Feb 6, 2013 PHYSICS I Lecture 5

Transcription

1 Feb 6, 213 PHYSICS I Lecture 5 Course website: faculty.uml.edu/pchowdhury/95.141/ Course: UML95141SPRING213 Lecture Capture h"p://echo36.uml.edu/chowdhury213/physics1spring.html Register your i>clickers online at

2 Last Lecture Today Chapter 3 Beyond 1-D Vectors Relatie Velocity Chapter 3 Vector kinematics Projectile motion

3 2 Vector Kinematics: 1D From two measurements of time and position displacement = 2 " 1,t 1 1 2,t 2 = 2 " 1 a ector tip-to-tail 1 time interal t = t 2 " t 1 a scalar aerage elocity ag = t a ector

4 Vector Kinematics Kinematics in more than one dimension Preiously described 1D displacement as Δ, where motion could only be positie or negatie. In more than 1D, displacement is a ector r(t) (t) r

5 y Vector Kinematics : 2D r 1 r r2 Δ Δr = = 2 1 r r 2 1 displacement (in unit ector notation): r = ( 2 " 1 )î + (y 2 " y 1 ) ĵ + (z 2 " z 1 ) ˆk aerage elocity = r t

6 Aerage to Instantaneous Just like in 1D, let Δt get smaller and smaller. As Δt tends to zero, Δr tends to zero as well, but the ratio of Δr/Δt tends to a finite alue Instantaneous lim elocity r = = d r t " t dt r r

7 Instantaneous Velocity Deriatie of the position ector with respect to time: r = (t)î + y(t) ĵ + z(t) ˆk = d r dt = d dt î + dy dt = î + y ĵ + z ˆk ĵ + d dt ˆk

8 Acceleration Vector Aerage acceleration = t = 2 " 1 t 2 " t 1 Instantaneous acceleration a = lim t " t = d dt a = d dt = d dt î + d y dt ĵ + d y dt ˆk = d 2 dt 2 î + d 2 y dt 2 ĵ + d 2 z dt 2 ˆk

9 Eample Problem Gien position as a function of time, find position, instantaneous elocity and instantaneous acceleration at t=3s r(t) = (4t 2 1)î + 2 ĵ + (t 2 ) ˆk (t) = (8t)î + (2t) ˆk a(t) = 8î + 2 ˆk Plug in t=3s to get the final answers

10 Eample Problem By splitting the equations of motion into component form, we can sole problems one direction at a time An object starts from rest at the origin, with a constant acceleration in the + direction of +2 m/s 2 and in the y direction of 1 m/s 2. What is the displacement of the object at t = 4 s? = t a t 2 y = y t a y t 2

11 Motion with constant acceleration Vector equations shorthand for separate scalar equations in and y coordinates at at t r r + = + + = t a t a t y y t a t a t y y y y y + = + + = + = + + = For projectile motion, a =, a y = -g gt gt t y y t y y y = + = = + = Vector Kinematics

12 Projectile Motion: Clicker Quiz A helicopter moing at a constant horizontal elocity to the right drops a package when at position A. Which of the marked trajectories is closest to that obsered by a stationary person on the ground?

13 Projectile Motion Red ball fired horizontally at blue ball Graity OFF Graity ON Red ball fired at an angle towards blue ball

14 Eample (Rescue Helicopter) Helicopter flying horizontally at 7m/s wants to drop supplies on mountain top 2m below. How far in adance (horizontal distance) should the package be dropped? Draw diagram, choose coordinate system Knowns and unknowns Diide equations into and y Sole, noting that in the and y calculations the common parameter is the time interal t

15 Eample (Rescue Helicopter) = y = = 7 m/s y = a = a y = 9.8 m/s 2 y = y o + oy t a y t 2 2 = (9.8)t 2 =? y = 2 m t = 6.39s common parameter is time t Sole y-equations to find t Plug t into -equations to find = o + o t a t 2 = + 7(6.39)+ = 447m

16 Relatie Motion: Clicker Quiz A helicopter moing at a constant horizontal elocity to the right drops a package when at position A. Which of the marked trajectories is closest to that obsered by a person on the helicopter?

17 Eample (Golf Ball) A golf ball is hit with initial elocity at an angle θ aboe the horizontal. Find Equations of motion Draw diagram and choose coordinate system Fill in knowns =, y = y = cos, y = sin a =, a y = g = o + o t a t 2 = ( o cos )t y θ y = y o + oy t a y t 2 y = ( o sin )t 1 2 gt 2 Again, common parameter is time t

18 Eample (Golf Ball) A golf ball is hit with initial elocity at an angle θ aboe the horizontal. The time of flight (how long the ball is in the air) This depends only on the y-component equations, as the acceleration due to graity is only in the y-direction. y = y = oy t 1 2 gt 2 o + oy t a yt 2 Since both y and y are zero = t( oy gt 2 ) Two solutions for t where y= t = and 2 y g The second is the time of flight t = 2 y g

19 Eample (Golf Ball) A golf ball is hit with initial elocity at an angle θ aboe the horizontal. Find Range (how far does ball trael on flat ground) