Rotational Dynamics 8.01t Nov 8, 2004Fixed Axis Rotation: Angular Velocity and Angular Acceleration Angle variable θ d θAngular velocity ω≡ dt 2d θAngular acceleration α≡ 2dt Mass element ∆miRadius of orbit r ⊥ ,iFixed Axis Rotation: Tangential Velocity and Tangential Acceleration • Individual mass elements ∆ mi• Tangential velocity vtan, i = r⊥ ,iω • Tangential acceleration atan,i = r⊥ ,iα 2 • Radial Acceleration vtan, i = r⊥ ,iω 2= , arad i r⊥ ,iNewton’s Second Law • Tangential force on mass element produces torque • Newton’s Second Ftan, i =∆ma tan, iLaw i iFtan, i =∆mr ⊥,iα GG • Torque τG Si =rSi ×Fi, ,Torque Torque about is S: GGGτSi = rSi×F , , i • Counterclockwise • perpendicular to the plane () α , ⊥,i iτSi = rFtan, i=∆ m r ⊥,i 2Moment of Inertia • Total torque is the sum over all mass elements iN iN iN= = = total S i =∑ rFtan, i =∑∆ m r ⊥τ S =τ S,1 +τ S,2 + ... =∑τ , (⊥ ,i i i= 1 i= 1 i= 1 • Moment of Inertia about S: iN= ()2IS =∑∆ m r ⊥ ,i i i= 1 2 • Unit: [kg − m ] total • Summary: τ S = ISα )α ,i 2Rotational Work G • tangential force Ftan, i = Ftan, iθˆ G • displacement vector ∆rSi =( rS,⊥) ∆θθˆ , i • infinitesimal work GG∆Wi = Ftan, i ⋅∆rSi = Ftan, iθˆ ⋅( rS,⊥) ∆θθˆ = Ftan, i( rS,⊥) ∆θ , i iRotational Work • Newton’s Second Law Ftan, i =∆ma i tan, i • tangential acceleration atan, i =(rS,⊥) α i 2 • infinitesimal work ∆Wi =∆m rS,⊥ )α∆θ i ( i • summation i ( 2 ⎞ 2 ⎞ ∆W =⎜⎛∑∆m rS,⊥)i ⎟⎠α∆θ= ⎛⎜⎜ ∫ dm rS,⊥) ⎟αθ=ISαθ ( ∆ ∆ ⎝ i ⎝body ⎟⎠Rotational Work =• infinitesimal rotational work ∆WIS α∆θ • torque τ=IS αS • infinitesimal rotational work ∆W =∆τS θ θθf θθf = = • Integrate total work W =∫ d W =τSd θ∫ θθ0 θθ0 = =212cm cm cmK Iω=Rotational Kinetic Energy Rotational kinetic energy 1 21 i ( 22Kcm i =∆m v , =∆m rcm,⊥)ω ,2 i cm i i cm2 , =⎜⎛∑2 ∆m rcm,⊥)⎟ω=⎜⎜2 body ( 2 ⎞ Kcm =∑Kcm i 1 i ( 2 ⎞ cm 2 ⎛1 ∫dm rcm,⊥)⎟ωcm 2 = 1Icmωcm i ⎝ii ⎠⎝ ⎟⎠ 2 • Total kinetic energy 2Ktotal = Ktrans + Krot = 1 mv 2 +1 I ωcm cm cm 2 2 2PRS - kinetic energy A disk with mass M and radius R is spinning with angular velocity ω about an axis that passes through the rim of the disk perpendicular to its plane. Its total kinetic energy is: 1. 1/4 M R2 ω2 4. 1/4M R ω2 2. 1/2M R2 ω2 5. 1/4 M R ω2 3. 3/4 M R2 ω2 6. 1/2 M R ωRotational Work-Kinetic Energy Theorem • angular velocity ω ddt ≡θ • angular acceleration α ddt ≡ω • infinitesimal rotational work dω dθαθ=IS dθ=Idω=IdωωdW rot =I d S dt SdtS • integrate rotational workωωf ωωf = = 1Wrot =∫ dWrot =∫ I d ωω= ISω2 f −12 ISω02 S 2ωω0 ωω0 = =Rotational Work-Kinetic Energy Theorem • Fixed axis passing through a point S in the body 122, 1 Icm ωcm ,0 = Krot , f − Krot ,0 ≡∆ Krot Wrot = 2 Icm ωcm f − 2 • Rotation and translation Wtotal =∆Ktrans +∆ Krot ⎛12 −12 ⎞ ⎛ 1 21 = + , Wtotal =Wtrans +Wrot ⎝⎜2 mvcm f 2 mvcm ,0 ⎟ ⎜ Icm ωf − Icm ω02 ⎟⎞ ⎠ ⎝ 2 2 ⎠Concept Question • 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 is moving faster at thebottom? 