Terminal velocity of spherical body.

  • #1
PrincePhoenix
Gold Member
116
2
In our textbook, the equation for terminal velocity has been derived from Stokes' law and it comes down as follows,

vt = (mg)/6(pi)(eta)r

(r is the radius of the spherical body)

then, by putting the value of 'm' from m=(rho)V [where V = 4/3 * (pi)r3]

we get,
vt = 2(rho)gr2 / 9 (eta)

So in one equation, vt is directly related to r2, but in the other it is inversely related. Can someone please explain this to me.
Thak you.
 

Answers and Replies

  • #2
2
0
The equation vt = (mg)/6(pi)(eta)r gives the relation between the terminal velocity and the radius for a spherical body of constant mass, that is, you are stretching the body by changing r (mass does not change).

However, in vt = 2(rho)gr2 / 9 (eta) the density is constant and hence as you change the radius, mass also changes.
 
  • #3
PrincePhoenix
Gold Member
116
2
So these are two equations for different situations?
 
  • #4
2
0
Exactly.
 
  • #5
PrincePhoenix
Gold Member
116
2
Thanks for the help.
 

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