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Chapter01 - Chapter 1 Dimensional Analysis Kinematics 1.1...

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Chapter 1 Dimensional Analysis, Kinematics 1.1 Dimensional Analysis 1.2 Vectors 1.3 Relative motion in one dimension 1.4 Relative motion in two dimensions 1.5 Motion with constant acceleration a (1-D) 1.6 Projectile 1.7 Range on an inclined plane 1.1 Dimensional Analysis The fundamental quantities used in physical descriptions are called dimensions. Length, mass, and time are examples of dimensions. You could measure the distance between two points and express it in units of meters, centimeters, or feet. In any case, the quantity would have the dimension of length. It is common to express dimensional quantities by bracketed symbols, such as [L], [M], and [T] for length, mass, and time, respectively. Dimensional analysis is a procedure by which the dimensional consistency of any equation may be checked. If I say, x = at , where x , a , and t represent the displacement, acceleration and time respectively. Is it correct? [ x ] = [L] [ at ] = ( [L][T] - 2 ) ( [T] ) = [L][T] - 1 So the dimension for the sides are not equal, i.e. x = at is invalid. Now look at the equation x = at 2 The L.H.S. is x , i.e. [L] The R.H.S. is at 2 , i.e. ( [L][T] -2 ) ( [T] 2 ) = [L], Hence the equation is dimensionally correct, but you should know that it is physically incorrect. The correct equation is, as you should remember, 2 2 1 at x = . The fraction 2 1 is a constant and has no dimension, just like π . 1
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Example 1-1 A planet moving in a circular orbit experiences acceleration. This acceleration depends only on the speed v of the planet and on the radius r of its orbit. How can we use dimensional analysis to determine the acceleration, a , and relate it to v and r ? Answer: We assume that the acceleration is proportional to some power of the speed, e.g. a v p . Similarly, we assume that the acceleration is proportional to some power of the orbit radius, e.g. a r q . Our assumed form for the acceleration can be expressed by a = v p r q . The exponents p and q are unknown numbers that we are going to find out by using dimensional analysis. [ a ]=[L][T] - 2 [ v ]=[L][T] - 1 [ r ]=[L] Hence the L.H.S. of the assumed form is [L][T] - 2 , and that for the R.H.S. is [L] p+q [T] - p . Equate the index of [L] and [T], we have two equations - 2 = - p , 1 = p + q . After solving, we obtain p = 2 and q = - 1. Accordingly, we can write a dimensionally correct relation r v a 2 = . Example 1-2 The period P of a simple pendulum is the time for one complete swing. How does P depend on the mass m of the bob, the length l of the string, and the acceleration due to gravity g ? Answer: We begin by expressing the period P in terms of the other quantities as follows: P = k m x l y g z where k is a constant and x , y , and z are to be determined. Next we insert the dimensions of each quantity: P = ( [M] x ) ( [L] y ) ( [L] z [T] - 2 z ) = ( [M] x ) ( [L] y+ z ) ( [T] - 2 z ) 2
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And equate the powers of each dimension of either side of the equation. Thus, [T]: 1 = - 2 z [M]: 0 = x [L]: 0 = y + z These equations are easily solved and yield x = 0, z = - 1/2, and y = 1/2. Thus, g l k P = .
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