The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.

And it also says that W=10 kg. The text of the problem is confusing. ehild

 

[Saitama]

Same here. I think the correct answer for a 10 kg garbage can (and calling this "weight" in a physics text is not good) is zero. 20 m/s * 0.5 kg/s = 10 Newtons. Double that for a pure elastic collision and you get 20 Newtons, which is not enough to lift a 10 kg object off the ground.

Yes, giving weight as 10 kg is incorrect but the problem is from a very good book, errors may creep in despite the best efforts by the authors. :)

The source of the problem is "An Introduction to Mechanics by Daniel Kleppner and Robert Kolenkow".

What does maximum height mean? Does the can raise higher from that?

Change in momentum of water at maximum height is $2\Delta m\sqrt{v_0^2-2gh}$. The force applied on can in time $\Delta t$ is $2(\Delta m/\Delta t)\sqrt{v_0^2-2gh}$. Taking the limit $\Delta t\rightarrow 0$, the force acting on can is $2(dm/dt)\sqrt{v_0^2-2gh}$. This is balanced by the weight of can. Hence,
$$2\frac{dm}{dt}\sqrt{v_0^2-2gh}=W$$
From here I can solve for h.

Looks correct?

 

[NascentOxygen]

Looks correct?

The method looks correct, yes. But as for the answer you'll get ...

 

[voko]

In the 2010 edition, the answer clue is $v_0 = 20 \ \text{m/s}, \ W = 8.2 \ \text{N}, \ dm/dt = 0.5 \ \text{kg/s}, \ h_{\text{max}} \approx 15 \ \text{m}$.

 

[Saitama]

The method looks correct, yes. But as for the answer you'll get ...

Yes, the given clue doesn't satisfy the final result as said by the other posters.

 

[Saitama]

In the 2010 edition, the answer clue is $v_0 = 20 \ \text{m/s}, \ W = 8.2 \ \text{N}, \ dm/dt = 0.5 \ \text{kg/s}, \ h_{\text{max}} \approx 15 \ \text{m}$.

Mine is a low priced Indian edition 2009.

 

[voko]

The text as you gave it is an exact copy from the original 1973 edition.

 

[Saitama]

The text as you gave it is an exact copy from the original 1973 edition.

I have checked my text again, I have copied down the same exact wordings.

Check this book on Google Books, this is the one sold in my country:
http://books.google.co.in/books?id=lYYL6qnDTGAC&printsec=frontcover#v=onepage&q&f=false

Look for the Momentum chapter.

Also, the clue you posted doesn't satisfy the result I reach. I have tried both values of g i.e 9.8 and 10.

 

[ehild]

Change in momentum of water at maximum height is $2\Delta m\sqrt{v_0^2-2gh}$. The force applied on can in time $\Delta t$ is $2(\Delta m/\Delta t)\sqrt{v_0^2-2gh}$. Taking the limit $\Delta t\rightarrow 0$, the force acting on can is $2(dm/dt)\sqrt{v_0^2-2gh}$. This is balanced by the weight of can. Hence,
$$2\frac{dm}{dt}\sqrt{v_0^2-2gh}=W$$
From here I can solve for h.

Looks correct?

Yes.

ehild

 

[Saitama]

Yes.

ehild

Thanks for the help ehild! :)

 

[ehild]

In the 2010 edition, the answer clue is $v_0 = 20 \ \text{m/s}, \ W = 8.2 \ \text{N}, \ dm/dt = 0.5 \ \text{kg/s}, \ h_{\text{max}} \approx 15 \ \text{m}$.

It is 17 m with these data.

 

[Saitama]

It is 17 m with these data.

And if I use 10 N instead of 10 kg, it comes out to be 15 m, lol.

 

[voko]

I have checked my text again, I have copied down the same exact wordings.

Is it not what I said?

Check this book on Google Books, this is the one sold in my country:
http://books.google.co.in/books?id=lYYL6qnDTGAC&printsec=frontcover#v=onepage&q&f=false

Look for the Momentum chapter.

I cannot seem to be able to look inside, but I think this is just a reprint of the original edition, with all of its errors.

Also, the clue you posted doesn't satisfy the result I reach. I have tried both values of g i.e 9.8 and 10.

I agree there is still a discrepancy, but at least now the statement of the problem is consistent.

 

[Saitama]

Is it not what I said?

I misinterpreted your statement, sorry.

 

[TSny]

I have checked my text again, I have copied down the same exact wordings.

Check this book on Google Books, this is the one sold in my country:
http://books.google.co.in/books?id=lYYL6qnDTGAC&printsec=frontcover#v=onepage&q&f=false

Look for the Momentum chapter.

Here's the problem in the 2010 edition

 

[D H]

That's still wrong. 8.2 Newtons is the weight that can be suspended at 17 meters. It's about 15 meters for the 10 Newton garbage can.

 

[Saitama]

That's still wrong. 8.2 Newtons is the weight that can be suspended at 17 meters. It's about 15 meters for the 10 Newton garbage can.

Yes, that was already pointed out and it is exactly 15 meters if g=10 m/s^2 is used. :)

 

[rude man]

That's still wrong. 8.2 Newtons is the weight that can be suspended at 17 meters. It's about 15 meters for the 10 Newton garbage can.

That's more like it.

So the can's weight is not 10 kg, it's 10 N. An 80 kg man weighs 80*9.81 N. Really confusing!


I have one question: the assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

 

[ehild]

That's more like it.

So the can's weight is not 10 kg, it's 10 N. An 80 kg man weighs 80*9.81 N. Really confusing!

When I started school, (a very long time ago) we used kg for weight. And then (in my country) we said kg-weight for the weight and kg for mass. And then we used kilopond for the force equal to the weight of 1 kg mass, which was equivalent to Newton.

I have one question: the assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

Why should it be zero?

The maximum possible height is needed. If the change of momentum is less, the force of the geyser is equal to the weight at higher speed, at lower height.

ehild

 

[D H]

i have one question: The assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

why should it be zero?

I suspect rude man was looking at things from a realistic rather than theoretical perspective. I would expect that if this was tried experimentally, one would see collisions that are much closer to purely inelastic than elastic. When the water hits the can it will bounce off slightly inelastically and immediately collide with incoming water. The end result will be a close to purely inelastic collision with the water dribbling down the sides of the can.

However, the question is asking for the maximum possible height, and to get that you need elastic collisions.

 

[rude man]

When I started school, (a very long time ago) we used kg for weight. And then (in my country) we said kg-weight for the weight and kg for mass. And then we used kilopond for the force equal to the weight of 1 kg mass, which was equivalent to Newton.



Why should it be zero?

The maximum possible height is needed. If the change of momentum is less, the force of the geyser is equal to the weight at higher speed, at lower height.

ehild

That was not my question. My question is why is dp/dt = 2v dm/dt instead of v dm/dt.

In other words, why is the collision of the water with the can bottom considered an elastic vs. an inelastic collision.

 

[D H]

In other words, why is the collision of the water with the can bottom considered an elastic vs. an inelastic collision.

Because the problem is to find the maximum height at which the garbage can rides, not the height at which the garbage can will ride realistically. Elastic collisions will yield the largest possible height.