PHYS 1443 – Section 003 Lecture #20AnnouncementsSimple Harmonic and Uniform Circular MotionsExample 13.7Damped OscillationMore on Damped OscillationNewton’s Law of Universal GravitationMore on Law of Universal GravitationFree Fall Acceleration & Gravitational ForceExample 14.2Example 14.3Kepler’s Laws & EllipseThe Law of Gravity and the Motion of PlanetsKepler’s Third LawExample 14.4Kepler’s Second Law and Angular Momentum ConservationMonday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #20Monday, Nov. 25, 2002Dr. Jaehoon Yu1. Simple Harmonic and Uniform Circular Motions2. Damped Oscillation3. Newton’s Law of Universal Gravitation4. Free Fall Acceleration and Gravitational Force5. Kepler’s Laws6. Gravitation Field and Potential EnergyToday’s homework is homework #20 due 12:00pm, Monday, Dec. 2!!Monday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements•Class on Wednesday•Remember the Term Exam on Monday, Dec. 9 in the class–Covers chapters 11 – 15–Review on Wednesday, Dec. 4Monday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Simple Harmonic and Uniform Circular MotionsUniform circular motion can be understood as a superposition of two simple harmonic motions in x and y axis.When the particle rotates at a uniform angular speed , x and y coordinate position becomeSince the linear velocity in a uniform circular motion is A, the velocity components arext=0xyOPAxyOPAQxyt=t=t+xyOPAQvvxxvxyOPAQaaxSince the radial acceleration in a uniform circular motion is v2/A=2, the components arexacosA tAcosysinA tAsinsinv tA sinyvcosv tA coscosa tA cos2yasina tA sin2Monday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Example 13.7A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to the right. A) Determine the x coordinate as a function of time.Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the angular frequency is 8.00rad/s. Therefore the equation of motion in x direction isSince x=2.00, when t=0However, since the particle was moving to the right =-48.2o, Using the displacementx cos00.300.2 m 2.4800.8cos00.3 tmxFind the x components of the particle’s velocity and acceleration at any time t.xvLikewise, from velocityxacosA tm 00.8cos00.32.4800.300.2cos1dtdx 2.4800.8sin/0.242.4800.8sin00.800.3 tsmtdtdv 2.4800.8cos/1922.4800.8cos00.80.242 tsmtMonday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Damped OscillationMore realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.xFThe angular frequency for this motion isThe solution for the above 2nd order differential equation is xWe express the angular frequency asThis equation of motion tells us that when the retarding force is much smaller than restoration force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops.Let’s consider a system whose retarding force is air resistance R=-bv (b is called damping coefficient) and restoration force is -kxmk0Where the natural frequency 0bvkx xmadtdxbkx 22dtxdm tAcostmbe222mbmk2220mbDamping TermMonday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6More on Damped OscillationkAbv maxAs the retarding force becomes larger, the amplitude reduces more rapidly, eventually stopping at its equilibrium positionThe motion is called Underdamped when the magnitude of the maximum retarding force Rmax = bvmax <kAThe system is Critically dampedHow do you think the damping motion would change as retarding force changes?0Under what condition this system does not oscillate?If the retarding force is larger than restoration forcekAbvR maxmaxThe system is OverdampedWhat do you think happen?Once released from non-equilibrium position, the object would return to its equilibrium position and stops.Once released from non-equilibrium position, the object would return to its equilibrium position and stops, but a lot slower than beforemb200mb 2mk2Monday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. 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?21221rmmFg1110673.6GG is the universal gravitational constant, and its value isThis constant is not given by the theory but must be measured by experiment.With G21221rmmGFgUnit?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.Monday, Nov. 25, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8It means that the force exerted on the particle 2 by particle 1 is attractive force, pulling #2 toward #1.More on Law of Universal GravitationConsider two particles exerting gravitational forces to each other.Gravitational force is a field force: Forces act on object without physical contact between the objects at all times, independent of medium between them.1222112ˆrrmmGF The gravitational force exerted by a finite size, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distributions was concentrated at the center.m1m2rF21F1212ˆrTwo objects exert gravitational force on each other following Newton’s 3rd law.Taking as the unit vector, we can write the force m2
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