Sand drops on a conveyor belt at constant rate

[Bruce_Pipi121]

Homework Statement

A conveyor belt is driven at velocity $v$ by a motor. Sand drops vertically on to the belt at a rate of $m~kg~s^{-1}$.

What is the additional power needed to keep the conveyor belt moving at a steady speed when the sand starts to fall on it?

Relevant Equations

$E_K=\frac{1}{2} m v^2$, $p=mv$, $F=\frac{\Delta p}{\Delta t}$

I can understand how it can be explained using the change of momentum to find the average force on the belt, then us $$P=Fv$$ to get the power.

What I didn't understand is why the energy solution won't work. If in 1 second, sand with mass m has fallen on the belt and accelerated to speed v, then the sand has gained kinetic energy $$E_K=\frac{1}{2} m v^2$$. As the mark scheme does confirm this is the amount of energy gained by sand, it states that this cannot be the only energy involved here, otherwise the acceleration of a grain of sand will be infinitely large.

I couldn't understand what does it have to do with the acceleration of sand, after all, if we assume the belt is way more massive and sticky, then the acceleration of sand in this case would indeed be very very large.

 

[NTesla]

When the sand grains are falling on the conveyor belt, there is inelastic collision as a consequence of which, the sand sticks to the surface of the conveyor belt and starts moving alongwith the belt. If, instead of inelastic collision, elastic collision would have happened, then the sand particles would have bounced back in the direction opposite to their initial direction.
Now, since there is inelastic collision, therefore some amount of energy has gone out of the system of sand and the belt. Therefore, energy equations will not give you right answer in this particular case.

 

[haruspex]

it states that this cannot be the only energy involved here, otherwise the acceleration of a grain of sand will be infinitely large.

That's a weird explanation.
Obviously no individual grain of sand can get to speed v instantly, but that has nothing to with elastic versus inelastic collision. Most likely, each grain tumbles backwards relative to the belt at first, taking a little while to match the belt's speed.

 

[NTesla]

but that has nothing to with elastic versus inelastic collision.

If the collision between sand grains and belt has nothing to do with elastic/inelastic collision, then by what mechanism are the sand particles sticking to the belt and start moving alongwith the belt ?

 

[haruspex]

If the collision between sand grains and belt has nothing to do with elastic/inelastic collision, then by what mechanism are the sand particles sticking to the belt and start moving alongwith the belt ?

I wrote that the impossibility of getting instantly to the speed of the belt has nothing to do with elasticity. This is in response to the quoted "infinitely large acceleration" argument. Whether they bounce or stick, the velocity change takes a little time.
Certainly elasticity would make it impossible to match the velocity of the belt at all. They would bounce in any number of directions.

 

[Bruce_Pipi121]

Cool, thanks for your explanations guys. So we should say whether the acceleration is infinite or not has nothing to do with the amount of extra energy needed here, right?

I tried to take the analysis of using inelastic model further.

Suppose the mass of belt is M moving with speed v, and sand particle m initially at rest. So after collision $$Mv=(M+m)v\prime$$ and we get $v\prime=(\frac{M}{M+m})v$.

In order to make them moving with speed v again, we need extra energy$$\Delta E=\frac{1}{2}(M+m)v^2 - \frac{1}{2}(M+m)(\frac{M}{M+m})^2v^2$$$$=[\frac{1}{2}(M+m)v^2 - \frac{1}{2}Mv^2 ]+[ \frac{1}{2}Mv^2 - \frac{1}{2}(M+m)(\frac{M}{M+m})^2v^2]$$
$$=\frac{1}{2}mv^2 + \frac{1}{2}Mv^2(1-\frac{1}{1+\frac{m}{M}}) $$
Since $\frac{m}{M}$ is very small, so $\frac{1}{1+\frac{m}{M}}=1-\frac{m}{M}+(\frac{m}{M})^2-(\frac{m}{M})^3+\cdots$ So the extra energy we need is
$$\Delta E=\frac{1}{2}mv^2 + \frac{1}{2}Mv^2(\frac{m}{M}-[\frac{m}{M})^2+(\frac{m}{M})^3-\cdots]= mv^2 - O(\frac{m}{M})$$
But if this is correct, then the extra energy is a little bit smaller than $mv^2$, and why the change of momentum method would imply an answer exactly equals to $mv^2$?

 

[NTesla]

I couldn't find anything wrong in your calculation.
However, the only explanation for this discrepancy, is that, since the collision is inelastic, therefore although in the first equation momentum was conserved, but energy wasn't conserved. During the inelastic collision, it may happen that some energy has gone into vibrating the atoms/molecules of the belt/sand, which we could sense as heat, however miniscule that may be.
So a difference in answer was to be expected.
However, it would be interesting to hear what other veterans on the subject have to say about this issue.

