Approaches to Sand on a Conveyor Belt [closed]

Question

Say we are given a conveyor belt with sand falling onto it at rate $\Omega$. I am trying to find the power it takes for the belt to operate if it goes forward with constant velocity $v$, but using two different approaches I get two different answers.

1. The first way, I say that in one second, $\Omega$ amount of sand falls onto the belt, thus every second $\frac 1 2 \Omega v^2$ is required to continue moving the belt forward (simply by looking at kinetic energy difference) and $P=\frac 1 2 \Omega v^2$.

2. The second way, I saw that $F=\frac {d(mv)} {dt}$, thus $F=\Omega v$. Finally, $P=Fv$ so $P=\Omega v^2$.

So which one is it: $P=\Omega v^2$ or $P=\frac {1} {2} \Omega v^2$, and where did I make an incorrect assumption in my two methods?

Answer 1

Your first answer gives you the rate at which the sand gains kinetic energy whereas your second answer gives you the rate at which work has to be done on the conveyor belt to keep it moving at constant speed.

When the sand falls on the belt in order to accelerate the sand to the same velocity as the belt there must be frictional forces between the sand and the belt.
There must also be relative movement between the belt and the sand during this acceleration phase as the sand cannot instantaneously to the velocity of the belt.
During the acceleration phase when there is slippage between the belt and the sand heat/thermal energy is generated and the rate of heat generation is the difference between your two answers.

So the motor which drives the belt increases the kinetic energy of the sand whilst heat is being generated due to the relative movement between the sand and the belt during the acceleration phase of the sand.

Answer 2

I think it is easier to accept that $P = \Omega v^2$, if one shows why the thermal energy generated happens to be excactly $\frac{1}{2}\Omega v^2$. Let an observer $O´$ be in a reference frame which is stationary with respect to the conveyor belt (i.e. $O´$ will be seen to have a velocity $+v$ in the lab - frame). To $O´$ a grain of sand with mass $dm$ will have a velocity $-v$ when falling onto the belt, and will start sliding backwards. Friction will oppose this, and the kinetic energy, $\frac{1}{2}dmv^2$, is converted to thermal energy. $O´$ will measure a temperature increase of the belt (and grain), consistent with this. Another observer $O$ in the lab - frame, will se the conveyor belt moving with velocity $v$, and the grain of sand starting out with zero velocity. Due to friction, it will start sliding along until it reaches velocity $v$, and its kinetic energy is $\frac{1}{2}dmv^2$. At the same time, thermal energy is generated. When $O$ measures the temperatrue increase of the belt, he will get the same readings as $O´$ (temperature being a scalar), and so this energy will be found to be $\frac{1}{2}dmv^2$. By conservation of energy, the energy that must be supplied to the conveyor belt to get a grain of sand up to speed, is $dmv^2$. To match the rate of sand falling onto the belt, the power that must be supplied is therefore $P = \Omega v^2$.

Answer 3

The answer by Farcher is right. I will comment further in hopes of increasing clarity and in order to clear up some mistakes introduced in other answers.

This kind of problem can be tackled by considering conservation of momentum. We also have conservation of energy, but that is more difficult to apply correctly because you have to take into account frictional heating.

Let's consider the sand falling during a time $t$ (you can make this a long time if you like). Before falling its momentum (in the direction of the conveyor) was zero. After falling and taking up the motion of the conveyor its momentum is $\Omega t v$. Now you might doubt that, and say that the last little bit of sand only just hit the conveyor so it is not yet moving at speed $v$. But that is a tiny correction which can be ignored in comparison to $\Omega t v$ (and we will argue in the end that it is correct to ignore it completely). The force is applied during the time $t$, so the required force is given by $$ F t = \Omega t v $$ so $F = \Omega v$ and $P =\Omega v^2$.

Now let's consider energy. The kinetic energy of the sand accounts for half the power; the other half has gone to heat. A further insight can be obtained by asking what would happen if there were no friction. In that case the sand would slip along the conveyor, it would not be accelerated. So instead let's have it fall into a sequence of containers like train trucks. Then as it falls into each truck, if there is no friction it will slip along the floor (we arrange for no vertical bounce). The end of the truck approaches and then hits the sand. Assuming an elastic collision the sand bounces off and moves off towards the other end. It continues bouncing to and fro inside the truck. This bouncing motion has more kinetic energy than the amount $(1/2) \Omega v^2 t$ where $v$ is the speed of the truck relative to the ground. How much more? The answer is: twice as much.

Now let's go back to the frictional case. The horizontal bouncing which I just described is then happening on a microscopic scale, generating sound waves in the sand and in the conveyor belt. These waves carry away the energy and the sand settles down until it is no longer in motion relative to the belt.

Answer 4

I believe the mistake arises from the expression $P = Fv$. Let's see how this equation was derived.

Consider a time interval $\mathrm{d}t$ in which the amount of mass falling is $\mathrm{d}m = \Omega \mathrm{d}t$.

$$F = \frac{\mathrm{d}p}{\mathrm{d}t} = \Omega v$$

Then consider the work done in this time $\mathrm{d}t$.

$$\mathrm{d}W = F \mathrm{d}s$$

Now here is where the mistake arises. If you take $\mathrm{d}s = v\mathrm{d}t$ then you get $P = Fv$.

However not all of the mass $\mathrm{d}m$ is travelling by a distance of $v\mathrm{d}t$. Suppose there are some number of particles falling in $\mathrm{d}t$ time. Then the first particle travels a distance of $v\mathrm{d}t$ however the last particle travels $0$ distance.

So the average distance travelled by the $\mathrm{d}m$ mass in this time interval is $\frac{1}{2}v\mathrm{d}t$. So we have

$$\mathrm{d}W = F \mathrm{d}s = \frac{1}{2}Fv\mathrm{d}t$$ $$P = \frac{1}{2}\Omega v^2$$

Answer 5

The work needed to accelerate a mass from rest to a constant speed $v$ is always twice the ultimate kinetic energy of the mass. So, work is $mv^2$ (not $\frac12mv^2$), or in your example $P=\Omega v^2$, since power is work per unit time, and $\Omega=m/t$.

Where did the extra energy go? If you think about it you realize that mass cannot be accelerated instantaneously because that would require an infinite force. There needs to be some stretching between the object that is moving at $v$ and the mass being accelerated. The interconnected objects act like a spring and are loaded with an energy equal to $\frac12mv^2$.

Why is this spring energy the same as the ultimate kinetic energy of the mass? Change your frame of reference from the mass (sand grain) to the object pulling the mass (the conveyor). In this frame the conveyor is stationary and the mass is moving backwards at $v$ and has $KE=\frac12mv^2$. When the mass stops moving relative to the conveyor, all this kinetic energy has been converted into spring energy. So back in the frame of reference of the sand, the total energy required is $mv^2$: half for the kinetic energy of the sand grain, and half for stretching the "spring" between the sand grain and the conveyor belt.

Answer 6

Your second approach F=Ωv; but the sand is gaining velocity from zero to v when it is falling onto belt, it is not a uniform velocity in this process. Therefore the power should be dP = Fdv, and hence the answer to your first approach $P=1/2 Ωv^2 $

Answer 7

In your second method don't forget f=ma so you still need a 1/2 term to get mass over time. All the sand isn't on the belt yet at t=0.