PHYS 1441 – Section 002 Lecture #18AnnouncementsSlide 3Slide 4Reminder: Extra-Credit Special ProjectCenter of MassMotion of a Diver and the Center of MassEx. 7 – 12 Center of MassVelocity of the Center of MassAnother Look at the Ice Skater ProblemRotational Motion and Angular DisplacementSI Unit of the Angular DisplacementExample 8 – 1Ex. Adjacent Synchronous SatellitesEx. A Total Eclipse of the SunWednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #18Wednesday, Nov. 17, 2010Dr. Jaehoon Yu•Center of Mass•Velocity of the Center of Mass•Fundamentals of Rotational Motion•Equations of Rotational Kinematics•Relationship Between Angular and Linear QuantitiesAnnouncements•Two colloquia this week–One at 4pm Wednesday, Nov. 17–Another at 4pm Friday, Nov. 19Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2Wednesday, Nov. 17, 20103PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Nov. 17, 20104PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5Reminder: Extra-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 12 of this lecture note in a far greater detail than the note.–20 points extra credit•Show mathematically what happens to the final velocities if m1=m2 and describe in words the resulting motion.–5 point extra credit•Due: Start of the class Monday, Nov. 29Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6Center of MassWe’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situations objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and represents the motion of the system as if all the mass is on that point. Consider a massless rod with two balls attached at either end.CMx �The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if the entire mass of the system is on the center of mass. ar= Fur∑/ MWhat does above statement tell you concerning the forces being exerted on the system?m1m2x1x2The position of the center of mass of this system is the mass averaged position of the systemxCMCM is closer to the heavier object1 1 2 2m x m x+1 2 m m+Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7Motion of a Diver and the Center of MassDiver performs a simple dive.The motion of the center of mass follows a parabola since it is a projectile motion.Diver performs a complicated dive.The motion of the center of mass still follows the same parabola since it still is a projectile motion.The motion of the center of mass of the diver is always the same.Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8Ex. 7 – 12 Center of MassThee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and x3=6.0m. Find the position of CM. Using the formula for CMiiiiiCMmxmx1.0M �12.03MM= =M M M+ +4.0( )m=5.0M+ �6.0M+ �Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9cmxD =cmv =Velocity of the Center of MassIn an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.1 1 2 21 2m x m xm mD + D+cmxtD=D1 1 2 21 2m x t m x tm mD D + D D=+1 1 2 21 2m v m vm m++Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu10Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a velocity of +2.5 m/s. What is the velocity of the center of mass of this system? Another Look at the Ice Skater Problem100v m s=12.5fv m s=+0cmv =cmfv =200v m s= m1v1+m2v2m1+m2=021.5fv m s=- m1v1 f+m2v2 fm1+m2 =54⋅ +2.5( )+88⋅ −1.5( )54+88=3142=0.02≈0m sWednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11In 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 perpendicularly 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, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu12 (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= �[email protected]@ �o2πrAnd one degrees is1o21360p= �oo1180p= �[email protected]@ �ooHow many degrees is in one radian?How radians is one degree?How many radians are in 10.5 revolutions?360o 2πrr=2 radpDimension? None10.5rev =Very important: In solving angular problems, all units, degrees or revolutions, must be converted to radians. 10.5rev ⋅2πradrev=( )21 radpWednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu13Example 8 – 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. Thus 3 ×10−4radl = = 3×10−4rad( )× 360o2π rad( ) =0.017o(b) Since l=rΘ and for small angle arc length is approximately the same as the chord length.rq = 100m ×3×10−4rad = 3 ×10−2m=3cmWednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu14Synchronous satellites are put into an orbit whose radius is 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 =( )( )74.23 10 m 0.0349 rad�61.48 10 m (920 miles)= �What do we need to find out?The Arc length!!!Wednesday, Nov. 17, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu15The diameter of the sun is about 400 times
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