Conveyor belt
A conveyor belt has speed v m/s. A stream of goods is being loaded onto it via a hopper, at a steady rate of m kg/sec. How much power has to be supplied to the conveyor belt to compensate for the load?
Neglect any unnecessary complications.
Sorry, a bit boring.
Neglect any unnecessary complications.
Sorry, a bit boring.
Labels: funphysics
posted by Chris at
1:11 PM
13 Comments:
Power=g*m*v
to neglect all possible complications we just load the stuff with the same speed as the belt is moving, and voilà, you don't need no power at all =)
and even when you don't supply that speed from the beginning, you still need to allow the belt so slow down a bit, else it would need infinite power since the acceleration would be infinite.
Dunno if thats what you meant SBA but you should clarify your variables. e.g. what is "g" ? gravitational accel.? who knows, and if yes why should be g important? You would always need the same power here or on moon or on the solar surface ^^
Well spotted t:b:H. I should have said that the goods were being dropped vertically. You are right, the actual value of g (accel of gravity) is irrelevant. Of course, some gravity is required. Also friction (or a "mechanism") is required so that the goods come to rest on the conveyor belt.
The goods won't come to rest on the belt instantly, so infinite accels and powers don't arise.
For simplicity, only consider the steady state situation, not what happens when the goods are first dropped.
P=Power, m=Mass, v=velocity k=mass/second (our drop rate)
kin. Energy T=(m/2)*v^2
thats what the belt has to deliver in order to keep things going.
Here we don't have a mas but rather a flow of mass k.
which tells us:
P=(k/2)*v^2
quick check on units-> [(kg*m^2)/(s^3)] which is joules per second which is power ;)
t:b:H, I defined m to be the flow of mass. i.e. my m is your k.
Even allowing for that, your answer is wrong (surprised!?). Care to try again? Greetz.
maybe you should point out what is wrong .. or what i forgot, by just making it again probably the same result will arise because i like conservation of energy ^^
i choose k because i dont like m/2 * v^2 to be power...
hm, dunno what energy i forgot, but have now other things to do. i allow other people to try first =)
greetz
t:b:H I don't want to give too much away. Your ^^ is where your problm is.
We also should satisfy momentum conservation. If M=M(t) is conveyor belt mass, v is its speed, momentum conservation gives
v*dM=-M*dv.
dM=mdt, hence v=v_0exp(-mt/M).
Therefore N=dT/dt=mv^2dt/2=mv_0^2*
*exp(-2mt/M)/2 is power.
Hi Anonymous. You got the key idea, but you tripped yourself up somewhere.
A clue - I posted the problem, in part, because calculus isn't required.
As nobody seems to be really trying - here's the answer.
Conservation of energy isn't particularly helpful as the goods have to be accelerated and will be sliding on the belt and dissipating energy through friction. Also the goods will have had some energy if they were e.g. being dropped from a hopper say. Too messy!
Conservation of momentum doesn't have that problem. Any initial momentum is perpendicular to the belt and is simply transferred (ultimately) to the Earth.
On average, during each second a mass m is added to the belt and is accelerated up to speed v. That m eventually gains momentum mv. So the belt must be providing momentum mv every second (on average), so the force is also mv (Newton's second law). Power is the rate of doing work => (mv)v = mv^2.
The curiosity is that CoE only predicts exactly half that power. This suggests that exactly half the power is wasted to friction as the goods come up to speed. I don't recall any text book dealing with that even though it seems to be a fact.
I went a bit fast there, explanation:
work = force times distance or W = F*d.
If the mass (m) is moving with speed v, then d=v*t where t is the time taken. So W = F*v*t. Power P = W/t = F*v = (mv)*v = mv^2.
Last para 3:21 PM post is a bit confusing. The meaning was intended to be that the CoE predicted power only accounts for the energy that is being converted to kinetic energy. You need to provide that much power again to allow for the energy that is being converted to frictional losses.
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