1) A 2) B 3) Both have the sameConcept Question • 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 has more total kinetic energy at the bottom? 1) A 2) B 3) Both have the sameRotational Power • rotational power is the time rate of doing rotational work dWrotProt ≡ dt • product of the applied torque with the angular velocity dθ≡ dWrot =τS,⊥ dt =τωProtdt S,⊥Class Problem A turntable is a uniform disc of mass m and a radius R. The turntable is spinning initially at a constant frequency f . The motor is turned off and the turntable slows to a stop in t seconds. Assume that the angularacceleration is constant. The moment of inertia of the disc is I. a) What is the initial angular velocity of the turntable? b) What is the angular acceleration of the turntable? c) What is the magnitude of the frictional torque acting on the disc? d) How much work is done by the frictional torque? e) What is the change in kinetic energy of the turntable? f) Graph the rotational power as a function of time.Simple Pendulum Pendulum: bob hanging from end of string • Pivot point • bob of negligible size • massless stringSimple Pendulum: Torque Diagram torque about the pivot point GGτS = r G , ×mg = lrˆ × mg (sin θθˆ+cos rˆ)=− lmg sin θkˆ− s m angular acceleration G 2 α = d θkˆ(vector quantity) 2dt • Points along axis • Positive or negativeSimple Pendulum: Rotational Equation of Motion • moment of inertial of a point mass about the pivot point 2IS = ml G • Rotational Law of Motion τG S = ISα • Simple pendulum oscillator equation 2 −lmg sinθ= ml 2 d θ dt2Simple Pendulum: Small Angle Approximation Angle of oscillation is small sinθ≅θ 2 • Simple harmonic oscillator ddt θ≅−gl θ2 2dx k • Analogy to spring equation =− mx2dt • Angular frequency of oscillation ωpendulum ≅ g l 2π l•Period T0 = ≅2πω gpSimple Pendulum: Mechanical Energy • released from rest at an angle θ0Extra Topic: Simple Pendulum: Mechanical Energy • Velocity dθ vtan = l dt 1 ⎛ dθ⎞2 • Kinetic energy Kf = 1 mv tan2 = m l ⎜2 2 ⎝ dt ⎠⎟ • Initial energy E0 = K0 + U = mgl (1 cos θ0)− 0 1 ⎛ dθ⎞2 • Final energy Ef = Kf + U = m l ⎟+ mgl (1 cos θ)f 2 ⎝⎜ dt ⎠− • Conservation of energy 1 ⎛ dθ⎞2 ml ⎟+ mgl (1 cos θ)= mgl (1 cos θ )− 02 ⎝⎜ dt ⎠−Simple Pendulum: Angular Velocity Equation of Motion • Angular velocity dθ 2g dt = l (cosθ− cosθ0) • Integral form dθ∫ (cos θ− cosθ ) =∫ 2 lgdt 0Simple Pendulum: Integral Form • Change of variables bsin a = sin (θ 2) b = sin (θ 2)0 • Integral form da 12 =∫ gdt 2 l∫(1− b sin2 a)• 12Power series approximation 2 − 2(1− b sin2 a)= 1+ 1 b2sin a + 3 b4 sin4
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