Incorrect; Try Again; 3 Attempts Remaining Enter Your Answer Using Dimensions of Acceleration.

i Units and Measurement

1.iv Dimensional Assay

Learning Objectives

Past the end of this department, you lot will exist able to:

Find the dimensions of a mathematical expression involving physical quantities.
Determine whether an equation involving physical quantities is dimensionally consequent.

The dimension of any physical quantity expresses its dependence on the base of operations quantities equally a product of symbols (or powers of symbols) representing the base quantities. (Figure) lists the base of operations quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension 50 or L¹, a measurement of mass has dimension M or Grand^ane, and a measurement of time has dimension T or T¹. Like units, dimensions obey the rules of algebra. Thus, area is the production of two lengths and so has dimension L², or length squared. Similarly, book is the production of 3 lengths and has dimension L³, or length cubed. Speed has dimension length over time, L/T or LT^–i. Volumetric mass density has dimension M/L³ or ML^–3, or mass over length cubed. In general, the dimension of any physical quantity tin can be written equally

${\text{L}}^{a}{\text{M}}^{b}{\text{T}}^{c}{\text{I}}^{d}{\text{Θ}}^{e}{\text{N}}^{f}{\text{J}}^{g}$

for some powers

$a,b,c,d,e,f,$

and g. We can write the dimensions of a length in this form with

$a=1$

and the remaining six powers all set equal to zilch:

${\text{L}}^{1}={\text{L}}^{1}{\text{M}}^{0}{\text{T}}^{0}{\text{I}}^{0}{\text{Θ}}^{0}{\text{N}}^{0}{\text{J}}^{0}.$

Any quantity with a dimension that tin can exist written and so that all seven powers are zero (that is, its dimension is

${\text{L}}^{0}{\text{M}}^{0}{\text{T}}^{0}{\text{I}}^{0}{\text{Θ}}^{0}{\text{N}}^{0}{\text{J}}^{0}$

) is called dimensionless (or sometimes "of dimension ane," because anything raised to the zero power is i). Physicists oftentimes call dimensionless quantities pure numbers.

Base of operations Quantities and Their Dimensions
Base Quantity	Symbol for Dimension
Length	L
Mass	M
Time	T
Electric current	I
Thermodynamic temperature	Θ
Amount of substance	North
Luminous intensity	J

Physicists ofttimes utilize square brackets around the symbol for a physical quantity to correspond the dimensions of that quantity. For example, if

$r$

is the radius of a cylinder and

$h$

is its tiptop, then we write

$[r]=\text{L}$

and

$[h]=\text{L}$

to indicate the dimensions of the radius and peak are both those of length, or L. Similarly, if we employ the symbol

$A$

for the expanse of a cylinder and

$V$

for its volume, then [A] = Lⁱⁱ and [Five] = 50ⁱⁱⁱ. If nosotros use the symbol

$m$

for the mass of the cylinder and

$\rho$

for the density of the material from which the cylinder is made, then

$[m]=\text{M}$

and

$[\rho ]={\text{ML}}^{-3}.$

The importance of the concept of dimension arises from the fact that whatsoever mathematical equation relating physical quantities must be dimensionally consistent, which means the equation must obey the following rules:

Every term in an expression must have the same dimensions; information technology does non make sense to add together or subtract quantities of differing dimension (think of the onetime proverb: "You tin't add together apples and oranges"). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
The arguments of whatever of the standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions crave pure numbers as inputs and give pure numbers every bit outputs.

If either of these rules is violated, an equation is non dimensionally consistent and cannot peradventure be a right argument of physical law. This simple fact can be used to cheque for typos or algebra mistakes, to assist think the various laws of physics, and even to advise the form that new laws of physics might accept. This last utilize of dimensions is across the telescopic of this text, only is something you will undoubtedly acquire subsequently in your academic career.

Example

Using Dimensions to Remember an Equation

Suppose we need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our mind when we recall of circles:

$\pi {r}^{2}$

and

$2\pi r.$

I expression is the circumference of a circle of radius r and the other is its expanse. But which is which?

Strategy

I natural strategy is to look it up, merely this could take time to find information from a reputable source. Also, even if we call back the source is reputable, we shouldn't trust everything we read. Information technology is nice to have a way to double-check only by thinking about it. Too, we might be in a situation in which we cannot await things upwardly (such as during a test). Thus, the strategy is to find the dimensions of both expressions by making utilize of the fact that dimensions follow the rules of algebra. If either expression does not have the same dimensions as area, so information technology cannot perchance be the correct equation for the area of a circle.

Solution

We know the dimension of area is Fifty². At present, the dimension of the expression

$\pi {r}^{2}$

$[\pi {r}^{2}]=[\pi ]·{[r]}^{2}=1·{\text{L}}^{2}={\text{L}}^{2},$

since the abiding

$\pi$

is a pure number and the radius

$r$

is a length. Therefore,

$\pi {r}^{2}$

has the dimension of area. Similarly, the dimension of the expression

$2\pi r$

$[2\pi r]=[2]·[\pi ]·[r]=1·1·\text{L}=\text{L,}$

since the constants

$2$

and

$\pi$

are both dimensionless and the radius

$r$

is a length. Nosotros see that

$2\pi r$

has the dimension of length, which means it cannot possibly be an area.

