Chapter 3: 2D Kinematics Tuesday January 20th Chapter 3: Vectors Review: Properties of vectors Review: Unit vectors Position and displacement Velocity and acceleration vectors Relative motion Constant acceleration in 2D and 3D Projectile motion Demonstrations and examples Reading: up to page 41 in the text book (Ch. 3)
Review: Components of vectors Resolving vector components a x = acosθ a y = a sinθ The inverse process a= a + a 2 2 x y tanθ = a a y x
y Review: Unit vectors z a y ĵ x î, ĵ and ˆk are unit vectors They have length equal to unity (1), and point respectively along the x, y and z axes of a right handed Cartesian coordinate system. a = a î + a ĵ x y a x î
y Review: Unit vectors z x î, ĵ and ˆk are unit vectors They have length equal to unity (1), and point respectively along the x, y and z axes of a right handed Cartesian coordinate system. a y ĵ a x î a = acosθî + asinθ ĵ Note: θ is usually measured from x to y (in a righthanded sense around the z- axis)
Adding vectors by components Consider two vectors: r = x î + y ĵ + z ˆk 1 1 1 1 & r = x î + y ĵ + z ˆk 2 2 2 2 Then r r = (x x )î + ( y y ) ĵ + (z z ) ˆk 2 1 2 1 2 1 2 1 & r + r = (x + x )î + ( y + y ) ĵ + (z + z ) ˆk 1 2 2 1 2 1 2 1
Example: Compute a + b and a b y a b θ a θ b x a = 3î + 4 ĵ b = 8î + 6 ĵ a = 5 m, b = 10 m θ a = 53.13 deg θ b = 36.87 deg
Multiplying vectors (very brief) Multiplying a vector by a scalar: This operation simply changes the length of the vector, not the orientation. a a 4 a = 3 a = a a a a a a
The scalar product, or dot product a b = abcosφ (a)(bcosφ) = (acosφ)(b) cosφ = cos( φ) a b = b a The scalar product represents the product of the magnitude of one vector and the component of the second vector along the direction of the first If φ = 0 o, then a b = ab If φ = 90 o, then a b = 0
The scalar product, or dot product a b = abcosφ (a)(bcosφ) = (acosφ)(b) cosφ = cos( φ) a b = b a The scalar product becomes relevant in Chapter 6 when considering work and power. There is also a vector product, or cross product, which becomes relevant in Chapter 11. I save discussion of this until later in the semester see also Appendix to this lecture.
Position and displacement Position vectors represent coordinates. The displacement between two position vectors is given as follows: Δ r = r r 1 2 2 1 r 1 r 2 O
Average and instantaneous velocity average velocity = displacement time interval = r 2 r 1 Δt = Δ r Δt Instantaneous velocity: Or: v = d r dt = dx dt î + dy dt ĵ + dz dt ˆk v x = dx dt ; v y = dy dt ; v z = dz dt
Average and instantaneous acceleration a avg = change in velocity time interval Instantaneous acceleration: = v 2 v 1 Δt = Δ v Δt Or: a = d v dt = dv x dt î + dv y dt ĵ + dv z dt ˆk a x = dv x dt ; a y = dv y dt ; a z = dv z dt
A bit like displacement Relative Velocity v 2 v 2,1 rel v 1 O Velocity of 2 relative to 1: v 2,1 = v v rel 2 1
A bit like displacement Relative Velocity v 2 v 2,1 rel v 1 O Velocity of 2 relative to 1: v 2,1 = v v rel 2 1
Relative Motion in Different Reference Frames Two observers O and O y ball y v b v ' b v O' O x O Velocity of ball observed in O: x v = v ' b b + v O'
Projectile motion This series of photographic images illustrates the fact that vertical motion is unaffected by horizontal motion, i.e., the two balls accelerate downwards at the same constant rate, irrespective of their horizontal component of motion. In all of the projectile motion problems that we will consider, we shall assume that the only acceleration is due to gravity (a=-g) which acts in the y direction.
Displacement The equations of motion that we introduced in chapter 2 apply equally well in two and three-dimensional motion. All we need to do is to break the motion up into components, and treat each component independently. Displacement: Δ r = ( x x )î + y 1 2 2 1 2 y 1 ( ) ĵ + ( z 2 z ) ˆk 1 Then we expect to have a different set of kinematic equations for each vector component.
Equations of motion for constant acceleration Equation number 3.8 3.9 Equation v = v + at 0 r r = v t + 1 0 0 2 at 2 Missing quantity ( r r 0 ) v v 2 = v 0 2 + 2 a ( r r 0 ) ( r r ) = 1 0 2 ( v + v)t 0 ( r r ) = vt 1 at 2 0 2 Important: equations apply ONLY if acceleration is constant. t a v 0
Equations of motion for constant acceleration These equations work the same in any direction, e.g., along x, y or z. Equation number 2.7 Equation v x = v 0x + a x t Missing quantity x x 0 2.10 2.11 2.9 x x 0 = v 0x t + 1 2 a x t 2 v 2 = v 2 + 2a (x x ) x 0x x 0 x x 0 = 1 2 (v 0x + v x )t v x t a x x x 0 = v x t 1 2 a x t 2 Important: equations apply ONLY if acceleration is constant. v 0
Equations of motion for constant acceleration These equations work the same in any direction, e.g., along x, y or z. Equation number 2.7 2.10 2.11 2.9 Equation v = v + a t y 0 y y 1 0 0y 2 y y = v t+ a t 2 = 2 + y 0y y 0 v v 2 a ( y y ) 1 0 0y y y = 2 ( v + vy ) t 1 0 y 2 y y = v t a t Important: equations apply ONLY if acceleration is constant. y y 2 2 Missing quantity y y 0 v y t a y v 0 y
Equations of motion for constant acceleration Special case of free-fall under gravity, a y = -g. g = 9.81 m/s 2 here at the surface of the earth. Equation Missing number Equation quantity v y = v 0 y gt y y 0 = v 0 y t 1 2 gt 2 y y 0 v y v 2 = v 2 2g( y y ) y 0 y 0 1 0 0y y y = 2 ( v + vy ) t y y 0 = v y t + 1 2 gt 2 t a y v 0 y
Demonstration
Appendices
The scalar product in component form a b = a x î + a y ĵ+ a z ˆk ( ) b x î + b y ĵ+ b z ˆk ( ) a b = a x b x + a y b y + a z b z Because: î î = ĵ ĵ = ˆk ˆk = 1 î ĵ = ĵ ˆk = ˆk î = 0 This is the property of orthogonality
The vector product, or cross product a b = c, where c = absinφ a b = b a ( ) Directionof c to both a and b î î = ĵ ĵ = ˆk ˆk = 0 î ĵ = ˆk ĵ ˆk = î ˆk î = ĵ ĵ î = ˆk ˆk ĵ = î î ˆk = ĵ
a b = a x î + a y ĵ + a z ˆk ( ) b x î + b y ĵ + b z ˆk ( ) î î ˆk +ve ĵj ˆk -ve ĵj a x î b y ĵ = a x b y ( î ĵ) = a x b y ˆk a b = ( a y b z b y a z )î + ( a b b a ) ĵ + ( a b a b ) ˆk z x z x x y y x