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Chapt. 10 Rotation of a Rigid Object About a Fixed Axis 10.1 Angular Position, Velocity, and Acceleration Translation: motion along a straight line (circular motion ?) Rotation: surround itself, spins rigid body: no elastic, no relative motion rotation: moving surrounding the fixed axis, rotation axis, axis of rotation s 360 , 1rev = 360 o = 2π (rad ) , 1rad = = 57.3 o r 2π Angular displacement: ∆θ = θ 1 − θ 2 Angular position: θ = An angular displacement in the counterclockwise direction is positive θ − θ 1 ∆θ = Angular velocity: averaged ω = 2 t 2 − t1 ∆t instantaneous: ω = dθ , rpm = rev per minute dt Angular acceleration: averaged α = instantaneous: α = ∆ω ∆t dω dt Sample Problem: The disk in Fig. 11-5a is rotating about its central axis like a merry-go-round. The angular position θ(t) of a reference line on the disk is given by θ = −1.00 − 0.600t + 0.250t 2 with t in seconds, q in radians, and the zero angular position as indicated in the figure. a) Graph the angular position of the disk versus time from t = -3.0 s to t = 6.0 s. Sketch the disk and its angular position reference line at t = -2.0 s, 0 s, and 4.0 s, and when the curve crosses the t axis. b) At what time tmin does θ(t) reach the minimum value? What is that minimum value? c) Graph the angular velocity ω of the disk versus time from t = -3.0 s to t = 6.0 s. Sketch the disk and indicate the direction of turning and the sign of ω at t = -2.0 s and 4.0 s, and also at tmin. 1 Right-hand rule Specify Your Axis 10.2 Rotational Kinematics: The Rigid Object Under Constant Angular Acceleration Sample Example: A grindstone rotates at constant angular acceleration α = 0.35 rad/s2. At time t = 0, it has an angular velocity of ω0 = -4.6 rad/s and a reference line on it is horizontal, at the angular position θ0 = 0. α= dω = 0.35rad / s 2 , dω = 0.35dt , ω − ω 0 = 0.35t , ω = −4.6 + 0.35t dt dθ = ω = −4.6 + 0.35t , θ − θ 0 = −4.6t + 0.175t 2 , θ = −4.6t + 0.175t 2 dt a) At what time after t = 0 is the reference line at the angular position θ = 5.0 rev? c) At what time t does the grindstone momentarily stop? 2 10.3 Angular and Translational Quantities The position: s = rθ The speed (velocity?): v = rω , T = 2πr v v2 The acceleration: a t = rα , a r = = rω 2 r Sample Example: Figure shows a centrifuge used to accustom astronaut trainees to high accelerations. The radius r of the circle traveled by an astronaut is 15 m. (a) At what constant angular speed must the centrifuge rotate if the astronaut is to have a linear acceleration of magnitude 11g? 11 × 9.8 = 2.68rad / s 15 (b) What is the tangential acceleration of the astronaut if the centrifuge accelerates at a constant rate from rest to the angular speed of (a) in 120 s? rω 2 = 15 ⋅ ω 2 = 11 × 9.8 , ω = a t = rα = rα avg = 15 ω −0 120 = 15 2.68 = 0.34m / s 2 120 10.4 Rotational Kinetic Energy K= 1 1 1 2 2 2 mi vi = ∑ miω 2 ri = ∑ mi ri ω 2 ∑ 2 2 2 I = ∑ mi ri (rotational inertia), K = 2 1 2 Iω 2 Example: The Oxygen Molecule Consider an oxygen molecule (O2) rotating in the xy plane about the z