 

[Frodo]

You should ignore the vertical direction - it is irrelevant. Any energy gained by dropping will be dissipated into heat and sound or whatever. As is any jumbling around on the belt or any consideration of elastic or inelastic - ignore it all. You could think of the sand being viscous treacle or axle grease.

It is often useful to simplify a problem to see what is happening and then go back to the original.

Consider a plate of mass 10m rolling on frictionless bearings and moving at v m/s.

Now drop a mass of m onto it.

The plate's velocity after the mass lands is, by conservation of momentum, 10v/11.

You now accelerate the whole 11m back to v in one second. The acceleration required is v/11 m/sec/sec.

Force = ma, so the force needed is 11m x v/11 = mv.

But if this is correct, then the extra energy is a little bit smaller than , and why the change of momentum method would imply an answer exactly equals to ?

Without going through your calculation ...

... you need m to tend to zero for most accuracy and hence, m/M will also tend to zero. If you integrate, with an infinitely small variation; instead of using a finite difference, with a finite variation; the m/M term will be zero.

 

[NTesla]

@Frodo
The question that OP had asked in post#6 was: Why the difference is creeping up in calculation when the calculation is done in two different way: one by using momentum equation and the other way by using energy equation. The question was: why. And you've written again the momentum equation. That doesn't explain, nor makes any attempt at the why part of the question. Your solution of "ignore it all" is not helpful. The considerations that you are willing to ignore is precisely what I think is making the difference. The other half of what you've written is quite obvious, but that is just mathematical reasoning, it still doesn't answer why the difference is coming up in the first place.
Any idea on the why part of OP's question in post#6 would be welcome.

 

[Frodo]

@Frodo
The question that OP had asked in post#6 was: Why the difference is creeping up in calculation when the calculation is done in two different way: one by using momentum equation and the other way by using energy equation. ...
Any idea on the why part of OP's question in post#6 would be welcome.

If you use the energy equation and calculate the extra kinetic energy acquired by the added mass, you are only calculating the energy given to the mass m to accelerate the mass m from zero speed to v. But this in not the only thing which happens so that calculation is completely irrelevant. **

When the mass drops on the belt, it slows the belt to a speed less than v. This is your starting position.

First, you use conservation of momentum to calculate the belt speed immediately after the load lands.

The motor now requires additional power to accelerate the "belt with its existing previous load plus the mass dropped onto it, all moving at less than v", back to v.

Note how, at the beginning, the belt has no load, so the belt will be slowed a lot. The power is required to accelerate a small total mass by a large velocity.

Later on, the belt will have a large existing load, so it won't be slowed much. The power is now required to accelerate a much larger total mass by a much smaller velocity.

As all the equations are linear these two required powers: "small total mass by a large velocity" and "large total mass by a small velocity", are equal so the additional power will be constant over time.

** I suppose you could calculate this, and then calculate the energy required to accelerate the "belt and existing load" from the "slowed to below v" velocity back to v. Add this to the KE acquired by the dropped mass being accelerated to v and you will get the same answer.

 

[NTesla]

If you use the energy equation, you are only calculating the energy given to the mass m to accelerate it from zero speed to v. But this in not what happens so that calculation is completely irrelevant.

Could you please elaborate on the the part where you've said: this is not what happens.
What does happen ? Is part of the total energy of the system going into something other than the sand and the belt ?

 

[Frodo]

I was editing to clarify this point and our posts crossed. See my ** addition.

 

[NTesla]

calculate the energy required to accelerate the belt and existing load" from the "slowed to below v" velocity back to v. Add this to the KE acquired by the dropped mass being accelerated to v and you will get the same answer

In post #6, OP has already calculated the energy required to accelerate the belt and existing load" from the "slowed to below v" velocity back to v.
$$ mv^2 - O(\frac{m}{M})$$

Now, KE acquired by the grain of sand is $0.5mv^2$. Adding these 2 does not give the answer as is calculated by using momentum equations.

 

[Frodo]

I think you are misunderstanding what the OP is saying.

He calculates the energy required to be $mv^2$ which is correct. Note it is twice the extra energy acquired by the mass alone, which is $\frac{1}{2}mv^2$.

So, one half of the power goes to accelerate the mass and one half the power goes to accelerate the belt and load back up to speed.

The question he asks is: why does the last term in $mv^2 - O(\frac{m}{M})$ exist? Why isn't the answer just $mv^2$ without the $O(\frac{m}{M})$ last term? He is not asking why his answer $mv^2$ is different from $\frac{1}{2}mv^2$ for the mass alone.

The reason is because he has used an infinite expansion approximation to do the calculation which assumes $m/M$ is very small, and he discards later terms. Had he not used the approximation and instead worked through the reorganisation of the expression, he would not have got that term (which, in any case, tends to zero).

 

[haruspex]

Cool, thanks for your explanations guys. So we should say whether the acceleration is infinite or not has nothing to do with the amount of extra energy needed here, right?

