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# Study Guide for Test 3 PHYS 1600

AU

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This 6 page Study Guide was uploaded by Hayden Massey on Wednesday March 30, 2016. The Study Guide belongs to PHYS 1600 at Auburn University taught by Trevor Steinke in Spring 2016. Since its upload, it has received 17 views. For similar materials see Engineering Physics I in Physics 2 at Auburn University.

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Date Created: 03/30/16

Circular Motion: Even if you stay the same speed, but changed direction, you’ve changed velocity Changed velocity = acceleration 2 a= v r v = ωr ω = 2Π/T = 2Πf v = 2Πr/T = 2Πfr f = 1 revolution per second T = 1 second per revolution mv2 ΣF=T=ma= =mω r 2 r 2 ΣF=μmgcosθ+mgsinθ= mv r Swinging a toy vertically: 1 1] 2] 3] T mg 4 2 T T 4] T If I spin something, how can I figure out how many times around it goes?? mg 3 If it isn’t moving, Xi = Xf mg For rotations, look at angle it’s at to begin with θi If its not mg rotating, θi=θ f Rotational θ 2Π velocity (ω) = = s T For linear motion, Xf = Xi + vt For rotational motionθ f =θi+ωt 2Π rads = 1 revolution For linear motion, Xf = Xi + vt + ½ at2 For rotational motion, θ f =θi+ωt+ αt 2 2 2 2 For rotational motion, ω fω +2i(Δθ) Torque: T = Fr = Frsin θ ΣT=(?mass?)α MgR ΣTorque=Tsinθ∗R− 2 −mgr=0 Units for torque = kgm^2/s^2 Units for Angular Acceleration = rad/s^2 “angular mass” = moment of inertia (I) I = md^2 I point = MR^2 I=ΣMR= (MR )1+(MR )2 I = Integral of r^2 dm Conservation of Energy Translational KE= ½ mv^2 Gravitational PE = mgh Spring PE= ½ kx^2 Energy loss due to friction = W =μN (Δ x) Rotational Kinetic Energy = ½ Iω^2 Angular Momentum: L = Iω Energy is conserved unless acted upon by outside force Momentum is conserved unless acted upon by outside force Momentum is conserved separately in each dimension Angular Momentum is conserved unless acted upon by an outside torque Angular Momentum is conserved separate from linear momentum Linear Momentum conserved: (mvi)1 + (Mvi)2 = (mvf)1 + (Mvf)2 Angular Momentum conserved: (Iωi)1 + (Iωi)2 = (Iωf)1 + (Iωf)2 Total Kinetic Energy Conserved: 1mv i = 1mv f + 1 M vf+ I ω 2 (2) ( 1( 2 1 2 2 )2 Sum of all torque = Moment of Inertia * Angular acceleration Static Equilibrium No acceleration sum of torque = 0 MgR ΣTorque=Tsinθ∗R− 2 −mgr=0 Leaning Ladder Problem: A A = Normal force between wall and ladder B = Gravity acting on ladder = mg D C = Frictional force between floor and ladder D = Normal force between floor and ladder B C First look at Forces: 1. Decide which way is positive 2. Write appropriate equation ΣForce=A−C=0 Horizontal ΣForce=D−B=D−mg=0 Vertical Thus, F friction max = μN = μmg Then look at Torque: 3. Choose a point of rotation – choose bottom (its usually easier) 4. Write an appropriate equation L F g() cosθ−F wall sinθ=0 2 F g ( ) cosθ=F wallsinθ 2 sinθ F g cosθ= 2 F wall θ=tan −1 mg 2μmg Center of Mass: center of mass:Σ1r1+m2r2+m3r 3 M Density x area = mass Missing pieces: Look at the size of the piece cut out Use diameter or length as hypotenuse to find length of sides Find area of figure Multiply area of missing by density to find Mass Subtract out missing mass from total Elastic Properties of Solids: V 2 f −V 1 f K= where k=1 is elastic and k=0 is inelastic V 1i−V 2i Tensile stress = F/A Strain: result of a stress Tensile strain: ratio of change in length to initial length = ΔL/Li Modulus: combining stretch and strain ' TensileStress F /A Young sModulus= = TensileStrainΔL/Li Young’s Modulus measures the resistance of a solid to a change in its length Shear Stress F/A Shear Modulus = = ShearStrain Δ X/h Shear Modulus measures the resistance to the motion of the planes within a solid parallel to each other VolumeStress F /A Bulk Modulus = = Volume Strain ΔV /Vi Bulk Modulus is the resistance of solids or liquids to changes in their volume. Volume Stress: a ratio of total force exerted on the area of the surface

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