Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)

Transcription

1 Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position t, r, v, a Mass m force F Energy: kinetic, potential gravity, spring + work (friction) System of particles Rotation: τ = I α Special case: α = constant Conservation of energy Conservation of momentum R cm = constant F net,x = ma x, F net,y = ma y, τ net,z = Iα z Rolling motion Summary: University Physics I Physics 231 (Chapters 1-11) Compiled by Prof. Godunov Equilibrium F net,x = 0 F net,y = 0 τ net,z = 0 1

2 1D Kinematics (definitions) v avg = x 2 x 1 t 2 t 1, a avg = v 2 v 1 t 2 t 1, 3D Kinematics (definitions) v avg = r 2 r 1 t 2 t 1 v = lim Δt 0 x t + Δt x(t) Δt = dx dt v t + Δt v(t) a = lim = d2 x Δt 0 Δt dt 2 Δ r v = lim Δt 0 Δt d r = dt = v x i + v y j + v z k components: v x = dx dt, v y = dy dt, v z = dz dt a avg = v 2 v 1 Δ v, a = lim t 2 t 1 Δt 0 Δt Relative motion r PA = r PB + r BA v PA = v PB + v BA a PA = a PB Kinematics d v = dt = d2 r dt 2 1D Motion a = constant x = x 0 + v 0 t at2 v = v 0 + at v 2 = v a x x 0 If two objects same place x 1 = x 2 (same time) 2D Projectile Motion x = x 0 + v 0 cos θ 0 t y = y 0 + v 0 sin θ 0 t 1 2 gt2 v x = v 0 cos θ 0 v y = v 0 sin θ 0 gt Special cases: 1. y f = y 0 (same ground) 2. θ 0 = 0 0 (horizontal launch) 3. θ 0 = 90 0 (vertically launch) Uniform Circular Motion a = v2 r, T = 2π v 2

3 Dynamics Newton s Second Law: For any particle of mass m, the net force F on the particle is always equal to the mass m times the particle s acceleration: F = m a Newton s Third Law: If object 1 exerts a force F 21 on object 2, then object 2 always exerts force F 12 on object 1 given by F 12 = F 21 Forces Gravity F g = mg toward the ground Normal N = mg cos θ ± F external, perpendicular to the surface (and away) Tension T along the string Static friction f s μ s N parallel to surface & opposite the external force Kinetic friction f k = μ k N always opposite to direction of velocity 2D Linear (Translational) Motion F net,x = ma x F net,y = ma y Centripetal force F net,r = mv2 R (toward the center of rotation) Problem Solving: For every object we have to draw a free-body diagram. 1) Include ALL forces acting on the body matter. 2) If a problem includes more than one body - draw a separate free-body diagram for each body. 3) Not to include: any forces that the body 3 exerts on any other body.

4 Energy Kinetic Energy K = 1 2 mv2 Potential Gravitational U g y = mgy Potential Elastic U s x = 1 2 kx2 Work W i f = F x x f x i W i f = F d = Fd cos θ Power P avg = W Δt, + F y y f y i dw P = dt = F v Principle of Conservation of Energy If all of the n forces F i (i = 1, n) acting on a particle are conservative, each with its corresponding potential energy U i ( r), the total mechanical energy is constant in time K i + U i1 + U i2 + + U in = K f + U f1 + U f2 + + U fn Conservation of Energy with nonconservative forces (specifically with friction) K i + U i1 + U i2 + + U in = K f + U f1 + U f2 + + U fn + f k d 4

5 Systems of Particles Newton s Second law for a systems of particles d P dt = F ext or Ma CM = F ext Principle of Conservation of Linear Momentum: If the net force external force F ext on an N- particle system is zero, the system s total mechanical momentum is constant P = m 1 v 1 + m 2 v m N v N If the net force F ext acting on a system is zero, then a system moves with constant velocity v CM = contant And if initially v CM = 0, then the position of the center of mass R CM does not change despite individual positions of particles may change. R CM = m 1 r 1 + m 2 r m n r n + + m n Collisions: Elastic: p 1i + p 2i = p 1f + p 2f (momentum), E 1i + E 2i = E 1f + E 2f (energy conserved) v 1f = m 1 m 2 v 1i + 2m 2 v 2i v 2f = 2m 1 v 1i + m 2 m 1 v 2i Inelastic: p 1i + p 2i = p 1f + p 2f (momentum), Completely inelastic collision: m 1 v 1i + m 2 v 2i = v f E 1i + E 2i E 1f + E 2f (energy - NOT conserved) 5

6 Rotation Rotational kinetic energy K = 1 2 Iω2 Kinetic energy of rolling K = 1 2 Mv cm I cmω 2 v cm = ωr Rotational Inertia (for a single particle) I = mr 2 Rotational Inertia for common objects I solid cylinder = 1 2 MR2, I solid sphere = 2 5 MR2 Translational and Rotational Dynamics (Newton s Second law) F net = m a τ = I α translational motion rotational motion (α = a/r) for rotation in (x, y) plane F net,x = ma x, F net,y = ma y, τ net,z = Iα z Angular variables: θ, ω, α Relating linear and angular variables s = θr, v = ωr, a = αr Rotational kinematics with α = const θ = θ 0 + ω 0 t αt2 ω = ω 0 + αt Torque τ = r F = rf sin φ τ z = xf y yf x for rotation in (x, y) plane Angular Momentum L = r p (a particle), L = Iω (rigid body) τ net = I α = dl dt Equilibrium (translational + rotational) F net,x = 0 F net,y = 0 τ net,z = 0 Newton s second law if τ net = 0 then L = constant or I i ω i = I f ω f balance of forces balance of forces balance of torques 6

7 Scalar (dot) product Math and more c = a b = ab cos θ c = a b = a x b x + a y b y Trigonometry x 2 + y 2 = r 2 α + β = 90 0 x = r cos α y = r sin α Vector (cross) product c = a b = ab sin φ For vectors in (xy) plane c z = a x b y a y b x Vectors (components) a x = a cos θ, a y = a sin θ a = a x 2 + a y 2, θ = arctan a y a x Unit conversion (example) 130 km h = = 36.1 m s 130 km 1 h 1000 m 1 km 1 h 3600 s Quadratic equation ax 2 + bx + c = 0, x = b ± b2 4ac 2a 7