We rule out

$2\pi r$

because it is not dimensionally consistent with being an area. We see that

$\pi {r}^{2}$

is dimensionally consistent with being an surface area, and so if we have to choose between these two expressions,

$\pi {r}^{2}$

is the one to choose.

Significance

This may seem similar kind of a silly example, simply the ideas are very general. As long every bit nosotros know the dimensions of the individual concrete quantities that appear in an equation, nosotros tin bank check to see whether the equation is dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this will not help united states remember dimensionless factors that appear in the equations (for example, if yous had accidentally conflated the 2 expressions from the example into

$2\pi {r}^{2},$

then dimensional analysis is no assist), just it does aid united states retrieve the correct bones form of equations.

Check Your Understanding

Suppose we want the formula for the book of a sphere. The 2 expressions commonly mentioned in simple discussions of spheres are

$4\pi {r}^{2}$

and

$4\pi {r}^{3}\text{/}3.$

One is the volume of a sphere of radius r and the other is its surface expanse. Which 1 is the volume?

[reveal-answer q="fs-id1168328152709″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1168328152709″]

$4\pi {r}^{3}\text{/}3$

[/hidden-answer]

Example

Checking Equations for Dimensional Consistency

Consider the physical quantities

$s,$

$v,$

$a,$

and

$t$

with dimensions

$[s]=\text{L},$

$[v]={\text{LT}}^{-1},$

$[a]={\text{LT}}^{-2},$

and

$[t]=\text{T}.$

Determine whether each of the following equations is dimensionally consistent: (a)

$s=vt+0.5a{t}^{2};$

(b)

$s=v{t}^{2}+0.5at;$

and (c)

$v=\text{sin}(a{t}^{2}\text{/}s).$

Strategy

By the definition of dimensional consistency, we need to bank check that each term in a given equation has the aforementioned dimensions as the other terms in that equation and that the arguments of any standard mathematical functions are dimensionless.

Solution

There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need but look at the dimensions of each term appearing in the equation. There are three terms, i in the left expression and ii in the expression on the right, so nosotros wait at each in turn:
$\begin{array}{l} [s]=\text{L} \\ {[vt]}=[v]\cdot[t]=\text{LT}^{-1}\cdot\text{T}=\text{LT}^0=\text{L} \\ {[0.5at^2]}=[a]\cdot[t]^2=\text{LT}^{-2}\cdot\text{T}^2=\text{LT}^0=\text{L}. \end{array}$

All three terms take the same dimension, and so this equation is dimensionally consistent.
Once again, at that place are no trigonometric, exponential, or logarithmic functions, and so we but need to look at the dimensions of each of the three terms actualization in the equation:
$\begin{array}{l} [s]=\text{L} \\ {[vt^2]}=[v]\cdot[t]^2=\text{LT}^{-1}\cdot\text{T}^2=\text{LT} \\ {[at]}=[a]\cdot[t]=\text{LT}^{-2}\cdot\text{T}=\text{LT}^{-1}. \end{array}$

None of the three terms has the aforementioned dimension as any other, so this is nigh equally far from beingness dimensionally consistent as you tin go. The technical term for an equation like this is nonsense.
This equation has a trigonometric function in it, so first we should bank check that the argument of the sine role is dimensionless:
$\begin{array}{c} \left[\dfrac{a{t}^{2}}{s}\right]=\dfrac{[a]\cdot{[t]}^{2}}{[s]}=\dfrac{{\text{LT}}^{-2}\cdot{\text{T}}^{2}}{\text{L}}=\dfrac{\text{L}}{\text{L}}=1.\hfill \end{array}$

The argument is dimensionless. So far, so good. At present we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:

$\begin{array}{c} [v]={\text{LT}}^{-1}\hfill \\ {\left[\text{sin}\left(\dfrac{a{t}^{2}}{s}\right)\right]}=1.\hfill \end{array}$

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another instance of "nonsense."

Significance

If nosotros are trusting people, these types of dimensional checks might seem unnecessary. Merely, rest assured, any textbook on a quantitative subject such equally physics (including this i) nigh certainly contains some equations with typos. Checking equations routinely past dimensional assay save us the embarrassment of using an wrong equation. As well, checking the dimensions of an equation nosotros obtain through algebraic manipulation is a great mode to make sure we did not make a mistake (or to spot a mistake, if we made 1).

Check Your Understanding

Is the equation v = at dimensionally consequent?

[reveal-answer q="fs-id1168328194212″]Show Solution[/reveal-respond]

[hidden-answer a="fs-id1168328194212″]

yes

[/hidden-respond]

One farther point that needs to be mentioned is the consequence of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, simply like units, but what happens when nosotros have the derivative of 1 physical quantity with respect to another or integrate a concrete quantity over another? The derivative of a role is just the slope of the line tangent to its graph and slopes are ratios, so for concrete quantities v and t, we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t:

$[\frac{dv}{dt}]=\frac{[v]}{[t]}.$

Similarly, since integrals are just sums of products, the dimension of the integral of five with respect to t is simply the dimension of five times the dimension of t:

$[\int vdt]=[v]·[t].$

By the same reasoning, analogous rules agree for the units of concrete quantities derived from other quantities by integration or differentiation.

