Tuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #15Tuesday, June 27, 2006Dr. Jaehoon Yu• Angular Momentum & Its Conservation• Similarity of Linear and Angular Quantities• Conditions for Equilibrium• Mechanical Equilibrium• How to solve equilibrium problems?• Elastic properties of solidsTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements• Reading assignments– CH12 – 7 and 12 – 8 • Last quiz tomorrow– Early in the class– Covers Ch. 10 – 12• Final exam– Date and time: 8 – 10am, Friday, June 30– Location: SH103– Covers: Ch 9 – what we cover tomorrow– No class this ThursdayTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3Angular Momentum of a Rotating Rigid BodyLet’s consider a rigid body rotating about a fixed axisiiiivrmL =Each particle of the object rotates in the xy plane about the z-axis at the same angular speed, ωατIdtdLzext==∑Thus the torque-angular momentum relationship becomesWhat do you see?Since I is constant for a rigid bodyMagnitude of the angular momentum of a particle of mass miabout origin O is mivirixyzOpφL=rxprmSumming over all particle’s angular momentum about z axis∑=iizLL()ω∑=iiizrmL2dtdLzα is angular accelerationThus the net external torque acting on a rigid body rotating about a fixed axis is equal to the moment of inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.ω2iirm=()∑=iiirmω2ωI=dtdIω=αI=Tuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4Example for Rigid Body Angular MomentumA rigid rod of mass M and length l is pivoted without friction at its center. Two particles of mass m1and m2are attached to either end of the rod. The combination rotates on a vertical plane with an angular speed of ω. Find an expression for the magnitude of the angular momentum.IThe moment of inertia of this system isαFirst compute net external torqueθτcos21lgm=1m1gxyOlm1m2θm2gIf m1= m2, no angular momentum because net torque is 0. If θ=+/−π/2, at equilibrium so no angular momentum.⎟⎠⎞⎜⎝⎛++==212314mmMlILωωFind an expression for the magnitude of the angular acceleration of the system when the rod makes an angle θ with the horizon.2τττ+=1extThus α becomes21mmrodIII++=2112Ml=+⎟⎠⎞⎜⎝⎛++=212314mmMl cos222θτlgm−=()2cos21mmgl−=θIext∑=τ()⎟⎠⎞⎜⎝⎛++−=21211314cos21mmMlglmmθ()lgmmMmm/31cos22111⎟⎠⎞⎜⎝⎛++−=θ2114ml+2214mlTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Conservation of Angular MomentumRemember under what condition the linear momentum is conserved?Linear momentum is conserved when the net external force is 0.ifLL=GGThree important conservation laws for isolated system that does not get affected by external forcesAngular momentum of the system before and after a certain change is the same.By the same token, the angular momentum of a system is constant in both magnitude and direction, if the resultant external torque acting on the system is 0. iLGpconst=Gextτ=∑GWhat does this mean?Mechanical EnergyLinear MomentumAngular MomentumL=G0dpFdt==∑GGdLdt=G0fL=Gconstant=ifpp=GGii f fKU K U+=+constTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Example for Angular Momentum ConservationA star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star. ω=What is your guess about the answer?The period will be significantly shorter, because its radius got smaller.fiLL=Let’s make some assumptions:1. There is no external torque acting on it2. The shape remains spherical3. Its mass remains constantThe angular speed of the star with the period T isUsing angular momentum conservationThusfωffiIIωωι=fiIIιω=ifiTmrmrπ222=fTfωπ2=iifTrr⎟⎟⎠⎞⎜⎜⎝⎛=22days30100.10.324×⎟⎠⎞⎜⎝⎛×=days6107.2−×=s23.0=2π TTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.MomentumRotationalKineticKinetic EnergyPowerWorkWorkWorkTorqueForceForceAccelerationSpeedAngle (Radian)DistanceLength of motionMoment of InertiaMassMassRotationalLinearQuantities2Imr=rvt∆=∆tθω∆=∆vat∆=∆tωα∆=∆Fma=GGIτα=GGWFd=⋅GGPFv=⋅GGτω=P221mvK =221ωIKR=LMθWτθ=pmv=GGLIω=GGTuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Conditions for EquilibriumWhat do you think the term “An object is at its equilibrium” means?F=∑GThe object is either at rest (Static Equilibrium) or its center of mass is moving with a constant velocity (Dynamic Equilibrium). Is this it? When do you think an object is at its equilibrium?Translational Equilibrium: Equilibrium in linear motion The above condition is sufficient for a point-like particle to be at its translational equilibrium. However for object with size this is not sufficient. One more condition is needed. What is it? Let’s consider two forces equal magnitude but in opposite direction acting on a rigid object as shown in the figure. What do you think will happen?CMddF-FThe object will rotate about the CM. The net torque acting on the object about any axis must be 0. For an object to be at its static equilibrium, the object should not have linear or angular speed. τ=∑G0=CMv 0=ω00Tuesday, June 27, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu9More on Conditions for EquilibriumTo simplify the problem, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case? The six possible equations from the two vector equations turns to three equations.What happens if there are many forces exerting on the object?0F =∑G0τ=∑G∑= 0xF∑= 0zτOF1F4F3F2F5r5O’r’If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis.∑= 0yFWhy is this true?Because the object is not movingnot moving, no matter what the rotational axis is, there should not be in motion. It is simply a matter of mathematical calculation.ANDTuesday, June 27, 2006 PHYS 1443-001, Summer
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