Rigid Bodies - MIT
Rigid BodiesTwo-Dimensional RotationalKinematicsA rigid body is an extended object in which thedistance between any two points in the object isconstant in time.Springs or human bodies are non-rigid bodies.8.01W10D1Rotation and Translationof Rigid BodyRecall: Translational Motion ofthe Center of MassDemonstration: Motion of a thrown baton Total momentum of system of particlespsys m total Vcm External force and acceleration of centerof massTranslational motion: external force of gravity actson center of masstotalFext Rotational Motion: object rotates about center ofmassdVdpsys m total cm m total A cmdtdt1
Main Idea: Rotation of RigidBodyTorque produces angular acceleration about center ofmasstotalτ cm I cmα cmI cm is the moment of inertial about the center of massα cm is the angular acceleration about center of massRotational Kinematicsfor Fixed Axis RotationA point like particle undergoing circular motion ata non-constant speed has(1)An angular velocity vector(2) an angular acceleration vectorTwo-Dimensional Rotation Fixed axis rotation:Disc is rotating about axispassing through thecenter of the disc and isperpendicular to theplane of the disc. Plane of motion is fixed:For straight line motion,bicycle wheel rotatesabout fixed direction andcenter of mass istranslatingFixed Axis Rotation: AngularVelocityAngle variableSI unit:θ[rad]dθ ˆω ω kˆ kAngular velocitydt 1SI unit: rad s dθVector:ω Componentdtdθω dtmagnitudeω 0, direction kˆω 0, direction kˆdirection2
Concept Question: AngularSpeedFixed Axis Rotation: AngularAccelerationd 2θObject A sits at the outer edge (rim) of a merry-go-round, andobject B sits halfway between the rim and the axis of rotation. Themerry-go-round makes a complete revolution once every thirtyseconds. The magnitude of the angular velocity of Object B isAngular acceleration: α α kˆ 2 kˆdtSI unit rad s 2 Vector:Component:α d 2θ dω dtdt 2α Magnitude:dωdt1.2.3.4.dω 0, direction kˆdtdω 0, direction kˆdtDirection:Rotational Kinematics: ConstantAngular AccelerationThe angular quantitiesare exactly analogous to the quantitiesx, vx , and axfor one-dimensional motion, and obey the same type of integral relationst0Table Problem: RotationalKinematicsA turntable is a uniform disc of mass m and a radius R. Theturntable is spinning initially at a constant frequency f . The motor isturned off and the turntable slows to a stop in t seconds withconstant angular deceleration.θ , ω , and αω (t ) ω0 α (t ′) dt ′ ,half the magnitude of the angular velocity of Object A .the same as the magnitude of the angular velocity of Object A .twice the the magnitude of the angular velocity of Object A .impossible to determine.a) What is the direction and magnitude of the initial angular velocity ofthe turntable?tθ (t ) θ 0 ω (t ′) dt ′ .0b) What is the direction and magnitude of the angular acceleration ofthe turntable?Constant angular acceleration:ω (t) ω 0 α t12θ (t) θ 0 ω 0 t α t 2 (ω (t ) )2 ω 02 2α (θ (t ) θ 0 ) .c) What is the total angle in radians that the turntable spins whileslowing down?3
Angular Momentum of a PointParticle Point particle of mass m moving with avelocity v Momentum Fix a point S Vector rS from the point S to thelocation of the object Angular momentum about the point S SI UnitDirectionp mvRight Hand RuleL S rS p[kg m 2 s-1 ]Cross Product: AngularMomentum of a Point ParticleL S rS pMagnitude:b)Example Problem: AngularMomentum and Cross ProductA particle of mass m 2 kg moveswith a uniform velocityL S rS p sin θa)Cross Product: AngularMomentum of a Point Particlev 3.0 m s-1 î 3.0 m s-1 ĵmoment armrS , rS sin θAt time t, the position vector of theparticle with respect ot the point S isL S rS , prS 2.0 m î 3.0 m ĵPerpendicularmomentumpS , p sin θL S rS p Find the direction and the magnitudeof the angular momentum about theorigin, (the point S) at time t.4