axis. The rotation axis passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is 2.66 x 10-26 kg, and at room temperature the average separation between the two atoms is d = 1.21 x 10-10 m. particles.) (The atoms are modeled as (a) Calculate the moment of inertia of the molecule about the z axis. 2 d d I = ∑ mi ri 2 what is the radius ri ? I = m + m 2 2 2 (b) If the angular speed of the molecule about the z axis is 4.60 X 1012 rad/s, what is its rotational kinetic energy? z 1 1 Ek = mv 2 -> ER = Iω 2 2 2 M Example: Four Rotating Object a m b y m M x 3 Calculate the moment of inertia if the rotation is about the z axis and about the y axis. I = ∑ mi ri 2 what is the radius ri ? I z = ma 2 + ma 2 + Mb 2 + Mb 2 I y = Mb 2 + Mb 2 z 10.5 Calculation of Moments of Inertia Cartesian Coordinate: x, y, z Spherical Coordinate: r ,θ , φ z = r cos θ x = r sin θ cos φ , y = r sin θ sin φ x Cylindrical Coordinate: r ,θ , z z z y y x x = r cosθ , y = r sin θ (a) M = 2πRρ I = ∫ R 2 ρRdθ = MR 2 (b) M = π ( R2 − R1 ) ρ 2 2 R2 2π R2 R1 0 R1 I = ∫ r 2 ∫ rρdθ dr = ∫ 2πρr 3 dr = M 2 2 ( R1 + R2 ) 2 (c) M = πR 2 L ρ L R 2π 0 0 0 I = ∫ ∫ r 2 ∫ rρdθ drdz R I = L ∫ 2πr 3 ρdr = 0 1 MR 2 2 (d) 4 M = πR 2 Lρ I= L 2 R 2π ∫ ∫ ∫ (z − 2 ) + r 2 sin 2 θ ρrdθdrdz = L 0 0 2 (e) from (d), R << L , I = 1 1 MR 2 + ML2 4 12 1 ML2 12 (f) M = 4π 3 R ρ 3 R π 2π I = ∫ ∫ ∫ r 2 sin 2 θρr sin θdφ rdθ dr = 0 0 0 2 MR 2 5 (g) π M I = σ ∫ (R sin θ ) 2πR sin θdθ = 4πR 2 0 2 2 π M ∫0 (R sin θ ) 2πR sin θdθ = 2 2 2 2 ∫ (1 − cos θ )R d (cosθ ) = 3 MR 1 2 2 −1 (h) M = 2πRρ I = ∫ R 2 sin 2 θρRdθ = MR 2 = 1 MR 2 2 (i) ( ) b/2 a/2 M 1 I= x 2 + y 2 dxdy = M (a 2 + b 2 ) ∫ ∫ ab −b / 2 −a / 2 12 Parallel axis theorem: I = I CM + Mh 2 [ ] I = ∫ r 2 dm = ∫ ( x − a ) + ( y − b ) dm ( ) 2 2 ( ) I = ∫ x 2 + y 2 dm − 2a ∫ xdm − 2b ∫ ydm + ∫ a 2 + b 2 dm I = I cm + Mh 2 5 2 Sample Example: (a) What is the rotational inertia Icom of this body about an axis through its center of mass, perpendicular to the rod as shown? 2 2 1 L L I = ∑ mi ri = m + m = mL2 2 2 2 2 (b) What is the rotational inertia I of the body about an axis through the left end of the rod and parallel to the first axis? I = ∑ mi ri 2 = m ⋅ 0 2 + mL2 = mL2 2 use parallel axis theorem: I = I CM 2 1 L L + M = mL2 + 2m = mL2 2 2 2 Example: Rotating Rod A uniform rod of length L and mass M is free to rotate on a frictionless pin through one end. The rod is released from rest in the horizontal position. (a) What is the angular speed of the rod at its lowest position? 1. I cm = 1 L 1 1 ML2 , I = I cm + M ( ) 2 = ML2 , Energy conservation: Iω 2 = Mgh , 12 2 3 2 3g 11 L 1 ML2ω 2 = Mg , ω = , vcm = Lω 23 2 L 2 2. Mgh = 1 1 2 I cmω 2 + Mvcm 2 2 10.6 Torque r r Torque is a vector: N = τ = r × F = rF sin φ , τ = τ 1 + τ 2 + τ 3 + ... Torque <-> Force, What is the difference? Torque can increase or decrease the angular velocity. What about force? Example: The net torque on a cylinder τ = (R2T2 − R1T1 )kˆ r 6 10.7 The Rigid Body Under a Net Torque Ft = Ftan gent , Fr = Falong _ the _ radius Ft = Ftan gent is used for angular acceleration Ft = mat = mrα , τ = rFt = mr 2α = Iα The torque acting on the particle is proportional to its angular acceleration. Example: Rotating Rod A uniform rod of length L and mass M is attached to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. initial angular acceleration? What is the 2 I CM 1 ML2 L = ML2 , I pivot = I CM + M = 12 3 2 τ= L 3g 3 Mg = Iα , α = , at = Lα = g at right end 2 2L 2 Example: Angular Acceleration of a Wheel A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless horizontal axle. A light cord wrapped around the wheel supports an object of mass m. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord. I= 1 MR 2 , mg − T = ma , τ = RT = Iα , a = Rα 2 mg − I a = ma , a = RR mg m+ I R2 , T= Iα Ia mgI = 2 = , α= R R mR 2 + I 7 Example: Atwood Machine Revisited Two blocks having masses m1 and m2 are connected to each other by a light cord that passes over two identical frictionless pulleys, each having a moment of inertia I and radius R. Find the acceleration of each block and the tension T1, T2, and T3 in the cord. (No slipping) m1 g − T1 = m1a , R(T1 − T2 ) = Iα , R(T2 − T3 ) = Iα T3 − m2 g = m2 a , Rα = a a= (m1 − m2 )g m1 + m2 + 2I R2 10.8 Energy Considerations in Rotational Motion r r dW = F ⋅ ds = (F sin φ )rdθ dW = τ ⋅ dθ P= dW dθ =τ = τω dt dt ∑τ = Iα = I dω dω dθ dω =I =I ω dt dθ dt dθ work-kinetic energy theorem: ∑τ ⋅ dθ = Iωdω , ωf ωi Rotational motion θf W = ∫τ ⋅ dθ θi KR = 1 ∑W = ∫ Iωdω = 2 Iω 2 f 1 − Iωi2 2 Linear motion xf W = ∫ Fx dx xi 1 2 Iω 2 P = τω p = Iω 1 K = mv 2 2 P = Fv p = mv dL ∑τ = dt dp ∑ F = dt 8 Example: Rotating Rod Revisited A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. Determine the tangential speed of the CM and the tangential speed of the lowest point on the rod when it is in the vertical position. Mgh = Mg Mg L 1 2 1 L2 = Iω , I CM = ML2 , I pivot = I CM + M 2 2 12 4 3g L 1 ML2 2 = ω , ω= 2 2 3 L vCM = 3 gL L ω= 2 2 vend = Lω = 3gL 10.9 Rolling Motion of a Rigid Object Rolling as rotation and translation combined θ vcm = Rω , vedge = Rω sin θ ⋅ iˆ − Rω cos θ ⋅ ˆj vedge = Rω (1 + sin θ )iˆ − Rω cos θ ⋅ ˆj Rolling as pure rotation 9 1. K= 1 1 Mv 2 + Iω 2 2 2 2. K= 1 1 1 I pω 2 , I P = I CM + MR 2 , K = I CM ω 2 + MR 2ω 2 2 2 2 Sample Problem: A uniform solid cylindrical disk, of mass M = 1.4 kg and radius R = 8.5 cm, rolls smoothly across a horizontal table at a speed of 15 cm/s. What is its kinetic energy K? 1 1 2 K = I CM ω 2 + MvCM 2 2 2 1 MR 2 vCM 1 3 2 2 = + MvCM = MvCM = 24 2 2 R 2 4 mg sin θ − f s = ma , a = Rα Iα = R ⋅ f s , f s = I a= a R2 mg sin θ I m+ 2 R Sample Example: A uniform ball, of mass M = 6.00 kg and radius R, rolls smoothly from rest down a ramp at angle θ = 30.0°. (a) The ball descends a vertical height h = 1.20 m to reach the bottom of the ramp. What is its speed at the bottom? 1 1 2 2 MvCM + I CM ω 2 = Mgh , I = MR 2 , Rω = vCM 2 2 5 2 v 10 gh 1 12 2 MvCM + MR 2 CM2 = Mgh , vCM = = 4.1m / s 2 25 7 R (b) What are the magnitude and direction of the friction force on the ball as it rolls down the ramp? 10 a= g sin θ 4.9 I 2 = = 3.5m / s 2 , f s = 2 a = Ma = 8.4 N I 2 5 R 1+ 1+ 2 5 MR 11