# Decreasing diameter of a spehere and drag

• wkfrst

#### wkfrst

Hey all,
I have been mulling this around in my head for awhile and cannot make sense logically.
In thinking about a sphere falling through some random liquid, how does decreasing the diameter of the sphere affect the rate at which it falls?

I am thinking of two formulas and the seem to contradict each other in terms of their results.
First would be Stoke's law dealing with drag force:
F(d)= 3(pi)μVd
Here if I reduce the radius, the force of drag decreases which makes me think the sphere would fall faster because it has less resisting its fall.

The next would be a derived equation using the force of buoyancy, drag force, and weight. This assuming these are the only three forces on a falling sphere. It is solved for velocity.
V= [2r^2(ρ(sphere)-ρ(fluid))g]/(9μ)
Here it appears that if the radius is reduced, the velocity is also reduced and it therefore falls slower.
My "logic" tells me a smaller sphere would fall faster, but it has failed me many times in physics.
Thanks for the help.

Dust (even spherical dust) takes almost forever to settle out, whether in air or water.

Do you think that a glass cricket ball would sink through honey faster or slower than a marble?

If you are near an engineering workshop, you might be able to pick up a huge ball bearing and compare its descent through syrup with that of a smaller one.