Summary

The dimension of a physical quantity is just an expression of the base quantities from which information technology is derived.
All equations expressing concrete laws or principles must be dimensionally consistent. This fact can be used as an assistance in remembering physical laws, as a way to check whether claimed relationships betwixt physical quantities are possible, and even to derive new physical laws.

Issues

A student is trying to call back some formulas from geometry. In what follows, assume

$A$

is area,

$V$

is book, and all other variables are lengths. Decide which formulas are dimensionally consistent. (a)

$V=\pi {r}^{2}h;$

(b)

$A=2\pi {r}^{2}+2\pi rh;$

(c)

$V=0.5bh;$

(d)

$V=\pi {d}^{2};$

(e)

$V=\pi {d}^{3}\text{/}6.$

Consider the physical quantities southward, v, a, and t with dimensions

$[s]=\text{L},$

$[v]={\text{LT}}^{-1},$

$[a]={\text{LT}}^{-2},$

and

$[t]=\text{T}.$

Decide whether each of the following equations is dimensionally consistent. (a)

${v}^{2}=2as;$

(b)

$s=v{t}^{2}+0.5a{t}^{2};$

(c)

$v=s\text{/}t;$

(d)

$a=v\text{/}t.$

[reveal-respond q="fs-id1168328201713″]Prove Solution[/reveal-answer]

[hidden-answer a="fs-id1168328201713″]

a. Yep, both terms have dimension L²T^-2 b. No. c. Yep, both terms take dimension LT^-1 d. Yes, both terms take dimension LT^-2

[/hidden-answer]

Consider the physical quantities

$m,$

$s,$

$v,$

$a,$

and

$t$

with dimensions [m] = M, [south] = L, [v] = LT^–1, [a] = LT^–ii, and [t] = T. Bold each of the following equations is dimensionally consistent, find the dimension of the quantity on the left-hand side of the equation: (a) F = ma; (b) K = 0.fivemv ²; (c) p = mv; (d) W = mas; (e) L = mvr.

Suppose quantity

$s$

is a length and quantity

$t$

is a time. Suppose the quantities

$v$

and

$a$

are defined by v = ds/dt and a = dv/dt. (a) What is the dimension of five? (b) What is the dimension of the quantity a? What are the dimensions of (c)

$\int vdt,$

(d)

$\int adt,$

and (due east) da/dt?

[reveal-answer q="fs-id1168328204280″]Show Solution[/reveal-answer]

[hidden-respond a="fs-id1168328204280″]

a. [v] = LT^–ane; b. [a] = LT^–ii; c.

$[\int vdt]=\text{L;}$

$[\int adt]={\text{LT}}^{-1}\text{;}$

$[\frac{da}{dt}]={\text{LT}}^{-3}$

[/hidden-respond]

Suppose [V] = L³,

$[\rho ]={\text{ML}}^{-3},$

and [t] = T. (a) What is the dimension of

$\int \rho dV?$

(b) What is the dimension of dV/dt? (c) What is the dimension of

$\rho (dV\text{/}dt)?$

The arc length formula says the length

$s$

of arc subtended by bending

$Ɵ$

in a circle of radius

$r$

is given by the equation

$s=rƟ.$

What are the dimensions of (a) s, (b) r, and (c)

$\text{Ɵ?}$

[reveal-reply q="fs-id1168327948973″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1168327948973″]

a. Fifty; b. L; c. L⁰ = 1 (that is, it is dimensionless)

[/subconscious-answer]

Glossary

dimension

expression of the dependence of a physical quantity on the base of operations quantities as a production of powers of symbols representing the base quantities; in general, the dimension of a quantity has the form

${\text{L}}^{\text{a}}{\text{M}}^{\text{b}}{\text{T}}^{\text{c}}{\text{I}}^{\text{d}}{\text{Θ}}^{\text{e}}{\text{N}}^{\text{f}}{\text{J}}^{\text{g}}$

for some powers a, b, c, d, due east, f, and k.

dimensionally consistent: equation in which every term has the same dimensions and the arguments of any mathematical functions appearing in the equation are dimensionless

dimensionless

quantity with a dimension of

${\text{L}}^{0}{\text{M}}^{0}{\text{T}}^{0}{\text{I}}^{0}{\text{Θ}}^{0}{\text{N}}^{0}{\text{J}}^{0}=1;$

also chosen quantity of dimension ane or a pure number

kinglielf1940.blogspot.com

Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/1-4-dimensional-analysis/

Incorrect; Try Again; 3 Attempts Remaining Enter Your Answer Using Dimensions of Acceleration.

1.iv Dimensional Assay

Learning Objectives

Example

Using Dimensions to Remember an Equation

Strategy

Solution

Significance

Check Your Understanding

Example

Checking Equations for Dimensional Consistency

Strategy

Solution

Significance

Check Your Understanding

Summary

Issues

Glossary

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