lissie Posted August 30, 2013 Share Posted August 30, 2013 (edited) I'm not much of a physicist and was wondering if someone could explain to me how terminal speed is reached and whether or not the weight of the object effects it...thanks x Edited August 30, 2013 by lissie Link to comment Share on other sites More sharing options...
pears Posted August 30, 2013 Share Posted August 30, 2013 (edited) As I understand it terminal speed is reached when two forces cancel each other out, so you have gravity in one direction (weight) and air resistance in the other. The object stops accelerating at the point when the force of gravity and air-resistence in the opposite direction acheive equilibrium. http://en.wikipedia.org/wiki/Terminal_velocity The actual mass of the objects doesn't affect terminal velocity, so two objects with different masses would fall to the earth at the same time. Edited August 30, 2013 by pears Link to comment Share on other sites More sharing options...
swansont Posted August 30, 2013 Share Posted August 30, 2013 As I understand it terminal speed is reached when two forces cancel each other out, so you have gravity in one direction (weight) and air resistance in the other. The object stops accelerating at the point when the force of gravity and air-resistence in the opposite direction acheive equilibrium. http://en.wikipedia.org/wiki/Terminal_velocity The actual mass of the objects doesn't affect terminal velocity, so two objects with different masses would fall to the earth at the same time. You might want to rethink this, given that your link has an expression for vt that has a mass term in it. You have F = mg-kvn (n is probably 2, but we don't really care.) Setting F = 0, just by inspection you can see mass does not cancel. An object with a greater mass feels a larger gravitational force and need more drag to cancel, thus a higher speed Link to comment Share on other sites More sharing options...
pears Posted August 30, 2013 Share Posted August 30, 2013 Oh poo did I get it wrong? I had f1/m - f2/m = 0 which got rid of the mass from my equation. I will have another think about this. Link to comment Share on other sites More sharing options...
Iggy Posted August 30, 2013 Share Posted August 30, 2013 An interesting consequence is that you can drop any animal about the size of a mouse and smaller from any height and it'll walk away. Larger animals have roughly the same density, but a larger terminal velocity with their larger mass and radius. Link to comment Share on other sites More sharing options...
pears Posted September 1, 2013 Share Posted September 1, 2013 OK I see where I've gone wrong now. I started out talking about terminal velocity, and two opposing forces, then in my final sentence I am talking about free-fall (i.e. there is only one force in play - no air-resistance). So the last sentence in my first post ought to be ignored, as it's an entirely different scenario and completely wrong in this context.When an object falls through a fluid, such as something dropping through the air on earth, the terminal velocity depends on the two forces, as well as the mass of the object falling. So two parachutists at different weights, would indeed fall at different speeds (whereas in the absence of air-resistance they would reach the earth at the same time, and I presume, never reach terminal velocity but continue to accelerate all the way down.)How embarrassing to discover knowledge one thought one had remembered is all just a distant hazy jumble! Link to comment Share on other sites More sharing options...
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