# How Do You Solve a Dimensional Analysis Problem for a Block in Viscous Fluid?

• lichenguy
In summary: The time constant means dropping the initial value to ##1/e## percent or increasing the value to ##1-1/e## percent. Is that happening here?Yes, except that it takes a time ##\tau## to do that.
lichenguy

## Homework Statement

A block of mass ##m = 1.00 kg## is being dragged through some viscous fluid by
an external force ##F = 10.0 N##. The resistive force can be written as ##R = -bv##,
where ##v## is the speed and ##b = 4.00 kg/s## is a phenomenological constant. You
may ignore gravity (we imagine that the block is floating inside the fluid). The
following integrals may come in handy:
$$\int \frac 1 { z } dz = ln(z), \int e^{-at} dt = \frac 1 a e^{-at}$$
a) Using the three physical quantities provided, use dimensional analysis to
find the terminal velocity of the motion ##v_T## and the characteristic time ##τ## .
b) Assuming that the block starts from rest, find ##v(t)##.
c) Find the distance traveled as a function of time ##x(t)##.

Some of the work done by the force F on the system becomes internal energy,
some becomes kinetic energy.
d) How much internal energy has been generated by the time the
block reaches half the terminal velocity, ##v(t) = v_T/2##? How big a fraction of
the total work is that?

## The Attempt at a Solution

##m=M##
##F=\frac {ML} {T^2}##
##R=-\frac {LM} {T^2}##

##v_T = \frac L T = m^aF^bR^c##

##M=0=a+b+c##
##L=1=b+c##
##T=-1=-2b-2c##
This doesn't work. So I'm kinda stuck.
I also don't know what is meant by "characteristic time".

#### Attachments

• problem 3.png
4.2 KB · Views: 506
Last edited:
You need to separate ##R## into ##b## which has some dimensions (what?) and ##v## which has dimensions ##LT^{-1}##. What kind of equation do you get then?
The exponential ##e^{-at}## can be rewritten as ##e^{-t/\tau}## where ##\tau = 1/a## is the time constant or characteristic time. It is the time required for the exponential to drop to ##1/e## of its initial value.

On edit: To elaborate a bit. Since the magnitude of ##R## is ##bv##, when terminal velocity is reached, ##R## reaches its terminal value ##R_T=bv_T##. You can't have dimensions of ##v_T## on both sides of the equation.

On second edit: Part (d) says "How much internal energy has been generated by the time the block reaches half the terminal velocity, ##v(t)=v_T=2##?" Shouldn't it be "How much internal energy has been generated by the time the block reaches half the terminal velocity, ##v(t)=v_T/2##?"

Last edited:
lichenguy
kuruman said:
You need to separate ##R## into ##b## which has some dimensions (what?) and ##v## which has dimensions ##LT^{-1}##. What kind of equation do you get then?
The exponential ##e^{-at}## can be rewritten as ##e^{-t/\tau}## where ##\tau = 1/a## is the time constant or characteristic time. It is the time required for the exponential to drop to ##1/e## of its initial value.

On edit: To elaborate a bit. Since the magnitude of ##R## is ##bv##, when terminal velocity is reached, ##R## reaches its terminal value ##R_T=bv_T##. You can't have dimensions of ##v_T## on both sides of the equation.

On second edit: Part (d) says "How much internal energy has been generated by the time the block reaches half the terminal velocity, ##v(t)=v_T=2##?" Shouldn't it be "How much internal energy has been generated by the time the block reaches half the terminal velocity, ##v(t)=v_T/2##?"
##b## has dimensions ##MT^{-1}## and together with ##v## i got the ##R=-\frac {LM} {T^2}## above. Yeah, i already did that. I guess i don't understand "You can't have dimensions of ##v_T## on both sides of the equation.".

I understand what the time constant does now. I found a cool video on the subject:
You're right, it's supposed to be ##v(t)=v_T/2##, i corrected it.

