Phys01 - Unit 1 Kinematics 1.1 1.2 1.3 1.4 1.5 1.6...

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1 Unit 1 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 (1-D) 1.6 Projectile 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 . Example 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 ?
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2 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 2z ) P = ( M x ) ( L y+ z ) ( T 2z ) 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 = 2 1 , and y = 2 1 . Thus, l Pk g = . 1.2 Vectors (a) Definition of a vector a A ˆ = A A : magnitude ± a : unit vector with direction pointed (b) Cartesian coordinate system 2-D (i.e. a plane) A =+ Ai A j xy ±± A x : Component of A in x -direction A y : Component of A in y -direction The magnitude of A is given by 2 2 y x A A A A + = = . Note that we have the components of A θ cos A A x = and sin A A y = O a ˆ A G A G origin y x ) , ( y x A A a ˆ A G y A x A j ˆ i ˆ Unit vectors O
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3 3-D A =++ Ai A j Ak xyz ±± ± B Bi B j Bk ± k B A j B A i B A z z y y x x ˆ ) ( ˆ ) ( ˆ ) ( + + + + + = + B A (c) Mathematics of vectors C = A + B D = A B (d) Examples of vectors (1) Position vector r xi yj zk ± (2) Displacement (change in position) rr r r r =−= 21 2 1 () () tt (3) Average velocity during time t where tt t =− Velocity : A vector ! Note the difference of velocity and speed: speed is a scalar! v r rr av t t t == +− () ( ) 11 x z y A G x A y A z A k ˆ i ˆ j ˆ z y x 2 r G 1 r G 2 t t = 1 t t = r G Trajectory of a particle y x A G O C G B G y x A G O D G B G
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4 (4) Instantaneous velocity 0 ˆ ˆˆ lim x yz t vi v j vk t ∆→ ≡= + + r v 0 lim t d td t = rr v= first order derivative (5) Acceleration a v vv
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This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.

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Phys01 - Unit 1 Kinematics 1.1 1.2 1.3 1.4 1.5 1.6...

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