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Physics LibreTexts

2.3: Dimensions of Commonly Encountered Quantities

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Many physical quantities are derived from the base quantities by a set of algebraic relations defining the physical relation between these quantities. The dimension of the derived quantity is written as a power of the dimensions of the base quantities. For example velocity is a derived quantity and the dimension is given by the relationship  dim velocity =( length )/( time )=LT1where L length ,T time . Force is also a derived quantity and has dimensiondim force=( mass )( dim velocity )( time )where Mmass. We can also express force in terms of mass, length, and time by the relationshipdim force =( mass )( length )( time )2=MLT2The derived dimension of kinetic energy isdim kinetic energy=( mass )(dim velocity )2,which in terms of mass, length, and time is dim kinetic energy =( mass )( length )2( time )2=ML2T2The derived dimension of work isdim work=( dim force )( length ),which in terms of our fundamental dimensions is dim work=( mass )( length )2( time )2=ML2T2So work and kinetic energy have the same dimensions. Power is defined to be the rate of change in time of work so the dimensions are dim power = dim work  time = (dim force) (length)  time =( mass )( length )2( time )3=ML2T3In Table 2.3 we include the derived dimensions of some common mechanical quantities in terms of mass, length, and time.

Dimensional Analysis

There are many phenomena in nature that can be explained by simple relationships between the observed phenomena.

Table 2.3 Dimensions of Some Common Mechanical Quantities M mass ,L length,T time 

Quantity

Dimension

MKS unit

Angle

dimensionless

Dimensionless = radian

Solid Angle

dimensionless

Dimensionless = sterradian

Area

L2 m2

Volume

L3 m3

Frequency

T1

s1=hertz=Hz

Velocity

LT1 ms1

Acceleration

LT2 ms2

Angular Velocity

T1 rads1

Angular Acceleration

T2 rads2

Density

ML3 kgm3

Momentum

MLT1 kgms1

Angular Momentum

ML2T1 kgm2s1

Force

MLT2 kgms2=newton =N

Work, Energy

ML2T2 kgm2s2=joule=J

Torque

ML2T2 kgm2s2

Power

ML2T3 kgm2s3=watt=W

Pressure

ML1T2 kgm1s2=pascal=Pa

Example 2.5 Period of a Pendulum

Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time interval (period of the pendulum) that it takes the bob to complete one cycle of oscillation. How does the period of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion?

Solution

What possible quantities are involved? The length of the pendulum l, the mass of the pendulum bob \{\text{m}\), the gravitational acceleration g, and the angular amplitude of the bob θ0 are all possible quantities that may enter into a relationship for the period of the swing. Have we included every possible quantity? We can never be sure but let’s first work with this set and if we need more than we will have to think harder! Our problem is then to find a function f such thatT=f(l,m,g,θ0)We first make a list of the dimensions of our quantities as shown in Table 2.4.

Name of Quantity

Symbol

Dimensional Formula

Time of swing

t T

Length of pendulum

l L

Mass of pendulum

m M

Gravitational acceleration

g LT2

Angular amplitude of swing

θ0

No dimension

Our first observation is that the mass of the bob cannot enter into our relationship, as our final quantity has no dimensions of mass and no other quantity has dimensions of mass. Let’s focus on the length of the string and the gravitational acceleration. In order to eliminate length, these quantities must divide each other when appearing in some functional relation for the period T If we choose the combination l/g , the dimensions are

dim[l/g]= length  length / (time) 2=( time )2

It appears that the time of swing may proportional to the square root of this ratio. Thus we have a candidate formula

T(lg)1/2

in the above expression, the symbol “” represents a proportionality, not an approximation). Because the angular amplitude θ0 is dimensionless, it may or may not appear. We can account for this by introducing some function y(θ0) into our relationship, which is beyond the limits of this type of analysis. The period is then

T=y(θ0)(lg)1/2

We shall discover later on that y(θ0) is nearly independent of the angular amplitude θ0 for very small amplitudes and is equal to y(θ0)=2π,

T=2π(lg)1/2


This page titled 2.3: Dimensions of Commonly Encountered Quantities is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

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