Terminal velocity

daught

Well-known member
Campfire argument:

If you drop two balls that are shaped the same and they have the same coefficient of friction, but different weights does the heavier one reach a higher terminal velocity? This is in regular still air NOT vacuum.
 
Campfire argument:

If you drop two balls that are shaped the same and they have the same coefficient of friction, but different weights does the heavier one reach a higher terminal velocity? This is in regular still air NOT vacuum.

Since when you make such random threads?

I am starting to think about math I did 15 yrs ago.
I'm going to hazard a guess and say the heavier one has a higher terminal velocity
 
Mathematically, terminal velocity—without considering buoyancy effects—is given by
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where



http://en.wikipedia.org/wiki/Terminal_velocity
 
Campfire argument:

If you drop two balls that are shaped the same and they have the same coefficient of friction, but different weights does the heavier one reach a higher terminal velocity? This is in regular still air NOT vacuum.

What Newman said.
 
Isn't terminal velocity terminal velocity. Both balls will reach it at some point.
 
They will both reach their terminal velocity, but the heavier ball will have a higher velocity as evident by the equation posted above.
 
They will both reach their terminal velocity, but the heavier ball will have a higher velocity as evident by the equation posted above.

Ya, the tip off is that there is a formula for terminal velocity with variables and stuff. I got confused by the campfire argument component of the original post. Flashbacks.
 
Oke good answer ugur.

Now, does weight have much influence in a car's top speed? I thought weight only affects a car's acceleration.

Y do I post this, rockerguy? I am busy skiing and hiking in Washington and my only access to the internet is rare off an ancient symbian phone.
 
Oke good answer ugur.

Now, does weight have much influence in a car's top speed? I thought weight only affects a car's acceleration.

Y do I post this, rockerguy? I am busy skiing and hiking in Washington and my only access to the internet is rare off an ancient symbian phone.

I haven't thought it thoroughly but, in the formula above, if you substitute gravity with the acceleration of the car, then it turns into;

Vt = [ (2ma) / (pACd) ] ^ (1/2)
which is same as the one below - by using F = ma
Vt = [ (2F) / (pACd) ] ^ (1/2)
where the F is the torque of the engine. No mass is involved in the final equation, so the answer is no as caboose said.
 
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