Understanding Work Calculation in an Engine: External vs. Internal Forces

[Nexus99]

Homework Statement

Determine power of the eninge

Relevant Equations

Upon a sloping roller and from a negligible height, sand is placed with a rhythm of dmdt. Thanks to an engine that provides a power W, the speed of the roller is kept constant (v0). The roller brings the sand up to a certain amount height, from which it then falls vertically with the initial velocity of magnitude v0 on a container placed at the base level of the roller, h lower. The container is placed upon a scale (see drawing). At a certain instant t, when the roller is already covered with sand and a mass M of sand has accumulated in the container, Determine:

1) the power provided by the engine W;

2) the momentum transferred to the sand by the scale in the unit of time dp/dt;

3) the weight indicated by the scale.

I'm really struggling with this problem. I can't understand which are the forces that i need to calculate work. Probabily when the sand falls the force of which i need to calculate work is weight force, but which is the force that carries the sand at the top of the roller?

 

[BvU]

Hello Pluto, !

Thanks to an engine that provides a power W, the speed of the roller is kept constant (v0).

which is the force that carries the sand at the top of the roller?

The engine is apparently hidden, e.g. behind the conveyor belt .

 

[Nexus99]

So, can you give me an hint to calculate the power ? I have been trying to do this exercise for 3 hours

 

[haruspex]

So, can you give me an hint to calculate the power ? I have been trying to do this exercise for 3 hours

Consider a small period of time dt.
What mass of sand is added at the bottom in that time? How much energy (KE+PE) does it bring with it?
What mass of sand leaves at the top in that time? How much energy (KE+PE) does it take away with it?

 

[Nexus99]

1) dm, 0
2) dm, ##\frac{1}{2}(dm)v_0^2 + (dm)gh##
So the power is ##\frac{Mgh}{t}## ?

 

[haruspex]

1) dm, 0
2) dm, ##\frac{1}{2}(dm)v^2 + (dm)gh##

Yes.

So, are you telling me that total work can be obtained by integrating ##(dm)gh##?

No... (what happened to the KE gain)?
In each time dt, how much work does the belt do on the sand? How can you deduce the power?

 

[Nexus99]

iI modified my previous one comment, but i think it's not correct anyway.
I had this idea: when the sand reach the container the speed is increased so i can obtain the work done by the work-energy principle and consequently the power by integrating
Is it right?

 

[haruspex]

iI modified my previous one comment, but i think it's not correct anyway.
I had this idea: when the sand reach the container the speed is increased so i can obtain the work done by the work-energy principle and consequently the power by integrating
Is it right?

The belt knows nothing about what happens to the sand after it leaves. Concentrate on the work done to get the sand from stationary at the bottom to moving at the top.
You have already found the work done in time dt. Deducing the power is trivial.

 

[Nexus99]

Isn't the work done in each dt equal to:
- dmgh ?
So the infinitesimal work done isn't ##-dmgh## ?

 

[haruspex]

Isn't the work done in each dt equal to:
- dmgh ?
So the total work done isn't ##-dmght## ?

Look at your expressions in post #5. In each dt, a dm enters with no useful energy and a dm leaves with ##\frac{1}{2}(dm)v_0^2 + (dm)gh##.
The sand along the belt looks the same as before.
What is the work done on the sand by the belt in time dt?

 

[Nexus99]

I'm not sure but i think the work done in time dt is - dmgh, if it's not true i don't know

 

[Delta2]

I'm not sure but i think the work done in time dt is - dmgh, if it's not true i don't know

Why don't you think that the work done in time dt is $$dW=\frac{dm}{2}v_0^2+dmgh$$ cause that's includes not only the change in the potential energy ##dmgh## of the mass dm but also the change in kinetic energy $$\frac{1}{2}dmv_0^2$$.

SO what do you think now?

