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Physics 1710—Warm-up Quiz Why does a diver rotate faster when she tucks in her arms and legs? A. B. C. D. E. She increases her angular momentum. She increases her moment of inertia. She decreases her moment of inertia. She pushes against her inertia. None of the above 65% 24% 6% A 4% B C D 2% E Physics 1710—Chapter 13 App: E & E Analysis: • Like an ice skater. Why does an ice skater increase her angular velocity without the benefit of a torque? L = r x p= r x ( m v) = r x ( m r x ⍵) Li = mi ri 2 ⍵ z Lz = (∑i mi ri 2 ) ⍵z Lz = Iz ⍵z ; & ⍵z = Lz / Iz Therefore, a decrease in I ( by reducing r) will result in an increase in ⍵ even if dL/dt = 0! Physics 1710—Chapter 13 App: E & E Rotating Platform Demonstration Physics 1710—Chapter 13 App: E & E Analysis: •Why does an ice skater increase her angular velocity without the benefit of a torque? L=rxp = r x ( m v) = r x ( m r x ⍵) Li = mi ri 2 ⍵ Lz = (∑i mi ri 2 ) ⍵ Lz = I ⍵; & ⍵ = Lz / I Therefore, a decrease in I ( by reducing r) will result in an increase in ⍵. Physics 1710—Chapter 13 App: E & E How does a ladder stay up? Think! No Talking! Confer! Peer Instruction Time Physics 1710—Chapter 13 App: E & E 1′ Lecture •Static equilibrium (no translational or rotational acceleration) requires that all forces and torques to balance. • If the acceleration due to gravity is the same for all particles comprising a body the center of mass is the center of gravity. CM = CG if g the same for all mi . • The moduli of elasticity (Y, E, B) characterizes the stress-strain relation: • stress= modulus • strain; σ = Y ε Physics 1710—Chapter 13 App: E & E One more turn on L—Angular Momentum: L=rxp The angular momentum is the vector product of the moment arm and the linear momentum. ∑ T = d L/dt The net torque is equal to the time rate of change in the angular momentum. Physics 1710—Chapter 12 App: E & E Second Law of Motion F=ma Or F = dp/dt Then: r x F = d (r x p)/dt Torque = τ = d L/dt L = r x p is the “angular momentum.” Physics 1710—Chapter 12 App: E & E Second Law of Motion Torque = τ = d L/dt If τ = 0, then L is a constant. L = constant means angular momentum is conserved. Physics 1710—Chapter 12 App: E & E Second Law of Motion & Gyroscopic Precession L=Iω Torque = τ = d L/dt = d( I ω)/dt r τ = r x F = d L/dt = I (dω/dt); τ F = mg dω/dt = I-1 τ Physics 1710—Chapter 13 App: E & E What will happen to a tilted spinning top when it is supported on one end only? Why? Think! Peer Instruction Time No Talking! Confer! Physics 1710—Chapter 11 App: E & E What will happen to a tilted, spinning top when it is supported on one end only? Why? A. B. C. D. E. It will fall over because of gravity. It will spin faster because of I is changing. It will precess clockwise due to torque. It will precess counterclockwise due to torque. None of the above. 54% 30% 11% 5% 0% A B C D E Physics 1710—Chapter 11 App: E & E Experiment: Spinning Top •Observe the direction of rotation (ω) • Observe the motion (dω/dt) Physics 1710—Chapter 11 App: E & E Second Law of Motion & Gyroscopic Precession L=Iω Torque = τ = d L/dt = d( I ω)/dt r τ = r x F = d L/dt = I (dω/dt); τ F = mg dω/dt = I-1 τ Physics 1710—Chapter 11 App: E & E Torque and the Right Hand Rule: r X F Physics 1710—Chapter 11 App: E & E Gyroscopic Precession Torque = τ = d L/dt = d( I ω)/dt = I ( dω/dt) (Top view) Physics 1710—Chapter 11 App: E & E Fundamental Angular Momentum Fundamental unit of angular momentum = ℏ ℏ = 1.054 x 10 -34 kg‧m/s2 ICM⍵ ≈ ℏ ⍵ ≈ ℏ / ICM = 1.054 x 10 -34 kg‧m/s2 / (1.95 x 10 -46 kg‧m) = 5.41 x 10 11 rad/s Physics 1710—Chapter 11 App: E & E MRI (a.k.a. NMR: Nuclear Magnetic Resonance) Magnetic Resonance Imaging Magnetic Field Torque=? Hydrogen atoms