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UT Arlington PHYS 1443 - Lecture Notes

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PHYS 1443 – Section 003 Lecture #10Exam ResultsSlide 3Exam problems 14 – 18Slide 5Slide 6Newton’s Law of Universal GravitationFree Fall Acceleration & Gravitational ForceExample for GravitationKepler’s Laws & EllipseThe Law of Gravity and the Motion of PlanetsKepler’s Third LawExample of Kepler’s Third LawKepler’s Second Law and Angular Momentum ConservationMotion in Accelerated FramesExample of Motion in Accelerated FramesWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #10•Exam review•Newton’s Law of Gravitation•Kepler’s LawsWednesday,Oct. 1, 2003Dr. Jaehoon YuNo homework today!!Homework #5 deadline extended till noon, next Wednesday, Oct. 10!!Wednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu2Exam Results•Class average: 53•Top score: 90.5•Remember that–There are three exams of which two bests are used for final grade–Each exam accounts for 25%  50% total•Other factors–Lab: 20%–Homework: 15%–Quiz: 15%–Extra credits: 10%•Final grades will be assigned on a sliding scale•Exam can be easily made up by other creditsWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Wednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Exam problems 14 – 18 A 5 kg block is placed on a frictionless inclined plane of angle 30o and pushed up the plane with a horizontal force of magnitude 30 N. 14. Draw a free-body diagram for this motion.30oMaFgn+x+yF=30NFree-bodyDiagram+x+ynF= -Mg=30o=30oF=30NWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5yFxFx comp.y comp.0yaoo30cos30sin FMgxMayMaoo30sin30cos FMgn016. What are the direction and magnitude of block’s acceleration? 2/30.0530cos3030sin8.95smoothusMFMgaxoo30cos30sinThus the block’s acceleration is up the incline. The magnitude is222/3.0 smaaayxWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu617. What is the magnitude of force exerted on the plane by the block? Since this force is the same magnitude as the normal force but opposite direction, one can obtain the magnitude from y-component of the net forceoo30sin30cos FMgnN5730sin3030cos8.95 oo18. What is the smallest horizontal force needed to move the block up the plane? In order for the block to be pushed up the plane, the component of the horizontal force up along the plane must overcome the component of the gravitational force downward.30cos30sin FMgFxNMgF 2830tan8.9530cos30sinoooNF 28030sin30cos FMgnFy0Wednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Newton’s Law of Universal GravitationPeople have been very curious about the stars in the sky, making observations for a long time. But the data people collected have not been explained until Newton has discovered the law of gravitation. Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.How would you write this principle mathematically?21221rmmFg1110673.6GG is the universal gravitational constant, and its value isThis constant is not given by the theory but must be measured by experiment.With G21221rmmGFgUnit?22/ kgmN This form of forces is known as an inverse-square law, because the magnitude of the force is inversely proportional to the square of the distances between the objects.Wednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8Free Fall Acceleration & Gravitational ForceWeight of an object with mass m is mg. Using the force exerting on a particle of mass m on the surface of the Earth, one can get•The gravitational acceleration is independent of the mass of the object•The gravitational acceleration decreases as the altitude increases•If the distance from the surface of the Earth gets infinitely large, the weight of the object approaches 0.What would the gravitational acceleration be if the object is at an altitude h above the surface of the Earth?mgWhat do these tell us about the gravitational acceleration?gF2EERmMGg2EERMG'mg2rmMGE 2hRmMGEE'g 2hRMGEEWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9Example for GravitationUsing the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.gSince the gravitational acceleration is So the mass of the Earth is GgRMEE2Therefore the density of the Earth is 2EERMG2111067.6EERMEEVM324EERGgR3EGRg4333611/1050.51037.61067.6480.93mkgWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu10Kepler’s Laws & EllipseKepler lived in Germany and discovered the law’s governing planets’ movement some 70 years before Newton, by analyzing data.Newton’s laws explain the cause of the above laws. Kepler’s third law is the direct consequence of law of gravitation being inverse square law.1. All planets move in elliptical orbits with the Sun at one focal point.2. The radius vector drawn from the Sun to a planet sweeps out equal area in equal time intervals. (Angular momentum conservation)3. The square of the orbital period of any planet is proportional to the cube of the semi-major axis of the elliptical orbit.F1F2bcaEllipses have two different axis, major (long) and minor (short) axis, and two focal points, F1 & F2 a is the length of a semi-major axisb is the length of a semi-minor axisWednesday, Oct. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu11The Law of Gravity and the Motion of Planets•Newton assumed that the law of gravitation applies the same whether it is on the Moon or the apple on the surface of the Earth.•The interacting bodies are assumed to be point like particles.Therefore the centripetal acceleration of the Moon, aM, isNewton predicted that the ratio of the Moon’s acceleration aM to the apple’s acceleration g would be gaMREMoonApplegaMv234/1070.280.91075.2 smaMNewton also calculated the Moon’s orbital acceleration aM from the knowledge of its distance from the Earth and its orbital period, T=27.32 days=2.36x106sMaThis means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration is reduced by the square of the ratio. This proves that the inverse square


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UT Arlington PHYS 1443 - Lecture Notes

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