Dimensional analysis problem

  • #1
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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".
 

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Answers and Replies

  • #2
kuruman
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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##?"
 
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  • #3
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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.
 
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  • #4
haruspex
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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.)
 
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  • #5
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So, ##R## is not an independent variable and that is why we can't use it?
Using ##b## instead yields: ##v_T=F/b##
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?
 
  • #6
haruspex
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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.
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.
 
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  • #7
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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?
 
  • #8
haruspex
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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.
 
  • #9
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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?
 
  • #10
kuruman
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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.
 
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