Just a question on terminal velocity

AI Thread Summary
Terminal velocity is defined as the constant velocity achieved when the net force acting on an object is zero, resulting in zero acceleration. This occurs when gravitational force is balanced by air resistance, which increases with speed. If air resistance becomes significantly strong, it can balance gravity at a lower speed, leading to a lower terminal velocity. The concept of terminal velocity implies that while the object approaches this speed, it never actually reaches it, as it asymptotically approaches the terminal value. Understanding these dynamics clarifies how forces interact to establish terminal velocity.
flyingpig
Messages
2,574
Reaction score
1

Homework Statement



It is a definition that the terminal velocity is the same as constant velocity where the net force is 0 and therefore the acceleration is also 0

Here is what I don't get, why must the terminal velocity must such that it is constant? I know that to be "terminal" it means it can no longer change it's velocity, but how come gravity will able to counteract it such that the sum of forces is 0? I mean what if air resistance is so strong that gravity can't counteract it? Wouldn't it mean the sum of forces is no longer 0 and the terminal velocity can never be reached or at that point, that is the terminal velocity?
 
Physics news on Phys.org
hi flyingpig! :wink:
flyingpig said:
… what if air resistance is so strong that gravity can't counteract it? Wouldn't it mean the sum of forces is no longer 0 and the terminal velocity can never be reached or at that point, that is the terminal velocity?

air resistance depends on speed

if air resistance is very strong, that means that it is very much stronger than usual for each particular speed

so air resistance will balance gravity at a lower speed …

ie terminal velocity is lower :smile:

(oh, and terminal velocity is never quite reached anyway … the speed gets exponentially close to terminal velocity :wink:)
 
tiny-tim said:
hi flyingpig! :wink:


air resistance depends on speed

if air resistance is very strong, that means that it is very much stronger than usual for each particular speed

so air resistance will balance gravity at a lower speed …

ie terminal velocity is lower :smile:

(oh, and terminal velocity is never quite reached anyway … the speed gets exponentially close to terminal velocity :wink:)

That is even more confusing now...

I just don't understand why or how gravity can balance out the terminal speed.
 
when v = 0, air resistance (let's call it R(v)) is 0,

so using a = F/m, we have a = mg/m = g

when R(v) is larger, a = (g - R(v))/m < g;

eventually R(v) = g, and then a = 0, ie the speed is constant :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top