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UT Arlington PHYS 1443 - Lecture Notes

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Slide 1AnnouncementsExtra-Credit Special ProjectExtra Credit: Two Dimensional CollisionsCenter of Mass and Center of GravityMotion of a Group of ParticlesRotational Motion and Angular DisplacementSI Unit of the Angular DisplacementExample 10 – 1Ex. Adjacent Synchronous SatellitesEx. A Total Eclipse of the SunAngular Displacement, Velocity, and AccelerationRotational KinematicsProblem Solving StrategyEx. 10 – 4: Rotational KinematicsExample for Rotational Kinematics cnt’dWednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #14Wednesday, June 29, 2011Dr. Jaehoon Yu•Motion of a Group of Particles•Rotational Motion•Rotational Kinematics•Relationship Between Angular and Linear quantities•TorqueWednesday, June 29, 2011 2Announcements•Planetarium Show extra credit–Must obtain the signature of the “Star Instructor” AFTER watching the show on the ticket stub–Tape one side of the ticket stubs on a sheet of paper with your name on it–Submit it on the last class Thursday, July 7•Late submissions will not be accepted!!!PHYS 1443-001, Summer 2011 Dr. Jaehoon YuTuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu3Extra-Credit Special Project•Derive the formula for the final velocity of two objects which underwent an elastic collision as a function of known quantities m1, m2, v01 and v02 in page 8 of lecture note on Tuesday, June 28, in a far greater detail than in the note.–20 points extra credit•Show mathematically what happens to the final velocities if m1=m2 and explain in detail in words the resulting motion.–5 point extra credit•NO Credit will be given if the process is too close to the note!•Due: Start of the class Tuesday, July 5Extra Credit: Two Dimensional Collisions•Proton #1 with a speed 5.0x106 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ. This must be done in much more detail than the book or on page 13 of lecture note on Tuesday, June 28.•10 points•Due beginning of the class Wednesday, July 6Tuesday, June 28, 2011 4PHYS 1443-001, Summer 2011 Dr. Jaehoon YuWednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu5The net effect of these small gravitational forces is equivalent to a single force acting on a point (Center of Gravity) with mass M.Center of Mass and Center of GravityThe center of mass of any symmetric object lies on the axis of symmetry and on any plane of symmetry, if the object’s mass is evenly distributed throughout the body.Center of GravityHow do you think you can determine the CM of an object that is not symmetric?ΔmiCMAxis of symmetryOne can use gravity to locate CM.1. Hang the object by one point and draw a vertical line following a plum-bob.2. Hang the object by another point and do the same.3. The point where the two lines meet is the CM. ΔmigSince a rigid object can be considered as a collection of small masses, one can see the total gravitational force exerted on the object as What does this equation tell you?The CoG is the point in an object as if all the gravitational force is acting on!Wednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu6Motion of a Group of ParticlesWe’ve learned that the CM of a system can represent the motion of a system. Therefore, for an isolated system of many particles in which the total mass M is preserved, the velocity, total momentum, acceleration of the system areVelocity of the systemCMvrTotal Momentum of the systemAcceleration of the systemCMarThe external force exerting on the systemIf net external force is 0System’s momentum is conserved.What about the internal forces?CMdrdt=r1iidmrdt M� �=� �� ��r1iidrmM dt=�riimvM=�riimvMM=�riimv=�rCMdvdt=r1iidmvdt M� �=� �� ��r1iidvmM dt=�riimaM=�rCMMa=riima=�rWednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu7In a simplest kind of rotation, points on a rigid object moves on circular paths about the axis of rotation.Rotational Motion and Angular DisplacementThe angle swept out by a line passing through any point on the body and intersecting the axis of rotation perpendicular is called the angular displacement.qD=qoq-It’s a vector!! So there must be directions…How do we define directions?+:if counter-clockwise-:if clockwiseThe direction vector points gets determined based on the right-hand rule.These are just conventions!!Wednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu8 (in radians)q =For one full revolution:2 radp = q=SI Unit of the Angular DisplacementArc lengthRadius=srSince the circumference of a circle is 1 radian is1 rad36012radp= �o1801radp= �o57.3@o2πrAnd one degrees is1o21360p= �oo1180p= �[email protected]@ �ooHow many degrees are in one radian?How radians is one degree?How many radians are in 10.5 revolutions?360o2 radpDimension? None10.5rev=Very important: In solving angular problems, all units, degrees or revolutions, must be converted to radians.( )21 radpWednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu9Example 10 – 1 A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m? (a) One radian is 360o/2π. Thusl =(b) Since l=rΘ and for small angle arc length is approximately the same as the chord length.rq=Wednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu10Synchronous satellites are put into an orbit of radius 4.23×107m. If the angular separation of the two satellites is 2.00 degrees, find the arc length that separates them.Ex. Adjacent Synchronous Satellites2.00degs= (in radians)q =Arc lengthRadius=srConvert degrees to radians2 rad360degp� �=� �� �0.0349 radrq=( )( )74.2310m0.0349 rad�61.4810m (920 miles)=�What do we need to find out?The Arc length!!!Wednesday, June 29, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu11The diameter of the sun is about 400 times greater than that of the moon. By coincidence, the sun is also about 400 times farther from the earth than is the moon. For an observer on the earth, compare the angle subtended


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UT Arlington PHYS 1443 - Lecture Notes

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