Dimensional Analysis - Comparison of exponents

In summary: M*L)/(C*M*L) has dimensions of velocity (v).In summary, the paper introduces classical mechanics concerning one dimension. There is a section which is not well explained, and I hope someone can explain it to me. Dimensionally, the falling tower has mass (M), length (L), and acceleration due to gravity (g). To slow the falling tower down to look more realistic, the film company needs to multiply t by the speed of fall (v).
  • #1
ajassat
55
0
I have been reading a paper which introduced classical mechanics concerning only one dimension. Before we get into anything about displacement and velocity, I read about dimensional analysis.

There is a section of the paper which isn't very well explained and I hope someone can explain it to me, so that I can continue reading.

"A film company wants to shoot the leaning tower of pisa falling. They have built a model tower. They cannot just let it drop otherwise it would look too unrealistic. Therefore they need to slow the video down so it looks more realistic by a factor t. This factor depends on mass of tower,m, the height of the tower,h, and acceleration due to gravity, g. In other words...

t = C m^x h^y g^z
Where C is a dimensionless constant.

The dimensions =

[T] = [M]^x [L]^y ([L]/[T-squared])^z

We can now compare the exponents of [L], [M] and [T].

0 = y + z
0 = x
1 = -2z


While I understand what the author is attempting to do, I do not understand how we gain the above equations, and hence the following.

x = 0
y = 0.5
z = -0.5

t = C (sqrt) h/g

Thank you in advance.

ps - I want to understand fully how we compare the exponents, get three equations and the formula.
 
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  • #2
I think what it's explaining is that the factor t is multiplied by the speed of fall, to slow it down.

M=mass
L=lenght
T=time

Therefore your end units of t*v (where v is you speed) should be units of v. Speed is distance divided by time, so it's dimensional units are L1 T-1. I assume that you understand that this is the same as writing Lenght / Time ;).

Hence:

t*v=v

so;

C Mx Ly gz * L1 T-1 = L1 T-1

Now g is not in fundamental units it's Newtons per Kilogram, but Newtons are Kilograms-Meters per second per second. So;

g => M1 L1 T-2 / M1

g => M1 L1 T-2 M-1

So

g = L1 T-2

Ok knowing that;C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

We can ignore C because it's not got a dimension and consider this as a set of equations. So basically we are going to count up the powers for each of the dimensions on each side of the equation. So for M, then L, then T; we have;C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

(1) x = 0
(2) y+z+1 = 1
(3) -2z+-1=-1

Equation 1, tells us that this factor t doesn't depend on the mass at all. Which should make sense as a tonne of feathers and a ton of iron fall at the same rate. 2 and 3 can be rewritten;

(2) y+z=0
(3) -2z=0

Errm different to what the last one is, but let's continue anyhow. These are a set of simultaneous equations, these we can solve as -2z=0, implies that z needs to be 0, but that would imply that y also equals zero, so somewhere in my adding up of the powers I've gone wrong. Ok using;

(3) -2z=1

We can see that z must be -1/2 because -2*-1/2 = 1 As we know y+z=0 and y+-1/2=0 so y=1/2

Going back to;

C Mx Ly (L1 T-2)z

We put our values of x,y and z into the powers to find out what our constant is. Like;

C M0 L1/2 (L1 T-2)-1/2

Hence;

t = C sqrt{h/g}

As the power of 1/2 is the same as a square root.

As for the problem with equation 3, it might be because t is not a scale factor for the speed, but rather the acceleration of the tower. So it would actually be;

t*a=v

C Mx Ly (L1 T-2)z * L1 T-2 = L1 T-1

Which would give; -2z-2=-1 which is -2z=1 so it's a scale factor for the acceleration of the tower. So the formula for the velocity of the tower would be;

C sqrt{h/g}*a = v

I think...
Haths
 
  • #3
Here's a different take on the problem statement, which I think could be worded better.

To my reading, it sounds like we want to make the tower take time t to fall (note that it is given dimensions of time (T) later on). That is, t itself is not the scale factor, but the desired scaled time.

Per the problem, we suppose it can depend in as-yet-unknown ways on the mass of the tower m, its height h, g, and an arbitrary dimensionless constant C. Turning that last sentence into an equation just gives
t = C mx hy gz,​
where x, y and z are constant exponents we want to determine.

Dimensional analysis makes this easy: t has dimensions of time (T), so we demand that C mx hy gz have dimensions of time as well. Since m has dimensions of mass (M), h has dimensions of length (L), g has dimensions of length per time-squared (L/T2 = LT-2), and C has no dimensions at all, we have
T = Mx Ly (LT-2)z = Mx L(y + z)T(-2z).​

Now all we have to do is match the exponent of each dimension on each side of the equation. Writing everything out explicitly,
M0 L0 T1 = Mx L(y + z)T(-2z).​
Separately matching the exponents of M, L, and T on each side of the equation immediately gives the three equations
0 = x
0 = y + z
1 = -2z,​
which have the solution
x = 0
z = -1 / 2
y = 1 / 2.​
Plugging x, y and z back into the original formula, we have
t = C m0 h1/2 g-1/2 = C(h / g)1/2,
[tex]t = C\sqrt{h / g}.[/tex]​
And that's all there is to it.
 
  • #4
Excellent. Both of your replies helped me to understand things better. Any confusions/doubts were gotten rid off when I read daschaich's reply.

Thank you.
 

1. What is dimensional analysis?

Dimensional analysis is a mathematical technique used to compare physical quantities and their units. It involves converting between different units of measurement using conversion factors based on the relationships between the units.

2. How is dimensional analysis used in science?

Dimensional analysis is commonly used in science to check the accuracy of equations and calculations, to convert between units of measurement, and to solve problems involving different units and dimensions.

3. What is the purpose of comparing exponents in dimensional analysis?

Comparing exponents in dimensional analysis allows for the cancellation of units and simplification of equations. It also ensures that the final result has the correct units and dimensions.

4. Can dimensional analysis be used in any field of science?

Yes, dimensional analysis can be used in any field of science, including physics, chemistry, biology, and engineering. It is a universal method for comparing and converting units and dimensions.

5. Are there any limitations to dimensional analysis?

While dimensional analysis is a powerful tool, it does have some limitations. It cannot account for numerical constants or other factors that may affect the result of a calculation. It also assumes that the equation being used is correct and that the units are being used appropriately.

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