Rigid Body: Rotational and Translational Motion; Rolling without Slipping 8.01 W11D1/W11D2 W11D1 and W11D2 Reading Assignment: MIT 8.01 Course Notes Chapter 20 Rigid Body: Translation and Rotational Motion Kinematics for Fixed Axis Rotation Sections 20.1-20.5 Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis Sections 21.1-21.5Announcements Sections 1-4 No Class Week 11 Monday Sunday Tutoring in 26-152 from 1-5 pm Problem Set 8 due Week 11 Tuesday at 9 pm in box outside 26-152 No Math Review Week 11 Exam 3 Tuesday Nov 26 7:30-9:30 pm Conflict Exam 3 Wednesday Nov 27 8-10 am, 10-12 noon Nov 27 Drop DateDemo: Rotation and Translation of Rigid Body Thrown Rigid Rod Translational Motion: the gravitational external force acts on center-of-mass Rotational Motion: object rotates about center-of-mass. Note that the center-of-mass may be accelerating syext total totalcmcmsddm mdt dt= = =VpF AFixed Axis Rotation and Translation For straight line motion, bicycle wheel rotates about fixed direction and center of mass is translatingOverview: Rotation about the Center-of-Mass of a Rigid Body The total external torque produces an angular acceleration about the center-of-mass is the moment of inertial about the center-of-mass is the angular acceleration about the center-of-mass is the angular momentum about the center-of-mass Icmextcmcm cm cmdIdtα= =Lτ αcmcmLDemo: Bicycle Wheel Rolling Without SlippingRolling Bicycle Wheel Motion of point P on rim of rolling bicycle wheel Relative velocity of point P on rim: Reference frame fixed to ground Center of mass reference frame vP=vcm, P+VcmRolling Bicycle Wheel Distance traveled in center of mass reference frame of point P on rim in time Δt: Δs = RΔθ= RωcmΔtDistance traveled in ground fixed reference frame of point P on rim in time Δt: ΔXcm=VcmΔtRolling Bicycle Wheel: Constraint Relations Rolling without slipping: Δs = ΔXcmRolling and Skidding: Rωcm= VcmΔs < ΔXcmRωcm< VcmΔs > ΔXcmRωcm> VcmRolling and Slipping:Rolling Without Slipping The velocity of the point on the rim that is in contact with the ground is zero in the reference frame fixed to the ground.Concept Question: Rolling Without Slipping If a wheel of radius R rolls without slipping through an angle θ, what is the relationship between the distance the wheel rolls, x, and the product Rθ? 1. x > Rθ. 2. x = Rθ. 3. x < Rθ.Angular Momentum for 2-Dim Rotation and Translation The angular momentum for a translating and rotating object is given by Angular momentum arising from translational of center-of-mass The second term is the angular momentum arising from rotation about center-of mass, sys,cm ,cmS S= ×L R p sys,cm cm,i cm,i1i NS S iim=== × + ×∑L R p r v Lcm= IcmωcmWorked Ex.: Angular Momentum for Earth What is the ratio of the angular momentum about the center of mass to the angular momentum of the center of mass motion of the Earth?Earth’s Orbital Angular Momentum • Orbital angular momentum about center of sun • Center of mass velocity and angular velocity • Period and angular velocity • Magnitude LSorbital=rS ,cm×ptotal= rs,emevcmˆk vcm= rs,eωorbitωorbit= (2π/ Torbit) = 2.0 × 10−7rad ⋅ s−1 LSorbital= mers,e2ωorbitˆk =mers,e22πTorbitˆk LSorbital= 2.67 × 1040kg ⋅ m2⋅ s−1ˆkEarth’s Spin Angular Momentum • Spin angular momentum about center of mass of earth • Period and angular velocity • Magnitude Lcmspin= Icmωspin=25meRe2ωspinˆnωspin=2πTspin= 7.29 × 10−5rad ⋅ s−1 Lcmspin= 7.09 × 1033kg ⋅ m2⋅ s−1ˆnEarth’s Angular Momentum For a body undergoing orbital motion like the earth orbiting the sun, the two terms can be thought of as an orbital angular momentum about the center-of-mass of the earth-sun system, denoted by S, Spin angular momentum about center-of-mass of earth Total angular momentum about S sys,cm ,cm ,ˆS S s e e cmr m v= × =L R p k spin 2cm cm spin spin2ˆ5e eI m Rω= =L nω LStotal= rs,emevcmˆk +25meRe2ωspinˆnRules to Live By: Kinetic Energy of Rotation and Translation Kinetic energy of rotation about center-of-mass Translational kinetic energy Kinetic energy is sum Krot=12Icmωcm2 K = Ktrans+ Krot=12mvcm2+12Icmωcm2 Ktrans=12mvcm2Demo B 113: Rolling CylindersTable Problem: Cylinder on Inclined Plane Energy Method A very thin hollow cylinder of outer radius R and mass m with moment of inertia I cm about the center of mass starts from rest and moves down an incline tilted at an angle from the horizontal. The center of mass of the cylinder has dropped a vertical distance h when it reaches the bottom of the incline. Let g denote the gravitational constant. The coefficient of static friction between the cylinder and the surface is . The cylinder rolls without slipping down the incline. Using energy techniques calculate the velocity of the center of mass of the cylinder when it reaches the bottom of the incline. θ µsConcept Question: Cylinder Race Two cylinders of the same size and mass roll down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which reaches the bottom first? 1) A 2) B 3) Both at the same time.Concept Question: Cylinder Race Different Masses Two cylinders of the same size but different masses roll down an incline, starting from rest. Cylinder A has a greater mass. Which reaches the bottom first? 1) A 2) B 3) Both at the same time.Rotational and Translational Motion Dynamics 8.01 W11D2 W11D1 and W11D2 Reading Assignment: MIT 8.01 Course Notes Chapter 20 Rigid Body: Translation and Rotational Motion Kinematics for Fixed Axis Rotation Sections 20.1-20.5 Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis, Sections 21.1-21.5Announcements Sections 1-4 No Class Week 11 Monday Sunday Tutoring in 26-152 from 1-5 pm Problem Set 8 due Week 11 Tuesday at 9 pm in box outside 26-152 No Math Review Week 11 Exam 3 Tuesday Nov 26 7:30-9:30 pm Conflict Exam 3 Wednesday Nov 27 8-10 am, 10-12 noon Nov 27 Drop DateAngular Momentum and Torque 1) About any fixed point S 2) Decomposition: LS=LSorbital+Lcmspin=rs,cm× mTvcm+Lcmspin τS=τS ,iext=dLSdti∑
View Full Document
Unlocking...