Solution: Angular Momentumand Cross ProductThe angular momentum vector of the particleabout the point S is given by:L S rS p rS m v()( 2.0 m ˆi 3.0 m ˆj (2kg) 3.0 m s 1ˆi 3.0 m s 1ˆjTable Problem: AngularMomentum and Circular MotionConsider a point particle of mass mmoving in a circle of radius R withvelocity v Rω θˆ . Find the directionand magnitude of the angularmomentum about the center of a circlein terms of the radius R, mass m, andangular speed ω.) 0 12 kg m 2 s 1 kˆ 18 kg m 2 s 1 ( kˆ ) 0 6.0 kg m 2 s 1 kˆ .The direction is in the negative kö direction, and themagnitude isL S 6.0 kg m 2 s 1.i j k,j i k ,i i j j 0Angular Momentum and CircularMotion of a Point ParticleFixed axis of rotation: z-axisAngular velocityω ω kˆRigid Body Kinematicsfor Fixed Axis RotationBody rotates with angular velocity ωacceleration αand angularVelocityv ω r ω kˆ R rˆ Rω θˆAngular momentum about thepoint SL S rS p rS mv Rmv kˆ RmRω kˆ mR 2ω kˆ5
Divide Body into Small ElementsBody rotates with angular velocity,ωRotational kinetic energy about axispassing through Sαangular accelerationIndividual elements of massΔmiRadius of orbitr ,iRadial AccelerationMoment of Inertia about S : I S arad,i 2vtan,ir ,iSI Unit:Continuous body:K rot K rot,iHow does moment of inertia compare to the total mass and the centerof mass?Different measures of the distribution of the mass. dmbodyR cm 1m total r dmbodyMoment of Inertia about axis passing through S: (nine possiblemoments)IS bodydm (r ,dm ) 2i ,i)2i NRotational Kinetic Energy:Discussion: Moment of InertiaCenter of Mass: vector (three components) Δm (rr ,i r ,dmΔmi dmIS r ,iω 2mtotal 1I cmωcm 22i N2 kg m iTotal mass: scalarK cm i 1atan,i r ,iαTangential acceleration( )211Δmi vtan,i 2 Δmi r ,i ω 222K rot,i vtan,i r ,iωTangential velocityRotational Kinetic Energy andMoment of Inertia i 1 bodydm (r ,dm )2body 1 2 11 Δmi r ,i ω 2 dm (r ,dm )2 ω 2 I Sω 22 i 2 2 body ( )Concept Question: CollisionAn object of mass m with velocity vmoving in a straight line collideselastically with an identical object. Thesecond object is initially at rest and isattached at one end to a string oflength l and negligible mass. The otherend of the string is fixed (at the point S).After the collision the second objectundergoes circular motion with angularspeed ω . The kinetic energy of thesecond object after the collision is1. (1/4) m l2 ω24. (1/4) m l ω22. (1/2) m l2 ω25. (1/2) m l ω23. (3/4) m l2 ω26. (1/4) m l ω6
Strategy: Calculating Moment ofInertiaExample: Moment of Inertia of aDiscStep 1: Identify the axis of rotationConsider a thin uniform disc of radius R and mass m.What is the moment of inertia about an axis that passperpendicular through the center of the disc?Step 2: Choose a coordinate systemda r dr dθStep 3: Identify the infinitesimal mass element dm.r ,dm rr ,dm , of the circular orbit ofStep 4: Identify the radius,the infinitesimal mass element dm.(r ,dm )2 dm bodyI cmStep 6: Explicitly calculate the integrals.M π R2 r Rr 0I cm Table Problem: Moment of Inertiaof a RodConsider a thin uniform rod of length L andmass m. Calculate the moment of inertiaabout an axis that passes perpendicularthrough the center of mass of the rod. I cm Step 5: Set up the limits for the integral over the bodyin terms of the physical dimensions of the rigid body.dm mtotalM da Area π R 2Mr dr dθdm σ r dr dθ θ 2ππ R2r 3 dθ drσ Mπ R2r R θr 0 θ 2π dθ r 3dr M θ 0 π R22MR2 r Rr 0r 3dr 2M r4R2 4 r Rr 0r R r 0 02π r 3dr 2MR2 r Rr 0r 3dr2 M R4 1 MR 22R2 4Parallel Axis Theorem Rigid body of mass m. Moment of inertia I cm aboutaxis through center of mass ofthe body. Moment of inertia I S aboutparallel axis through point S inbody. dS,cm perpendicular distancebetween two parallel axes.I S Icm mdS2,cm7
