Question

Terminal Velocity

Answer 1

A 6.5 cm -diameter ball has a terminal velocity of 28 m/s. What is the balls mass?

Answer 2

updated 7 days ago, so it might already be too late. anyway,

I think to be able to answer the question it would be important to know the viscosity of the fluid in which the ball is falling.

Stokes' law for drag is \[F_d = 6 \pi \eta r v \] with \(r = d/2\) the balls radius, v the velocity relative to the fluid, \(\eta\) the viscosity. Terminal velocity means that the acceleration \( a = 0 \).
For a body m falling in the earth's gravitational field we can set \[ m g = 6 \pi \eta r v \\ \Rightarrow m = \frac{6 \pi \eta r v}{g}\]
We know \(r = diameter / 2, \quad v = 28 m/s, \quad g = 9.81 m/s^2 \), now all we need is the viscosity.

Answer 3

it was not given

Answer 4

assume its air

Answer 5

Hey, I am sorry. It seems I got it wrong. when I used viscosity of the air I got a mass of about 30 micrograms. Obviously that's totally wrong.
Stokes' Law is only valid when the air is flowing laminarly. At a velocity of 28 m/s a body with diameter of 6.5 cm creates turbulance and therefore we need to use another equation.

\[ D = 1/2 C \rho A v^2\] \(\rho = 1.225 kg/m^3\) is the density of air, C = .4 is the drag coefficient, \(A = \pi d^2/4\) is the reference area of the ball.

the idea is the same. Once the body is at the terminal velocity gravitational force is equal to drag and we can solve for m
\[ m = \frac{1/2 C \rho A v^2}{g} \approx 0.065 kg \]

comparing the data with sources on the internet the object might be a tennis ball.

Answer 6

The answer said it was 80g. =( weird I got nearly the same number as you.

Answer 7

Oh I just found out I had the wrong textbook edition. C = 1.2 row = 0.5