I tried to take the analysis of using inelastic model further.

Suppose the mass of belt is M moving with speed v, and sand particle m initially at rest. So after collision $$Mv=(M+m)v\prime$$ and we get $v\prime=(\frac{M}{M+m})v$.

In order to make them moving with speed v again, we need extra energy$$\Delta E=\frac{1}{2}(M+m)v^2 - \frac{1}{2}(M+m)(\frac{M}{M+m})^2v^2$$$$=[\frac{1}{2}(M+m)v^2 - \frac{1}{2}Mv^2 ]+[ \frac{1}{2}Mv^2 - \frac{1}{2}(M+m)(\frac{M}{M+m})^2v^2]$$
$$=\frac{1}{2}mv^2 + \frac{1}{2}Mv^2(1-\frac{1}{1+\frac{m}{M}}) $$
Since $\frac{m}{M}$ is very small, so $\frac{1}{1+\frac{m}{M}}=1-\frac{m}{M}+(\frac{m}{M})^2-(\frac{m}{M})^3+\cdots$ So the extra energy we need is
$$\Delta E=\frac{1}{2}mv^2 + \frac{1}{2}Mv^2(\frac{m}{M}-[\frac{m}{M})^2+(\frac{m}{M})^3-\cdots]= mv^2 - O(\frac{m}{M})$$
But if this is correct, then the extra energy is a little bit smaller than $mv^2$, and why the change of momentum method would imply an answer exactly equals to $mv^2$?

As you observe, the energy loss is maximised by letting the sand grains tend to zero individual mass, while maintaining the mass rate.
To see why, we can compare these two models:
1. Sand grains are mass m, falling at interval Δt, with the belt being accelerated back to speed v after each grain.
2. As 1, except the belt speed is allowed to decline, only being restored every nth grain.
It should be clear that the second is equivalent to a grain of mass nm falling each nΔt.
In 2, because the belt is moving at a lower speed with each successive grain, the energy loss is reduced.

 

[NTesla]

Had he not used the approximation and instead worked through the reorganisation of the expression

While using the energy equations in post#6, expanding the expression is the only way to solve it, at the end of which we need to use the approximation of m/M tending to zero.
What other reorganisation of the expression is possible to get the answer exactly equal to $mv^2$ ?

 

[NTesla]

As you observe, the energy loss is maximised by letting the sand grains tend to zero individual mass, while maintaining the mass rate.
To see why, we can compare these two models:
1. Sand grains are mass m, falling at interval Δt, with the belt being accelerated back to speed v after each grain.
2. As 1, except the belt speed is allowed to decline, only being restored every nth grain.
It should be clear that the second is equivalent to a grain of mass nm falling each nΔt.
In 2, because the belt is moving at a lower speed with each successive grain, the energy loss is reduced.

If we are assuming no inelastic collision therefore no energy loss from the system when the grain falls on the belt, then will it matter if individual grain are being dropped on the belt or a lump of grain is being dropped after every finite time.?

 

[haruspex]

If we are assuming no inelastic collision therefore no energy loss from the system when the grain falls on the belt, then will it matter if individual grain are being dropped on the belt or a lump of grain is being dropped after every finite time.?

Matter in what way? It won't alter the energy loss, since it would be zero in both models.

 

[NTesla]

energy loss would be zero in both models.

In post#6, when using energy equation, the extra energy needed to bring the system back to the same velocity $v$ is $mv^2 - O(\frac{m}{M})$. When momentum equations are used, then the energy needed to bring the system back to the same velocity $v$ is $mv^2$.

What could be the underlying reason for difference in these 2 results; not just mathematical reasoning(value of m/M tending to zero) in the expansion of the expression.

 

[haruspex]

In post#6, when using energy equation, the extra energy needed to bring the system back to the same velocity $v$ is $mv^2 - O(\frac{m}{M})$. When momentum equations are used, then the energy needed to bring the system back to the same velocity $v$ is $mv^2$.

What could be the underlying reason for difference in these 2 results; not just mathematical reasoning(value of m/M tending to zero) in the expansion of the expression.

You seem to be confusing two things.

In post #1, @Bruce_Pipi121 tried to use conservation of work and found that the energy needed to get the sand grain to the speed of the belt was $\frac 12mv^2$. Using conservation of momentum in a continuous model ($F=\frac{dp}{dt}$) the energy needed was $mv^2$, twice as much.
(The book gave the latter as correct, but a nonsense explanation for the reason.)

In post #2, you explained that it is an inelastic collision, so useful energy is lost, accounting for the extra $\frac 12mv^2$ of work the belt has to do.

Bruce was happy with that, but in post #6 used conservation of momentum, not energy, in a discrete model and got a slightly different answer from the continuous model. For the reason the discrete model gives slightly less energy needed (tending to zero extra as the particle mass tends to zero) see post #15.