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CourseCourse OutlineOutline 1. MATLAB tutorial 2. Motion of systems that can be idealized as particles • Description of motion, coordinate systems; Newton’s laws; • Calculating forces required to induce prescribed motion; • Deriving and solving equations of motion 3. Conservation laws for systems of particles • Work, power and energy; • Linear impulse and momentum • Angular momentum 4. Vibrations Exam topics • Characteristics of vibrations; vibration of free 1 DOF systems • Vibration of damped 1 DOF systems • Forced Vibrations 5. Motion of systems that can be idealized as rigid bodies • Description of rotational motion • kinematics; gears, pulleys and the rolling wheel • Inertial properties of rigid bodies; momentum and energy • Dynamics of rigid bodies Particle Dynamics – concept checklist • Understand the concept of an ‘inertial frame’ • Be able to idealize an engineering design as a set of particles, and know when this idealization will give accurate results • Describe the motion of a system of particles (eg components in a fixed coordinate system; components in a polar coordinate system, etc) • Be able to differentiate position vectors (with proper use of the chain rule!) to determine velocity and acceleration; and be able to integrate acceleration or velocity to determine position vector. • Be able to describe motion in normal-tangential and polar coordinates (eg be able to write down vector components of velocity and acceleration in terms of speed, radius of curvature of path, or coordinates in the cylindrical-polar system). • Be able to convert between Cartesian to normal-tangential or polar coordinate descriptions of motion • Be able to draw a correct free body diagram showing forces acting on system idealized as particles • Be able to write down Newton’s laws of motion in rectangular, normal-tangential, and polar coordinate systems • Be able to obtain an additional moment balance equation for a rigid body moving without rotation or rotating about a fixed axis at constant rate. • Be able to use Newton’s laws of motion to solve for unknown accelerations or forces in a system of particles • Use Newton’s laws of motion to derive differential equations governing the motion of a system of particles • Be able to re-write second order differential equations as a pair of first-order differential equations in a form that MATLAB can solve Particle Kinematics Inertial frame – non accelerating, non rotating reference frame Particle – point mass at some position in space j Position Vector r(t) = x(t)i + y(t)j + z(t)k v(t) dt r(t) path of particle Velocity Vector v()t=++ vtxyz () i vt () jk vt () r(t+dt) d dx dy dz =( xyzi ++ jk) = i + j + k O dt dt dt dt k i dx dy dz ⇒=vt() vt () = vt () = xdt yz dt dt • Direction of velocity vector is parallel to path • Magnitude of velocity vector is distance traveled / time Acceleration Vector d dvx dvy dvz a()t= atx () i + at y () j + at z () k =( v xyz i ++ v jk v ) = i + j + k dt dt dt dt 22 2 dvx dx dvy dy dvz dz ⇒==atxy() at() ==at z() == dt dt22dt dt dt dt 2 dv dvy dv at()= x v at()= v at()= z v Also xdx xydy y zdz z ParticleParticle KinematicsKinematics • Straight line motion with constant acceleration 1 2 r=X00 ++ V t ativ =( V0 + at) i a = a i 2 • Time/velocity/position dependent acceleration – use calculus tt r=+=+X v() t dt iv V a() t dt i 00∫∫ 00 vt dv dv dv a==⇒= f( vgt ) () gtdt() = ax() dt ∫∫f() v V0 0 dt xt() t dv dx dv dx dx ⇒ =⇒=ax() v ax () v==⇒= f( xgt ) () gtdt() dt ∫∫f() x dx dt dx X0 0 vt() xt () ∫∫vdv= a() x dx V0 0 Particle Kinematics Circular Motion at const speed • sinθ θω=ts = R θ V = ω R t cosθ r=R(cosθθ ij + sin ) R j n v=−+ωRV( sin θθ i cos jt) = θ=ωt Rsin θ i V 2 a=−+==ω22RR(cos θ i sin θω j ) nn Rcosθ R • General circular motion ωθ=d/ dt αω = d // dt = d22 θ dt sinθ s= Rθω V = ds/ dt = R t cosθ R r=R cosθθ ij + sin ( ) j n v=−+ωRV( sin θθ i cos jt) = θ Rsin θ i 2 a=−+−RRαθ( sin i cos θωθ j ) (cos ij + sin θ ) Rcosθ dV V 2 =+=+αωRRt2 ntn dt R Particle Kinematics • Motion along an arbitrary path 1 t Radius of curvature R = 22 = dx22 dy vtV + R 22 ds ds dV V 2 atn= + n dt R s rij=xs( ) + y (s) j i • Polar Coordinates eθ e r dr dθ r ve=r + r eθ j dt dt θ 2 d22 r dθd θθ dr d i ae=−rr ++2 e 22r θ dt dt dt dt dt Using Newton’s laws Calculating forces required to cause prescribed motion • Idealize system • Free body diagram • Kinematics (describe motion – usually goal is to find formula for acceleration) • F=ma for each particle. • M0c = (for steadily or non-rotating rigid bodies or frames only) • Solve for unknown forces or accelerations (just like statics) Using Newton’s laws to derive equations of motion 1. Idealize system 2. Introduce variables to describe motion (often x,y coords, but we will see other examples) 3. Write down r, differentiate to get a 4. Draw FBD 5. Fa= m 6. If necessary, eliminate reaction forces 7. Result will be differential equations for coords defined in (2), e.g. d2 x dx m +λω +=kx kY0 sin t dt2 dt 8. Identify initial conditions, and solve ODE Motion