1.2 Dimensions of Commonly Encountered Quantities

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Supplementary Notes 1: Dimensional Analysis 1.1 International System of System of Units There are only four fundamental quantities (measurements) necessary to specify all physical phenomena: length, time, mass and charge. All other quantities are expressible in terms of these, constructed as a matter of convenience. The basic system of units used throughout science and technology today is the internationally accepted Système International (SI) (Table 1). It consists of four base quantities and their corresponding base units: length (meter), mass (kilogram), time (second), and electric current (ampere). The unit for electric charge, the coulomb, is defined in terms of the ampere, and hence is referred to as a derived unit. In addition, three other quantities, temperature, amount of substance, and luminous intensity are part of the SI base quantities with corresponding units shown in Table 1. Mechanics is based on just the first three of these quantities, the MKS or meterkilogram-second system. An alternative metric system to this, still widely used, is the socalled CGS system (centimeter-gram-second). For distance and time measurements, British Imperial units (especially in the USA) based on the foot (ft), the yard (yd), the mile (mi), etc., as units of length, and also the minute, hour, day and year as units of time. Table 1 Système International (SI) System of Units Base Quantity Length Mass Time Electric Current Temperature Amount of Substance Luminous Intensity Base Unit meter (m) kilogram (kg) second (s) ampere (A) Kelvin (K) mole (mol) candela (cd) We shall refer to the dimension of the base quantity by the quantity itself, for example dim length [length] = L, dim mass [mass] M, dim time [time] T. (1) 1.2 Dimensions of Commonly Encountered Quantities 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 1

2 derived quantity can always be written as a product of the powers of the dimensions of the base quantities. Example 1 Derived Dimensions of Mechanical Quantities (i) The dimensions of velocity are given by the relationship [velocity] = [length]/[time] = L T -1. (2) (ii) Force is also a derived quantity and using the definition of force F = m a and acceleration a = d v / dt, force has dimensions [force] = [mass][velocity] [time]. (3) We could express force in terms of mass, length, and time by the relationship [force] = [mass][length] [time] 2 = M L T -2. (4) (iii) The derived dimension of kinetic energy follows from the definition that K = 1 2 mv2, thus which in terms of mass, length, and time is [kineticenergy] = [mass][velocity] 2, (5) [kineticenergy] = [mass][length]2 [time] 2 = M L 2 T -2 (6) (iv) The derived dimension of work is which in terms of our fundamental dimensions is [work] = [force][length], (7) [work] = [mass][length]2 [time] 2 = M L 2 T -2 (8) So work and kinetic energy have the same dimensions. 2

3 (v) Power is defined to be the rate of change in time of work so the dimensions are [power] = [work] [time] = [force][length] = [mass][length]2 = M L 2 T -3 [time] [time] 3 (9) In Table 2 we list the derived dimensions of some common mechanical quantities in terms of mass, length, and time. Table 2 Dimensions of Some Common Mechanical Quantities M mass, L length, T time Quantity Dimension MKS unit Angle dimensionless 1 Dimensionless = radian Steradian dimensionless Dimensionless = radian 2 Area L 2 m 2 Volume L 3 m 3 Frequency T -1 s 1 = hertz = Hz Velocity L T -1 m s 1 Acceleration L T -2 m s 2 Angular Velocity T -1 rad s 1 Angular Acceleration T -2 rad s 2 Density M L -3 kg m 3 Momentum M L T -1 kg m s 1 Angular Momentum M L 2 T -1 kg m 2 s 1 Force M L T -2 kg m s 2 = newton = N Work, Energy M L 2 T -2 kg m 2 s 2 = joule=j Torque M L 2 T -2 kg m 2 s 2 Power M L 2 T -3 kg m 2 s 3 =watt=w Pressure M L -1 T -2 kg m 1 s 2 = pascal= Pa 1 Even though angle and steradian are dimensionless quantities, it is often helpful to carry around a unit associated with them, like the radian, to understand their role in an expression or to determine if a result makes sense. 3

4 1.3 Dimensional Analysis There are many phenomena in nature that can be explained by simple relationships between the observed phenomena. When trying to find a dimensional correct formula for a quantity from a set of given quantities, an answer that is dimensionally correct will scale properly and is generally off by a constant of order unity. Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time (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? What possible quantities are involved? The length of the pendulum l, the mass of the pendulum bob m, the gravitational acceleration g, and the initial angular amplitude of the bob 0 are all possible quantities that may enter into the formula 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 that T = f ( l,m,g, 0 ) (10) We first make a list of the dimensions of our quantities as shown in Table 3. Choose the set: mass, length, and time, to use as the base dimensions. Table 3 Dimensions of quantities that may describe the period of pendulum Name of Quantity Symbol Dimensional Formula Time of swing t T Length of pendulum l L Mass of pendulum m M Gravitational acceleration g L T -2 Angular amplitude of swing 0 No dimension We begin by writing the period as a product of these given quantities that have dimensions, (thus the initial angular amplitude of swing cannot enter into our expression), with each given quantity raised to a rational power, T = bl X m Y g Z (11) where b is a dimensionless constant. Our first observation is that since the period has only dimensions of time, the mass of the bob cannot enter into our relationship since neither length of the gravitational acceleration can remove the dimension of the pendulum mass. Therefore the power Y = 0, hence 4

5 T = bl X g Z (12) A dimensional analysis of Eq. (12) yields [T ] = [L] X [L T 2 ] Z = [L] X 2Z [T ] 2Z (13) Let s focus on the length. In order to eliminate length, we see that the condition 0 = X + Z (14) must be satisfied in Eq.(13). In a similar fashion, when we consider time, Eq.(13) requires that 1 = 2Z. (15) It is a rather straightforward exercise to solve these two equations to find that X = Z = 1/2 (16) Thus a dimensionally correct formula for the period of the pendulum is T = b l / g. (17) Since the angular amplitude 0 is dimensionless, it may or may not appear. We can account for this by introducing some function y( 0 ) into our relationship, which is beyond the limits of this type of analysis. Then the period for a complete swing is T = y( 0 ) l / g (18) We shall discover later on by solving the problem exactly that y( 0 ) is independent of the angular amplitude 0 for very small amplitudes and is equal to y( 0 ) = 2, Example 2 Work and Kinetic Energy T = 2 l / g (19) A constant force of magnitude F acts over a distance d on an object of mass m. Find the final speed v of the body. 5

6 Solution: Here the list of quantities that might be involved is given for us. We write the speed as v = bf X d Y m Z (20) where X, Y, and Z are rational numbers. A dimensional analysis yields [L T 1 ] = [M L T 2 ] X [L] Y [M] Z (21) Therefore we have the following set of algebraic relations Solving this set of equations we have that So the final speed is 1 = X + Y (22) 1= 2X (23) 0 = X + Z (24) X = 1/2, Z = 1/2, X = 1/2, Y = 1/2. (25) v= b Fd/ m. (26) Later on that using the work kinetic energy theorem we will discover that the constant b = 2, and therefore v = 2Fd / m. (27) 6