 

[haruspex]

I'm not sure but i think the work done in time dt is - dmgh, if it's not true i don't know

Each dm arrives on the belt with zero PE and zero KE. It leaves with the PE and KE you found in post #5. How much energy does it gain? Where did that energy come from?

 

[Delta2]

@haruspex how do we know that the dm in the base and the dm in the top are equal?

 

[Nexus99]

Ok i understood.
So we can calculate power:
##\frac{dm}{dt} = \lambda ##
## P = \frac{dW}{dt}## = ## \frac{1}{2}\lambda v_0^2 + \lambda gh##

As for the second and third question, I did it this way:

2)
##q_i = dmv_0##
##q_f = dm\sqrt{2gh + v_0^2}## (velocity calculated by conservation of energy)
##\frac{dq}{dt} = \lambda(\sqrt{2gh + v_0^2} - v_0^2)##

3) Weight registered by the scale = Mg + the weight of the sand that is flying
Where the weight of the sand that is flying is equal to:
##\lambda(\sqrt{2gh + v_0^2} - v)## where ## \frac{(\sqrt{2gh + v_0^2} - v)}{g} ## is the fall time

 

[Delta2]

Ok i understood.
So we can calculate power:
##\frac{dm}{dt} = \lambda ##
## P = \frac{dW}{dt}## = ## \frac{1}{2}\lambda v_0^2 + \lambda gh##

As for the second and third question, I did it this way:

I think you are correct here.

2)
##q_i = dmv_0##
##q_f = dm\sqrt{2gh + v_0^2}## (velocity calculated by conservation of energy)
##\frac{dq}{dt} = \lambda(\sqrt{2gh + v_0^2} - v_0^2)##

I think you should NOT include the term ##-\lambda v_0^2## in the above expression, because the momentum of the sand just before it hits the container is ##dm\sqrt{2gh+v_0^2}## and after it hits the container is zero.

3) Weight registered by the scale = Mg + the weight of the sand that is flying
Where the weight of the sand that is flying is equal to:
##\lambda(\sqrt{2gh + v_0^2} - v)## where ## \frac{(\sqrt{2gh + v_0^2} - v)}{g} ## is the fall time

I think (not very condifent myself here though ) that the correct answer here should be $$Mg+\frac{dp}{dt}$$

 

[haruspex]

@haruspex how do we know that the dm in the base and the dm in the top are equal?

It's steady state. The total on the belt is not changing.

 

[Delta2]

It's steady state. The total on the belt is not changing.

I don't disagree here, I just believe this should be explicitly given in the problem's description, that we are in the steady state.

 

[haruspex]

I don't disagree here, I just believe this should be explicitly given in the problem's description, that we are in the steady state.

"At a certain instant t, when the roller is already covered with sand"

 

[Delta2]

"At a certain instant t, when the roller is already covered with sand"

This might be good enough for you but for some reason it isn't good enough for me, i believe it should have been something along the lines of "Assume that the system reaches a steady state during which the amount of sand present in the roller does not change w.r.t time"
@Okpluto i might be wrong in my comments about 2) and 3) in post #16. Sorry this problem seems to confuse me alot!

 

[haruspex]

i might be wrong in my comments about 2) and 3) in post #16.

I endorse those comments.

 

[Nexus99]

I endorse those comments.

Ok i understood your correction on question 2, but why is that the correct answer on question 3?
Is it because ##\vec{F_{ext}} = \frac{d\vec{p}}{dt}## ?

 

[haruspex]

Ok i understood your correction on question 2, but why is that the correct answer on question 3?
Is it because ##\vec{F_{ext}} = \frac{d\vec{p}}{dt}## ?

Yes. The scale 'knows' nothing about the past history of the sand. It is only affected by the fact that sand is arriving at a certain mass rate and a certain velocity.

 

[Nexus99]

Ok thanks for your help, i didn't understand that Mg was an internal force

 

[haruspex]

Ok thanks for your help, i didn't understand that Mg was an internal force

No, it is an external force. Why did you write that? What answer do you have now?