Precession! MRI permits imaging of soft tissue due precession of the hydrogen in the water of the human body. Physics 1710—Chapter 11 App: E & E Teeter-totter: How does it balance? Fsupport = - Fg τnet = Στi F1 F2 For equilibrium: τnet = Στi = 0 Fnet = ΣFi = 0 Fg=F1+F2 Physics 1710—Chapter 11 App: E & E How does a ladder stay up? Good ideas? Physics 1710—Chapter 11 App: E & E Legends of the Fall—How a ladder stays up Fnet = 0 Tnet = 0 Fnet = 0 Physics 1710—Chapter 11 App: E & E Equilibrium: (equi= equal “=“; libium = scales “ ♎”) Equilibrium implies “balanced.” Fnet = d P/dt In equilibrium Fnet = 0 τnet = d L/dt In equilibrium τnet = 0 Physics 1710—Chapter 11 App: E & E Can an object be in equilibrium when the center of mass lies outside of the object? A. B. C. Yes. No. Depends. 56% 26% 18% A B C Physics 1710—Chapter 11 App: E & E Solution: In equilibrium Hollow Cone Fnet = 0 τnet = 0 Is it “stable?” i.e. does it recover from a small displacement ? Physics 1710—Chapter 11 App: E & E Center of Gravity If the acceleration due to gravity is the same for all parts of a body, then the center of mass corresponds to the center of gravity. CG =CM if g uniform Proof: Ti = Σxi Fgi = Σxi gmgi = Mg Σxi mgi /M= Fcg xcm Physics 1710—Chapter 11 App: E & E Center of Gravity (comparison) CM Ball Fg = - mg CM CG Moon Fg = - GmM/r2 Physics 1710—Chapter 11 App: E & E Elasticity Definitions: • Stress σ : the deforming force per unit area. • Strain ε : the unit deformation. Stress = modulus x strain σ = F/A = Y ε Physics 1710—Chapter 11 App: E & E Elasticity Stress σ – Strain ε “Curve” Elastic limit Stress σ (N/m2) • σ=Yε Strain ε = ΔL/L (%) Failure Physics 1710—Chapter 11 App: E & E Elasticity • • Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Tensile/Compressive Stress Young’s Modulus E Stress = modulus x strain σ = F/A = E ε = E ΔL/L L ΔL σ=Eε Physics 1710—Chapter 11 App: E & E Elasticity • • Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Shear Modulus G Stress = modulus x strain σ = F/A = G ε = G Δx/L L Δx σ Physics 1710—Chapter 11 App: E & E Elasticity • • Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Hydraulic Stress: Bulk Modulus B Stress = modulus x strain σ = F/A = p = B ε = B ΔV/V ΔV V p Physics 1710—Chapter 11 App: E & E Elasticity • • Stress: the deforming force per unit area. Strain: the unit deformation. – – – – – Tensile: “stretch” Compressive: “squeeze” Shear: “lean” Hydraulic: pressure Yield: permanently deformed Physics 1710—Chapter 11 App: E & E Summary • Static equilibrium implies that all forces and torques balance. • The center of mass is often the center of gravity. • The moduli of elasticity characterizes the stress-strain relation: • stress= modulus x strain Stress = modulus x strain σ = F/A = Y ε Physics 1710—Chapter 11 App: E & E Physics 1710—Chapter 11 App: E & E Why does the platform spin faster when he brings his arms in? Think! Peer Instruction Time No Talking! Confer! Physics 1710—Chapter 11 App: E & E Why does the platform spin faster when he brings his arms in? A. B. C. D. E. He increases his angular momentum. He increase his moment of inertia. He decrease his moment of inertia. He pushes against the inertia of the weights. None of the above 0% 0 of 1 10 Answer Now ! 0% 0% B A 0% 0% 0% E D C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Physics 1710—Chapter 11 App: E & E 10 Where should the fulcrum be place to balance the teeter-totter? A. B. C. 0% 0% 0 of 1 0% A 0% B C Answer Now ! 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Physics 1710—Chapter 11 App: E & E 10 Which way will the torque ladder move? A. Clockwise B. Counterclockwise C. Will stay balanced 0% 0% 0 of 1 0% A 0% B C Answer Now ! 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40