EXAM 3 SOLUTION

# EXAM 3 SOLUTION - midterm 03 – LEE JAE HO – Due Apr 3...

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Unformatted text preview: midterm 03 – LEE, JAE HO – Due: Apr 3 2008, 1:00 am 1 Mechanics - Basic Physical Concepts Math: Circle: 2 π r , π r 2 ; Sphere: 4 π r 2 , (4 / 3) π r 3 Quadratic Eq.: ax 2 + bx + c = 0, x = − b ± √ b 2 − 4 a c 2 a Cartesian and polar coordinates: x = r cos θ, y = r sin θ , r 2 = x 2 + y 2 , tan θ = y x Trigonometry: cos α cos β + sin α sin β = cos( α − β ) sin α + sin β = 2 sin α + β 2 cos α − β 2 cos α + cos β = 2 cos α + β 2 cos α − β 2 sin 2 θ = 2 sin θ cos θ, cos 2 θ = cos 2 θ − sin 2 θ 1 − cos θ = 2 sin 2 θ 2 , 1 + cos θ = 2 cos 2 θ 2 Vector algebra: vector A = ( A x ,A y ) = A x ˆ ı + A y ˆ Resultant: vector R = vector A + vector B = ( A x + B x ,A y + B y ) Dot: vector A · vector B = AB cos θ = A x B x + A y B y + A z B z Cross product: ˆ ı × ˆ = ˆ k , ˆ × ˆ k = ˆ ı , ˆ k × ˆ ı = ˆ vector C = vector A × vector B = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı ˆ ˆ k A x A y A z B x B y B z vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle C = AB sin θ = A ⊥ B = AB ⊥ , use right hand rule Calculus: d dx x n = nx n − 1 , d dx ln x = 1 x , d dθ sin θ = cos θ , d dθ cos θ = − sin θ , d dx const = 0 Measurements Dimensional analysis: e.g. , F = ma → [ M ][ L ][ T ] − 2 , or F = m v 2 r → [ M ][ L ][ T ] − 2 Summation: ∑ N i =1 ( ax i + b ) = a ∑ N i =1 x i + bN Motion One dimensional motion: v = d s dt , a = d v dt Average values: ¯ v = s f − s i t f − t i , ¯ a = v f − v i t f − t i One dimensional motion (constant acceleration): v ( t ) : v = v + at s ( t ) : s = ¯ v t = v t + 1 2 at 2 , ¯ v = v + v 2 v ( s ) : v 2 = v 2 + 2 as Nonuniform acceleration: x = x + v t + 1 2 at 2 + 1 6 j t 3 + 1 24 st 4 + 1 120 k t 5 + 1 720 pt 6 + ... , (jerk, snap, ... ) Projectile motion: t rise = t fall = t trip 2 = v y g h = 1 2 g t 2 fall , R = v ox t trip Circular: a c = v 2 r , v = 2 π r T , f = 1 T (Hertz=s − 1 ) Curvilinear motion: a = radicalBig a 2 t + a 2 r Relative velocity: vectorv = vectorv ′ + vectoru Law of Motion and applications Force: vector F = mvectora, F g = mg, vector F 12 = − vector F 21 Circular motion: a c = v 2 r , v = 2 π r T = 2 π r f Friction: F static ≤ μ s N F kinetic = μ k N Equilibrium (concurrent forces): ∑ i vector F i = 0 Energy Work (for all F): Δ W = W AB = W B − W A F bardbl s = Fs cos θ = vector F · vectors → integraltext B A vector F · dvectors (in Joules) Effects due to work done: vector F ext = mvectora − vector F c − vector f nc W ext | A → B = K B − K A + U B − U A + W diss | A → B Kinetic energy: K B − K A = integraltext B A mvectora · dvectors , K = 1 2 mv 2 K (conservative vector F ): U B − U A = − integraltext B A vector F · dvectors U gravity = mg y , U spring = 1 2 k x 2 From U to vector F : F x = − ∂ U ∂x , F y = − ∂ U ∂y , F z = − ∂ U ∂z F...
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## This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas.

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EXAM 3 SOLUTION - midterm 03 – LEE JAE HO – Due Apr 3...

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