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UT Arlington PHYS 1441 - Lecture Notes

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Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #18Wednesday, Apr. 7, 2004Dr. Jaehoon Yu• Torque• Moment of Inertia• Rotational Kinetic Energy• Angular Momentum and its conservation• Conditions for Equilibrium• ElasticityToday’s homework is #10 due noon, next Wednesday, Apr. 13, 2004!!Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Exam Result and Announcements• Mid-term grade one-on-one discussion– I had only 12 students so far.– In my office, RM 242-A, SH– During office hours: 2:30 – 3:30 pm, Mondays and Wednesdays• Next Monday: Last name starts with A – M• Next Wednesday: Last name starts with N – ZWednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3TorqueTorque is the tendency of a force to rotate an object about an axis. Torque, τ, is a vector quantity.≡τMagnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.FφdLine of ActionConsider an object pivoting about the point Pby the force F being exerted at a distance r. PrMoment armThe line that extends out of the tail of the force vector is called the line of action.The perpendicular distance from the pivoting point P to the line of action is called Moment arm.When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. d2F221τττ+=∑2211dFdF−==φsinrF FdWednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Moment of Inertia Rotational Inertia:What are the dimension and unit of Moment of Inertia?∑≡iiirmI22mkg⋅[]2MLMeasure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion.Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building.dmrI∫≡2For a group of particlesFor a rigid bodyWednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Torque & Angular AccelerationLet’s consider a point object with mass m rotating on a circle.What does this mean?The tangential force Ftand radial force FrThe tangential force FtisWhat do you see from the above relationship?mrFtFrWhat forces do you see in this motion?tFThe torque due to tangential force FtisrFt=τατI=Torque acting on a particle is proportional to the angular acceleration.What law do you see from this relationship?Analogs to Newton’s 2ndlaw of motion in rotation.αmr=rmat=α2mr=αI=tma=Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Rotational Kinetic EnergyWhat do you think the kinetic energy of a rigid object that is undergoing a circular motion is? Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object isSince moment of Inertia, I, is defined asKinetic energy of a masslet, mi, moving at a tangential speed, vi, isrimiθOxyviiKRK∑=iiirmI22=ωIKR21The above expression is simplified as221iivm=2=ω221iirm∑=iiK∑2=iiirmω2212⎟⎠⎞⎜⎝⎛=∑ωiiirm221Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Kinetic Energy of a Rolling SphereSince vCM=RωLet’s consider a sphere with radius R rolling down a hill without slipping.=KRxhθvCMω212CMCMvIR⎛⎞=⎜⎟⎝⎠Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hillWhat is the speed of the CM in terms of known quantities and how do you find this out?K2221CMCMvMRI⎟⎠⎞⎜⎝⎛+=2221CMCMvMRI⎟⎠⎞⎜⎝⎛+=Mgh=2/12MRIghvCMCM+=212CMIω2212MRω+212CMMv+Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Conservation of Angular MomentumRemember under what condition the linear momentum is conserved?Linear momentum is conserved when the net external force is 0.ifLL=JGJGThree 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. iLJGconstp ==∑extτWhat does this mean?Mechanical EnergyLinear MomentumAngular MomentumconstL =dtpdF ==∑0=dtLd0fL=JGconstant=ifpp=JGJGii f fKU K U+=+Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Example for Angular Momentum ConservationA star rotates with a period of 30days 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 start of radius 3.0km. Determine the period of rotation of the neutron star. Tπω2=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 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=Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Similarity 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=JG GIτα=GJGcosWFdθ=cosPFvθ=τω=P221mvK =221ωIKR=LMθWτθ=vmp =ωIL =Wednesday, Apr. 7, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu11Conditions for EquilibriumWhat do you think does the term “An object is at its equilibrium” mean?∑= 0FThe 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 static 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 opposite direction acting on a rigid object as


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