# Sanity check on falling steel ball in water

• Sherwood Botsford
In summary, the conversation discusses the speed of a 5 mm steel ball falling through water at 10 C. Different equations are used to calculate the terminal velocity, including the laminar flow equation and the turbulent drag equation. The density of the medium being displaced by the ball, in this case water, is also taken into consideration. After correcting an algebra mistake, the calculated terminal velocity is 0.14 m/s, which is a more reasonable result.
Sherwood Botsford

## Homework Statement

What is the speed of a 5 mm steel ball falling through water at 10 C[/B]

Viscosity of water: 1.3059 * 10-3 Pa*s (Various online sources)

## Homework Equations

Laminar flow:

Vt = gd2 (ρp - ρm)/18μ

Vt = terminal velocity
g = gravity (10m/s2)
d = diameter of ball
ρp = density of ball
ρm = density of medium
μ = viscosity of medium

Turbulent drag

Fd=12ρCdAv2

Fd = drag
Cd = drag coefficient = 1
A = cross section area
v = velocity.

Force of gravity on ball.
Fg = mg

Mass of ball
m = ρ*4πr3/3
I used 8000kg/m3 for density.

At this point I'm ignoring buoyancy effects.

## The Attempt at a Solution

I initially did this using the stokes law calculator here:

http://www.meracalculator.com/physics/fluid-mechanics/stokes-law.php

But this gives me an answer of 75 m/sec. Which seems to me to be absurd, even for laminar flow.

This equation is for stokes law laminar flow. Units error for viscosity?

So try it for turbulent flow.

Mass of ball = .0005 kg = .5 gm this seems reasonable.
Fg = .005 N

Equating the above two formulas Fg = Fd and solve for v

v = √(12ρCdA/Fd)

Plugging in numbers I get 6.7 m/s While closer this still seems high. I would expect something on the order of a meter/s

I've been unable to find an online calculator to check on this[/B]

Sherwood Botsford said:
v = √(12ρCdA/Fd)
That would mean a steel ball with a larger surface area and a smaller mass would fall faster. Is that plausible?
The square root doesn't have units of speed either.

Sherwood Botsford said:
Fd=12ρCdAv2
Define ρ here.

Crap. Comes from doing things in my head.

Cross section of ball = πd2/4
= 3.14 * 0.0052/4
= 20e-6 m2v2 = mg/(12ρCdA)

v=√(mg/(12ρCdA))

=√(.005/12*1000kg/m3* 1* 20e-6 m2))

=0.14 m/sec

Algebra mistake. This looks more reasonable. I think.

haruspex said:
Define ρ here.
Density of the medium being displaced by the falling sphere -- in this case water.

Yes?

That looks good.

## 1. How does the density of the steel ball affect its rate of descent in water?

The density of the steel ball does not affect its rate of descent in water. In a vacuum, objects with different densities fall at different rates due to the effects of air resistance. However, in water, the density of the object is not a major factor in its rate of descent.

## 2. What is the relationship between the size of the steel ball and its rate of descent in water?

The size of the steel ball does not significantly affect its rate of descent in water. As long as the ball is small enough to avoid turbulence, its size does not have a significant impact on its descent.

## 3. Will the steel ball reach terminal velocity in water?

No, the steel ball will not reach terminal velocity in water. Terminal velocity is the maximum speed an object can reach when falling due to the balance of gravitational force and air resistance. In water, the density and viscosity of the liquid create more resistance than air, so the steel ball will not reach terminal velocity.

## 4. How does the temperature of the water affect the falling rate of the steel ball?

The temperature of the water does not have a significant effect on the falling rate of the steel ball. While temperature can affect the viscosity of a liquid, the change in temperature is not enough to have a noticeable impact on the descent of the steel ball in water.

## 5. Can other factors, such as shape or material, affect the falling rate of the steel ball in water?

Yes, other factors such as shape and material can affect the falling rate of the steel ball in water. Objects with a streamlined shape experience less resistance and may fall faster in water. Additionally, objects made of materials with high densities, such as gold, may fall faster due to their weight. However, these factors are not as significant as the effects of density and viscosity of the water.

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