Phys 1501Q 1st Edition Lecture 6Outline of Last Lecture: Projectile MotionI. Projectile MotionII. Effect of Gravity on Projectile MotionIII. Example: Shooting a Monkey in a TreeIV. Path Followed By ProjectileV. Horizontal Motion = constantVI. Vertical Motion = changingVII. Relating Horizontal to Vertical MotionsHorizontal Range EquationVIII. Horizontal Range CalculationIX. Angle That Optimizes RangeX. Baseball Horizontal Range ExampleOutline of Current Lecture: Relative Motion & Uniform Circular MotionI. Relative MotionII. Reference FrameIII. Reference Frame ExampleIV. 2-D Relative Motion ExampleV. Moving Sidewalk ExampleVI. Crossing a River ExampleThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.VII. Uniform Circular Motion/Centripetal Accelerationa. Velocity Vector b. Finding accelerationVIII. Example of Circular MotionIX. Getting Accelerationa. Using Derivativeb. Finding magnitudec. Finding directionX. Circular Motion Important Notesa. Angles/Angular Frequencyb. Periodc. CircumferenceXI. Calculating Centripetal Acceleration ExampleCurrent Lecture: Relative Motion & Uniform Circular Motion I. Relative MotionRelative motion is the calculation of the motion of an object with regard to some other moving object or frame.II. Reference Frame- Surroundings which are perceived to be at rest- Motions are universally agreeable in a reference frame- Interpretation of motion depends on reference frameo Inertial Reference FrameNon accelerating reference frameIII. Reference Frame Example- Observer on ground sees planes motion relative to ground (VPlane ,Ground)- Passenger on plane sees plane as stationarySuppose someone throws a ball at VBall ,Planetowards the front of the plane.o Passengers see ball at VBall ,Planeo Observer sees VBall ,Ground(if the plane were clear to see through technically)oVBall ,Ground=VBall ,Plane+VPlane ,Ground (Planes cancel out)IV. 2-D Relative Motion Example- Position and velocity vectors are related by vector additionAssume Frame B moves at constant velocity VBA with respect to A.Position of PRPA in frame ARPB in frame BRBA = Position of B with respect to A Vector AdditionRPA=RPB+RBA(Bs cancel out)V. Moving Sidewalk ExampleA man is walking to a hydrant on the other side of a moving sidewalk. The man’s velocityis 1m/s. The sidewalk’s velocity is 2m/s and is 4m wide.1. Find time to get acrosst=∆ xV=4 m1 m/s=4 sec2. Find velocity of man with respect tohydrantVMan , Hydrant=VMan , Street+VStreet , Hydrant3. Find how far away from hydrant man ends up after crossing the street∆ y=VStreet , Hydrant∗t  2m/s * 4 m = 8m away from hydrantVI. Crossing a River Example- River flows with respect to Earth Vℜ- Boat moves with respect to river VBR- Boat moves with respect to Earth Vℜ+VBR = VBE (R’s cancel)Part I- Assume a river flows at 1m/s in the positive x direction. The boat travels with respectto ater at 2m/s in the positive y direction.Vℜ+VBR = VBE  VBE = 2j + 1iVBE=√12+22 = 2.24 m/sθ=tan−121 = 63.4°Part II- If the river is 100 m across, how long will it take to get from one side to the other?time=distancerate  100 m2.24 m/ s∗sinθ = 50sPart III- If we want to head straight across, we have to turn into the currento What velocity of the boat with respect to river should we choose?o What is the time to get across?We know VBR is 2m/s, find VBE.√VBE2+12 = 2VBE = 1.73 m/ssinθ=VBEVBR = 120°Find time to cross:time=distancerate  1 m1.73 m/ s = 57.7 sec*This is longer than 50 sec going straight as you are being slowed by the currentVII. Uniform Circular Motion / Centripetal Acceleration- Velocity is constant- Involves acceleration EVEN THOUGH velocity is constanto Why?Velocity is composed of magnitude and direction. The magnitude is constant but the direction is changing as it always points to the center of the circle.AKA CENTRIPETAL ACCELERATIONVIII. Example of Circular MotionA ball on the end of a string is being moved at a constant speed around a circle.Velocity Vector - Velocity vector is tangent to circle- It is perpendicular to the acceleration- X component: Vx = -sinΘ- Y component: Vy = cos ΘAcceleration- X component = Rcos Θ- Y component = Rsin ΘIX. Getting Accelerationa. Using DerivativeTake derivative of velocity to get accelerationb. Finding magnitudeV2R=magnitudec. Finding direction- “a” is directed inward towards radius- “a” is perpendicular to VX. Circular Motion Important Notesa. Angles/Angular Frequencyo Θ increases as you go around the circle at a constant rateo Θ = ωto ω (omega) = angular frequency (radians/sec)b. Periodo Period is the time it takes to go around the circle, Tc. Circumferenceo 2πr = vto 1 revolution  T = 2π/ωo ω = v/rXI. Calculating Centripetal Acceleration ExampleSwing a ball on a string in a circle with a .5 m radius and 2 revolutions per second.o Find speed2 revolutions/ 1 secPeriod = 1 revolution/.5 secT= ½ ½ = 2π/ ωω = 12.6 radians/secV = ω*rV = 6.28 m/s o Find centripetal accelerationV2R = a  6.282.5  a =