How much power is needed to suspend a mass in the air?
The four parts below deal with the same problem. I post them all here, should there be some subtle things I didn't see.
The textbook problem
A 9th edition college physics textbook has this as an example problem in its energy chapter (details cleaned-up):
An elevator car ("a lift") and its passengers has a mass of 1800 kg and negligible friction force in all its moving part. How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 3.00 m/s?
Answer: Speed is constant, so $a=0$. Now let $F$ be the total force and $T$ be the force exerted by the motor. Then,
$$
\sum F = T-Mg = 0 \qquad \Rightarrow \qquad T = Mg
$$
with power = force x velocity, we then have
$$
P = T\cdot v = Mg \cdot v
$$
$$
P = 1800 \,kg\cdot9.8\,m/{s^2}\cdot 3.00\,m/s=5.29\times10^4\,{\textrm W}
$$
And, what power must the motor deliver at the instant the speed of the elevator is $v$ if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s per second? Answer: Now $a=1.00 \,m/{s^2}$, and then
$$
\sum F = T-Mg = Ma \qquad \Rightarrow \qquad T = M(a+g).
$$
Proceeding as before, we have
$$
P = T\cdot v = M(a+g) \cdot v
$$
$$
P = 1800 \,kg\cdot (1.00+9.8)\,m/{s^2}\cdot 3.00\,m/s= 5.83 \times10^4\,{\textrm W}.
$$
That's the end of the example.
My questions. We might as well ask these:
- How much power must the motor deliver to suspend the elevator and its passengers so it's kept at the constant height? That is, zero velocity? Certainly it can't be zero right? Because otherwise the elevator car will free-fall. But how much power is required?
- What power must the motor deliver at the instant the speed of the elevator is $v=3.00$ (as before) if the motor is designed to provide the elevator car with a downward acceleration of 1.00 m/s per second?
I've been looking for an answer but to no avail.
The helicopter problem
This is the real question that motivates this post.
I've been asked by a friend who's going to build a model-helicopter. "Ok, I have a 2000 W engine for my 50 kg model-helicopter. Assuming 100% efficiency, will it able to float 1 m above ground for at least 1 minute?"
I can't answer that.
"Ok then, what is minimum power required for the engine if I want it to float 1 m above ground, assuming 100% efficiency?"
I'll ask physics.se
The pulley problem
This is a simpler version of the above two problems. Using a rope and a pulley, what is the minimum power required for a man to suspend a 1 kg mass in the air 1 m above ground for 1 second?
The table or string problem
There's not much problem here. Suspending a 1 kg mass 1 m above the ground? Simple. Put it on a table, or hang it with a string. No movement. No power required.
What's happening here?
Best Answer
A few thoughts to help you on your way.
When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work / time = force * velocity ). If you are changing the velocity of an object, you are changing its kinetic energy: if it's slowing down, it gives energy back to you; if it's speeding up, you need to give it more energy.
If the elevator car is not moving, no WORK needs to be done. You still need a FORCE - but you could tie a knot in the cable and turn off the power: the elevator car will stay in place, without electricity, without heat being generated...
The helicopter example is different. The only reason a helicopter can hover is because it is pushing air down. Every second that it hovers, it needs to move a volume of air at a certain speed. In this case, the helpful equation is
$$F\Delta t = \Delta p$$
The change in momentum of the air tells you the force that you can get. This can be done by moving a large volume of air a little bit, or moving a little bit of air by a lot. Both situations will give you the same momentum, but since energy goes as velocity squared, the larger blades will be more efficient (up to the point where the drag of the blades becomes an important factor).
To solve the problem you stated, you need to know the size of the blades of the helicopter. Making some really simplified assumptions (there is at least a factor 2 missing in this - but just to get the idea): if you have a helicopter blade that sweeps an area $A$ and pushes air of density $\rho$ down with velocity $v$ then the force is
$$F = m \cdot v = (A\rho v)\cdot v = A\rho v^2$$
and the power needed (kinetic energy of the air pushed down per second) is
$$E = \frac12 m v^2 = \frac12 (A \rho v) v^2 = \frac12 A \rho v^3$$
If we assume $A=3m^2$ (roughly 1 m radius), and $\rho=1 kg/m^3$, then for a force of $500 N$ we need a velocity
$$v = \sqrt{\frac{F}{A\rho}} = \sqrt{\frac{500}{3}} = 13 m/s$$
and the power required is
$$E = \frac12 A \rho v^3 = \frac12\cdot 3 \cdot 13^3 = 3.3 kW$$
This is a bit higher than the 2kW you have available. The solution would be to increase the size of the rotors - the larger the area, the lower the velocity of the air, and the better off you are.
As for the pulley and string, or fixed string - see the comments I made about the elevator. When nothing moves, no work is done. In the case of the helicopter, although the helicopter doesn't move, the wings (blades of the rotor) do - and so does the air that is being moved (and whose motion provides the force needed to keep the helicopter in the air).
I hope that clears up your understanding.