A model helicopter is hovering. The rotor thrust is doubled. I assume a=1G. what is the terminal v and when?

Question

Getting old, and need a refresher course! I would appreciate it if anybody could help me understand this a bit more.
The assumption is starting at zero velocity at hover, where the thrust is equaling the downward force due to its mass.

Answer 1

a = (v - u) / t

We know that "u" is zero.
We know that "a" is 9.81m/s/s
What we need to know is time.

But to tell you the truth, you don't need that to find out the terminal velocity.

You need, the helecopters mass, the gravitational acceleration (which in this case is upward), the drag coefficient, the density of the fluid the object is falling through, and the object's cross-sectional area.

And you need this equation:

Vt = square root of [(2 x m x g) / (ρ x A x Cd)]

where

Vt is the terminal velocity,
m is the mass of the falling object,
g is gravitational acceleration,
Cd is the drag coefficient,
ρ is the density of the fluid the object is falling through, and
A is the object's cross-sectional area

I'd do more for you but i dont have the info.
Hope this helps.

Answer 2

double the thrust and the helicopter goes up. :-)

Terminal V is when the frictional force equals the new additional thrust vector.

It doesn't seem to me you have enough information.

(Also, of course, as the helicopter climbs, the thrust decreases for the same rotation due to lower air pressue, at some altitude you end up back in a hover...)

Still, "Not enough information" would seem to be the answer

Answer 3

The terminal velocity is

(2T-mg)*t/m,

Where T is the original rotary thrust
m = mass
g = accelaration due to gravity
t = time

I am assuming that there is no additional resistance due to the air