Chapter 8: Collision equations have not been given here. Just the general form. p = mulp; = pj IF = $|Ap = FavgAt = JIJ = fy Fat | P = Pa + Pb + Pc t ... - mava + mbVb + mcVct ... mitmat ... Item = y fadm| vf - vi = Veln( M, Chapter 9: Translational Motion equations and Rotational Motion equations below. 0 = $10 = de |a = dw d'e dt at As = Aor lut = wr | at = arl I= E mir, I = fram | I = Icm + Md2 Emech = K + U|U, = mgh | U. = } kx? | Kun =} mul | Knot = } Iw2 Chapter 10: Uses lots of equations from Chapter 9, and the following equations. 7 =RXFAXB = [A| |B|sin@) | T = Io = IN = d( Iw) dt L= 107 = L L =FXp = [r|musin(B) IL = LF Graphics from Slides: Here are several still images from slides of different useful equations. Vf = V; + alt W . = W. + ant As = v.At + 1/2a(At ) 40 = W.At + 1/za(At) V = V.+ 2ads w; =w,+ 2040TABLE 9.2 Moments of Inertia of Various Bodies (a) Slender rod, (b) Slender rod, (c) Rectangular plate, (d) Thin rectangular plate, axis through center axis through one end axis through center axis along edge 1 = ML- 1 = -ML2 1 = 12M(a- + 62 ) 1 = -Ma2 (e) Hollow cylinder (f) Solid cylinder (g) Thin-walled hollow (h) Solid sphere (i) Thin-walled hollow cylinder sphere 1 = SM(R2 + R2) 1 = -MR 1 = MR2 I = =MR2 1 = =MR2 R2 R1 R R .R- -RUnit 2 Equations Chapter 4: This chapter is all about properly using Newton's Three Laws F = ma \\F, = ma = -mg | FAB = -FBA Chapter 5: Circular motion equations were removed to avoid confusion (slide below) fs,mar = Us FN I fk = UkFN | D = >CopAv | Vterm ~ V 4mg A EF, = map = = mw-r| \\ Fi = mat = marl > F. = 0 Chapter 6: Work is a transfer of energy; it has the units of energy W = F . Ar = FArcos(@) |W = J."' F . dr | Fsp = -kAs West = Kf - Ki = AK | P= - dt dW P = F . d Chapter 7: The differences in the work equation come from conservative and nonconservative forces Emech = K +U|U, = mgh | Uspr = - kx? | Fsp = - kx | AEmech = AK + AU = 0 F = - | Eint = fxAs | W = AK + AU + Eint Graphics from Slides: Here are several still images from slides of different useful equations.Free Body Diagrams Identify the different forces acting on a particle Represent the object as a particle at the center of a coordinate system Draw each force as a vector pointing in the correct direction If possible, identify the direction of the net force Uniform Circular Motion object moves with constant (tangential) speed Speed 2TR V = T 2n Rf a Centripetal Acceleration 2 a = R Centripetal Force EF, = mVT REssential concepts Particle acceleration, force, interaction Basic Goals How does a particle respond to a force? How do objects interact? Newton's First Law AV =0- a=0 -F=0 General Principles Newton's Second Law F= ma Newton's Third Law FAOn B = - FBonA Basic Problem Solving Strategy Linear Motion Trajectory Motion Circular Motion F. = max SF. =0 EF, = ma, my- - = mo-r r SF, =0 LF, = may LF, = ma, F, = ma, EF = ma. Linear & Trajectory Kinematics Circular Kinematics General Case Uniform Acceleration: T = - 2nr 251 Uniform V Vf = V. + alt Circular Motion: 0, =0, +wAt V = ds dt - = slope of position graph as = constant As = v.At + y/za(At ) 1,2 V, = Or dv a = - - = slope of velocity graph v7 = v2+ 2ads dt W / = w; +ant Av = [a dt = area under acceleration curve Uniform Motion: Nonuniform Circular Motion: 40 = W.At + 1/za(At) As = [v dt = area under velocity curve as =0 Vs = constant As = v.At W7 = W.+ 2040Unit 1 Equations Chapter 1: This chapter was setting up the math. Familiarity with vectors is big here A . B = B . A = |AB cos(0) | A . B = Ar Br + AyBy + A,B,I AXB - -BX A = [AB sin(0) | Ax B = (A,B, - A,By) i+ (A,Br - ArB.) j + (A,By - AB, ) k Chapter 2: Gradients are for variables that dont change with time, derivatives are for variables that change with time of = di + UpAt|v, = 2 At |Ax = fff vxdt | a = du Uf = vi + aAt laf = x; + v;At + zant? | (vf )? = (vi)2 + 20Ax Chapter 3: Several topics didn't have equations, like reference frames and some directions of relative velocity. Its worth looking those parts over. Ar = (Ax)i + (Ay)y = Tri try) | AT = ry - rix = ut + x' | Vang = A Tu= dy aavg = Au At At la = du | Range = sin(20) lac = 4old. = = wr Graphics from Slides: Here are several still images from slides of different useful equations. 1D Motion Physics is Calculus S position velocity a acceleration I time differentiate differentiate S a integrate integrate 5=0 V= 0 a=0 particle is particle is: . momentarily at rest particle is at the origin . instantaneously at rest at constant velocity . stationary . changing directions Minimum/ position set - dt "=0, solve for , plug back into S Maximum velocity set - = 0, solve for , plug back into E...y, = y, +VAI+ =a At- 2 Vy = Vita,At Vof = Viz+2a,Ay 2 Vif = Vita,At v =v 2 +2a Ax displacement Ax= X - X; Kinematics Equations (constant acceleration) Ax average velocity V.= Vxf = Vita,At At X, = X; +v At+ - a, At2 total distance 2 average speed total time ( V ) 2 = (V : )2 + 2a, Ax instantaneous velocity dx VE dt Free Fall instantaneous speed On Earth, when not touching anything, objects fall due to average acceleration Av VF gravity towards the Earth's At - t ; center with acceleration: instantaneous acceleration dv d'x g = 9.80 m/s2 dt dt 2 (neglecting air resistance)V= W = speed dt angular speed dt 2Tr v = 2arf = T W =2nf = T speed angular speed V = WT Angular speed does not depend on radius linear angular position Ar de velocity V = W= dt dt dw acceleration 0 = O = dt dt1. (10 Points) An astronaut hits a 46.0 g golf ball a distance of 400 m on an alien planet with mass 1.5 x 1020 kg. The golf ball leaves the club with a speed of 10 m/s and a launch angle of 45. What is the radius of the planet?2. (10 Points) A 4.00 kg block is pressed against a vertical wall with a force directed upwards at an angle of 30.0 with respect to the horizontal, as shown below. When the pushing force is 50.0 N, the block slides down the wall with an acceleration of 1.50 m/s. 30.09 What is the coefficient of kinetic friction between the block and the wall?3. (10 Points) A solid marble of mass m = 20.0 g and radius r = 1.00 cm is placed on a ramp at height H. Assuming that it rolls without slipping and referring to the image below. If R = 10.0 cm what is the minimum height H the ball needs to stay in contact with the loop? R H4. (10 points) A particle of mass 2.00 kg experiences the force shown below. At t = 0, the particle's velocity is 3.00 m/s in the positive x-direction. What is the velocity of the particle at t = 5.00 s? F. (N) A > t (s) 5.00 -4.00