of a projectile in earths gravity V0 r=XYZ ijk ++ 00 0 t = 0 k dr X0 =++VVVxyz000i jk dt j i 1 2 r=( X00 + Vxy t) ij ++( Y 00 V t) + Z 00 + V z t − gt k 2 vij=(Vxyz000) +( V) +−( V gt) k ak= −g Rearranging differential equations for MATLAB 2 d y dy 2 +20ζωnn += ω y • Example dt2 dt • Introduce v= dy/ dt d y v = • Then 2 dt v −−2ζωnnvy ω dw y • This has form = ft(,ww ) = dt v ConservationConservation laws for Laws particles: – concept Concept checklist Checklist • Know the definitions of power (or rate of work) of a force, and work done by a force • Know the definition of kinetic energy of a particle • Understand power-work-kinetic energy relations for a particle • Be able to use work/power/kinetic energy to solve problems involving particle motion • Be able to distinguish between conservative and non-conservative forces • Be able to calculate the potential energy of a conservative force • Be able to calculate the force associated with a potential energy function • Know the work-energy relation for a system of particles; (energy conservation for a closed system) • Use energy conservation to analyze motion of conservative systems of particles • Know the definition of the linear impulse of a force • Know the definition of linear momentum of a particle • Understand the impulse-momentum (and force-momentum) relations for a particle • Understand impulse-momentum relations for a system of particles (momentum conservation for a closed system) • Be able to use impulse-momentum to analyze motion of particles and systems of particles • Know the definition of restitution coefficient for a collision • Predict changes in velocity of two colliding particles in 2D and 3D using momentum and the restitution formula • Know the definition of angular impulse of a force • Know the definition of angular momentum of a particle • Understand the angular impulse-momentum relation • Be able to use angular momentum to solve central force problems/impact problems Work-Energy relations for a single particle k F P Rate of work done by a force P =Fv ⋅ v (power developed by force) O j i r t1 1 F(t) P = ⋅ r0 Total work done by a force W=∫ Fv ⋅ dt Wd∫ Fr k j 0 r0 O i r1 112 222 Kinetic energy T= mv = mv ++ v v 22( xyz) v r dT 0 k Power-kinetic energy relation P = P dt j r O 1 i r Work-kinetic energy relation 1 W=∫ Fr ⋅=− d TT0 r0 Potential Energy Type of force Potential energy F Potential energy of a m Gravity acting on a j conservative force (pair) particle near earths U= mgy y surface i r Ud()r=−⋅+ Frconstant ∫ F r 0 r m Gravitational force GMm exerted on mass m by U = − F = −grad(U ) mass M at the origin r r P Force exerted by a j r spring with stiffness k 1 2 F r0 and unstretched length U= kr( − L0 ) k F 2 L0 i j O 2 i r1 F +Q j +Q2 Force acting between QQ 1 = 12 i two charged particles U 4πε r r 1 Force exerted by one molecule of a noble gas 2 (e.g. He, Ar, etc) on 12 6 F another (Lennard Jones aa j UE= − 2 potential). a is the rr i equilibrium spacing 1 r between molecules, and E is the energy of the bond. Energy Relation for a Conservative System ext F m4 Internal Forces: (forces exerted by 1 ext F3 one part of the system on another) Rij m1 R13 R31 ext R12 m3 External Forces: (any other forces) Fi R32 R23 R21 System is conservative if all internal forces are m2 conservative forces (or constraint forces) ext F2 Energy relation for a conservative system External Power Pext () t t1 External work∆= Wext ∫ P() t dt t0 ∆=TOT + TOT − TOT + TOT m WTUTUext 1100( ) ext 4 m4 F1 ext ext F1 F3 ext m1 m1 F3 m3 ext m3 Special case – zero external work: F2 m2 m 2 TOT TOT TOT TOT ext +=+ F2 TUTU1100 tt= 0 tt= 1 TOT TOT Total KE T0 Total KE T1 KE+PE = constant TOT TOT Total PE U0 Total PE U1 Impulse-Momentum for a single particle Definitions F(t) t1 v Linear Impulse of a force IF= ∫ ()t dt k m t0 j O Linear momentum of a particle pv= m i Impulse-Momentum relations t=t0 dp p=p F = 0 F(t) v dt k m j Ip=10 − p O i t=t1 p=p1 ImpulseImpulse-Momentum-momentum for for a system a system of particles of particles ext F m4 1 ext R Force exerted on particle i by particle j F3 ij m1 R13 R31 ext R12 m3 Fi External force on particle i R32 R23 R21 m2 vi Velocity of particle i ext F2 Total External ForceFTOT () t Impulse-momentum for the system: t1 TOT TOT TOT dp Total External Impulse ℑ=∫ F ()t dt FTOT = t0 dt m ext 4 m4 F1 ext ext F1 TOT TOT TOT F3 ext = − m1 I pp10 m1 F3 m3 ext m3 F2 m2 m2 ext Special case – zero external impulse: F2 TOT TOT = tt= 0 tt1 pp10= TOT TOT Total momentum p0 Total momentum p1 (Linear momentum conserved) CollisionsCollisions A0 vB0 vx x AB11 A 0 B 0 Momentum mvAx+=+ mv Bx mv Ax mv Bx A B Restitution formula vBA11−=−− v ev( B 0 v A 0) m * vBB1=+ v 0 A (1 +−ev ) ( AB00 v) mm+ vA1 B1 AB x vx m vAA1=− v 0 B (1 +−ev ) ( AB00 v) A B mmAB+ A BA11 B 0 A 0 A0 B B0 Momentum mvv+= mm vv + m v v B AB A BA11 BA 00 BA00 Restitution formula (vv−) =( vv −) −+(1e ) ( vvnn −) ⋅ n m vAA10=− v B (1 +e ) (vAB00 −⋅ v) nn + A B mmBA v A1 B1 m v vBB10=+ v A (1 +e ) (vAB00 −⋅ v) nn + mmBA.
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