Last edited:
lichenguy said:
a) Using the three physical quantities provided, use dimensional analysis to
find the terminal velocity of the motion ##v_T## and the characteristic time ##τ## .
...
##R=-\frac {LM} {T^2}##
R is a variable within the process, i.e. a function of time. It cannot feature in the expression for terminal velocity. Find the more fundamental parameter. (Hint: to avoid confusion, you should also change your choice of unknowns for the powers.)

lichenguy
So, ##R## is not an independent variable and that is why we can't use it?
And solving for dimension ##T## gives: ##m/b##
But, what does this ##T## mean here? The time constant means dropping the initial value to ##1/e## percent or increasing the value to ##1-1/e## percent. Is that happening here?

lichenguy said:
R is not an independent variable and that is why we can't use it?
That's not quite it. The terminal velocity depends on certain parameters that describe the set-up. R is not one of those parameters. It is a function of time during the process, so its value changes with time. F, m and b are the constants that determine the behaviour.
lichenguy said:
what does this T mean here?
It tells you how the system behaves as you vary the parameters. If you increase m, keeping b constant, then the process slows down, i.e. it takes longer to reach a given fraction of its terminal velocity. If you were to plot the velocity as a fraction of terminal velocity against time, the time axis is stretched, but the graph otherwise stays the same. Conversely, increasing b for the same m speeds up the process. Changing F has no effect.

lichenguy
haruspex said:
That's not quite it. The terminal velocity depends on certain parameters that describe the set-up. R is not one of those parameters. It is a function of time during the process, so its value changes with time. F, m and b are the constants that determine the behaviour.

It tells you how the system behaves as you vary the parameters. If you increase m, keeping b constant, then the process slows down, i.e. it takes longer to reach a given fraction of its terminal velocity. If you were to plot the velocity as a fraction of terminal velocity against time, the time axis is stretched, but the graph otherwise stays the same. Conversely, increasing b for the same m speeds up the process. Changing F has no effect.
In this problem, if i compute ##τ=m/b## i get that ##τ=0.25## seconds. What fraction of ##v_T## does the block have when ##τ=0.25## seconds? Is that a bad question? Because it depends what ##v## is at the start, right?

lichenguy said:
In this problem, if i compute ##τ=m/b## i get that ##τ=0.25## seconds. What fraction of ##v_T## does the block have when ##τ=0.25## seconds? Is that a bad question? Because it depends what ##v## is at the start, right?
The assumption would be that initial speed is zero, i.e. choose start time such that speed is zero at time 0.
To find the fraction you would need to solve the differential equation.

Alright, finished, finally. Here are the results:
$$v(t) = v_T (1-e^{ - \frac t τ})$$ $$x(t) = v_T τ(\frac t τ + e^{ - \frac t τ} -1)$$ Where ##v_T = \frac F b## and ##τ = \frac m b##

On d:
I first found that velocity is ##v = \frac {v_T} 2## at time ##t=ln(2)τ##
Using ##W=ΔK+ΔE_{internal}##
##E_{internal} = Fτv_T(ln(2)- \frac 1 2 ) - \frac 1 8 mv_T^2##

Is there a different way of solving d?

There is more to it than what you have. The viscous force is responsible for ##\Delta E_{internal}##, but that does not account for all the work. The external force ##F## also does work. You can calculate that too*, but I would find what is asked directly using ##\Delta E_{internal}= \int_0^{t_{1/2}}{P dt},## where ##P=R~ v(t)=b~v^2(t)## is the rate (power) at which the internal energy is generated.

*On edit: I see that you tried to calculate the work done by ##F##. Your method is valid. You can do the integral, if you wish, and check for consistency.

Last edited:
lichenguy

## 1. What is dimensional analysis?

Dimensional analysis is a problem-solving technique used in science and engineering to convert units of measurement and check the consistency of equations. It involves manipulating units of measurement algebraically to arrive at a desired unit.

## 2. Why is dimensional analysis important?

Dimensional analysis is important because it allows scientists and engineers to convert between different units of measurement and check the consistency of equations. This helps to ensure accurate and precise calculations and prevents errors in experiments and engineering designs.

## 3. How do you perform dimensional analysis?

To perform dimensional analysis, you first need to identify the known and unknown units of measurement in the problem. Then, using conversion factors or unit relationships, you can manipulate the units algebraically until you arrive at the desired unit. Finally, you can perform the calculation using the converted units.

## 4. Can dimensional analysis be used in all scientific and engineering problems?

Yes, dimensional analysis can be used in most scientific and engineering problems that involve units of measurement. It is particularly useful in problems where conversions between different unit systems are required, or when checking the consistency of equations.

## 5. What are the benefits of using dimensional analysis?

The benefits of using dimensional analysis include increased accuracy and precision in calculations, the ability to convert between different units of measurement, and the ability to check the consistency of equations. It also helps to simplify complex problems by breaking them down into smaller, more manageable parts.

Replies
3
Views
1K
Replies
1
Views
162
Replies
3
Views
3K
Replies
7
Views
2K
Replies
1
Views
1K
Replies
33
Views
595
Replies
13
Views
4K
Replies
15
Views
2K
Replies
7
Views
891
Replies
7
Views
1K