# Trouble with Dimensional Analysis

1. Aug 24, 2013

### mileena

1. The problem statement, all variables and given/known data

Dimensional Analysis:

The following equation was given by a student during an examination:

(1/2)(mv2) = (1/2)(mv02) + √(mgh)

Do a dimensional analysis and explain why the equation can't be correct.

2. Relevant equations

m = grams
v = m/s2

3. The attempt at a solution

What is "g"?? Gravitational acceleration in m/s2?? That hasn't been introduced to us yet.

What is "h"? Plank's constant? That hasn't been introduced to us yet either!

Solution:

(1/2)(mv2) = (1/2)(mv02) + √(mgh)

(1/2)(g)(m/s2) ?=? (1/2)(g)(m/s2) + √[(g)(m/s2)(constant)]

(1/2)(mv2) ≠ (1/2)(mv02) + √(mgh)
unless m, g, or h = 0

Thanks!

2. Aug 24, 2013

### SteamKing

Staff Emeritus
'mgh' is the potential energy of mass 'm' raised a distance 'h' above a datum where the acceleration due to gravity is equal to 'g'. Planck is well down the road from your location yet.

(1/2)mv^2 is the kinetic energy of a mass 'm' traveling at a velocity 'v'.
(1/2)mvo^2 is the kinetic energy of a mass 'm' traveling ata velocity of 'vo'.

3. Aug 24, 2013

### rock.freak667

g is in m/s^2 as you stated and h would be height in metres.

m is in mass in which you should have m in kg.

v is velocity so it would be m/s.

For your equation to be correct, the units on the left must be same as the units on the right. What would the units on the left be using the corrected units I stated above?

What are the units for the two terms on the right?

4. Aug 24, 2013

### Emilyjoint

Strictly speaking dimensional analysis does not involve units directly. Dimensions are mass, M, length,L and time, T. So your quantities such as KE = 1/2 m v2 should appear as dimensions M L2T-2. Constants such as the 1/2 do not affect the logic.
The value of dimensions is that such things as speed and velocity are always expressed as LT-1 regardless of the units (m/s, mph, km/hr etc)
You then do the algebra and check that your equation is dimensionally (not necessarily units) correct.

5. Aug 24, 2013

### mileena

Hi,

First, thanks Emilyjoint! I went back and looked at my notes, and you are right. You must use base dimensions (L, M, T) to prove derived dimensions (e.g., [a] = acceleration, [v] = velocity, [A] = area) work in an equation!

Thanks also rock.freak667. You are right: I should have put down "kg" if I was going to use units, instead of "g". I got mixed up with the Gaussian system (or cgs).

And to SteamKing, thanks also. I got confused about the "h" for "height", as we learned in the book that L (length) is represented by [x] for distance.

Here is my revised equation:

(1/2)(mv2) = (1/2)(mv02) + √(mgh)

(1/2)(M)(L/T)2 = (1/2)(M)(L/T)02 + √[(M)(L/T2)(L)]

(1/2)(ML2)/T2 = (1/2)(ML02)/T02 + √(ML2/T2)

So, assuming L/T = (L/T)0, unless √(ML2/T2) = 0, the equation does not hold.

What I don't understand is why the text used "m" for mass when it is a base dimension, and thus should be "M" (capitalized).

Last edited: Aug 24, 2013
6. Aug 24, 2013

### mileena

Let me go further in my above post:

(1/2)(ML2)/T2 = (1/2)(ML02)/T02 + √(ML2/T2)

(1/2)(ML2)/T2 = (1/2)(ML02)/T02 + [(√M)(L)]/T

[STRIKE]So, assuming L/T = (L/T)0[/STRIKE], unless √(ML2/T2) = 0, the equation does not hold.

Replace with:

unless [(√M)(L)]/T = 0, the equation does not hold.

(Whether L/T = (L/T)0 is irrelevant, since we are not trying to prove the equation is numerically correct, but just dimensionally correct. So the dimensions in L/T = (L/T)0 do work.)

So I hope the above is correct.

7. Aug 24, 2013

### rock.freak667

Your left side has dimensions (you can leave out the 1/2 as it is unitless)

ML^2/T^2

the right side has two terms, for them to add up to give the same units as the left side, they both need to have the same dimensions (it's like how I cannot add apples and oranges). So just based on the fact that the dimensions of √(mgh) is not the same as the others is enough. You don't have to put that it should equal to zero for the equation to be true.

8. Aug 24, 2013

### SteamKing

Staff Emeritus
The problem presumably is asking you to determine if the first equation is dimensionally consistent. The terms in this equation are not expressed in terms of base units; 'm' is used frequently as a variable indicating the mass of an object. 'M' is used to indicate the base unit of mass, not the amount of the mass itself; you should not confuse the two.

9. Aug 25, 2013

### mileena

Ok, got it, and sorry for my late reply. I guess I was just going too far, in specifiying √(mgh) might be zero.

I dd not realize this. Thank you!

So both "M" and "m", for example, represent "mass", but one is a base dimension used to represent mass, and the other is a variable which could be replaced with a number, which represent the mass itself.

10. Aug 25, 2013

### SteamKing

Staff Emeritus
Correct.