How Do You Calculate Terminal Velocity for a Wooden Sphere Falling Through Air?

[booksrmylife]

Homework Statement

(a) Estimate the terminal speed of a wooden sphere (density 0.870 g/cm3) falling through air, if its radius is 8.00 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)
(b) From what height would a freely falling object reach this speed in the absence of air resistance?

Homework Equations

Vt = sqrt((2mg)/(DroeA))
v^2=2gx

The Attempt at a Solution

(A)
r=.08m
A=r^2pi=.020106
d=870kg/m^3
m=4/3pir^3*d=1.86585
so Vt should be sqrt((2(1.86585)g)/((.5)(1.2)(.020106))) or55.1
but it says that's wrong

can't start b without an a answer.

 

[jhae2.718]

Using those figures, I get the same answer.

I'm assuming you are using an online homework system? Does it specify what units it wants? Much of the problem is defined in terms of https://secure.wikimedia.org/wikipedia/en/wiki/CGS" . Perhaps try converting your answer to cm/s?

 

[booksrmylife]

nope, for sure mks... :( i was almost hoping i was doing something wrong.

 

[booksrmylife]

anyone else have an idea?

 

[PhanthomJay]

I agree with both of you that the answer is correct based on the values given. Using the density of air as 1.29 kg/m^3 will give a slightly different result, as in
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/fallq.html

 

[booksrmylife]

is it possible that I'm using the wrong equation?

 

[jhae2.718]

Probably not. I'm assuming this is an introductory physics course, so you're probably not being asked to model fluid flow with Navier-Stokes.

The dynamics of the system are, calling the drag force F and down positive:
m\mathbf{\ddot{x}} = m\mathbf{g} + \mathbf{F}.

Then, F is defined as:
\mathbf{F} = -C_D\frac{1}{2}\rho V^2 S \hat{\mathbf{V}} (i.e. opposite of velocity)

Since m\mathbf{\ddot{x}}=0 for terminal velocity, this leads to
0 = mg - C_D\frac{1}{2}\rho V^2 S

Solving for V_t:
V_t = \sqrt{\frac{2mg}{C_D\rho S}}

where C_D is coefficient of drag, and S is cross-sectional area orthogonal to the flow.

So, that eqn. should work.