Summary: Moment of InertiaMoment of Inertia about S:i NI S Δmi (r ,S ,i ) 2 i 1 r ,S 2 dmbodyExamples: Let S be the center of mass rod of length l and mass mI cm 1 2ml12 disc of radius R and mass mIcm 1mR 22I S Icm mdS2,cmParallel Axis theorem:Summary: Fixed Axis RotationKinematicsθAngle variableAngular velocityω dθ / dtAngular accelerationα d 2θ / dt 2Concept Question: KineticEnergyA disk with mass M and radius R is spinning withangular speed ω about an axis that passes throughthe rim of the disk perpendicular to its plane.Moment of inertia about cm is (1/2)M R2. Its totalkinetic energy is:1. (1/4)M R2 ω24. (1/4)M Rω22. (1/2)M R2 ω25. (1/2)M Rω23. (3/4)M R2 ω26. (1/4)M RωAngular Momentumfor Fixed Axis RotationAngular Momentum about the point SL S ,i rS ,i pi (r ,i rˆ zi kˆ ) ptan,i θˆL r p kˆ z p rˆS ,iΔmiMass elementRadius of orbiti Ni 1Parallel Axis Theoremtan,iitan,iptan,i Δmi vtan,i Δmi r ,iωr ,i2Moment of inertia I S Δmi (r ,i ) ,iTangential component of momentum bodyI S Md 2 I cmdm(r )2z-component of angular momentumabout S:LS , z ,i r ,i ptan,i r ,i Δmi r ,iω Δmi r ,i 2ωi Ni Ni 1i 1LS , z LS , z,i Δmi r ,i 2ω I Sω8
Concept Question: AngularMomentumA dumbbell is rotating at a constant angularspeed about its center. Compared to thedumbbell's angular momentum about its centerA, its angular momentum about point B (asshown in the figure) is1.2.3.bigger.the same.smaller.Concept Question: AngularmomentumA disk with mass M and radius R is spinning with angularvelocity ω about an axis that passes through the rim of thedisk perpendicular to its plane. The magnitude of its angularmomentum is:1.1M R 2ω 244.1M R 2ω42.1M R 2ω 225.1M R 2ω23.3M R 2ω 226.3M R 2ω29
Kinematics 8.01 W10D1 Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid bodies. Rotation and Translation of Rigid Body Demonstration: Motion of a thrown baton Translational motion: external force of gravity acts on center of mass
kinetics of rigid bodies, i.e., relations between the forces acting on a rigid body, the shape and mass of the body, and the motion produced. Results of this chapter will be restricted to: plane motion of rigid bodies, and rigid bodies consisting of plane slabs or bodies which are symmetrical with respect to the reference plane.
- plane motion of rigid bodies, and - rigid bodies consisting of plane slabs or bodies which are symmetrical with respect to the reference plane. D’Alembert’s principle is applied to prove that the external forces acting on a rigid body are equivalent a vector attached to the mass center and a couple of moment ma r Iα.
Ch.5 Plane Kinematics of Rigid Bodies Rigid body: Distances between the particles remain unchanged, Changes in shape are very small compared with the body movement Kinematics of particle: Only the positions of particles are interested Kinematics of rigid body: Movement of every part of rigid body is concerned (include rotational motion)
Rigid Body Modeling Store an object space triangulated surface to represent the surface of the rigid body Store an object space implicit surface to represent the interior volume of the rigid body Collision detection between two rigid bodies can then be carried out by checking the surface of one body against the interior volume of another
Rigid Body Motion and Rotational Dynamics 13.1 Rigid Bodies A rigid bodyconsists of a group of particles whose separations are all fixed in magnitude. Six independent coordinates are required to completely specify the position and orientation of a rigid body. For example, the location of the first particle is specified by three coordinates. A
Planar Kinetics of Rigid Bodies: Part 1 Planar Equations of Motion of a Rigid Body The general plane motion of a 2D rigid body consists of rotation and translation in x- and y- directions, i.e. a system with _. This directly suggests that there are _.
Kinematics of Two-Dimensional Rigid Body Motion Even though a rigid body is composed of an infinite number of particles, the motion of these particles is constrained to be such that the body remains a rigid body during the motion. In particular, the only degrees of freedom of a 2D rigid body are translation and rotation. Parallel Axes
EMC standards generally cover the range from 0 Hz to 400 GHz. Currently, however, not all frequency ranges are completely regu-lated. The first important frequency range is the range around the power network frequency, which in Europe is 50 Hz. Most loads connected to the power network are non-linear loads, i.e., they draw a current that does not follow the sinusoidal voltage. Non-linear loads .
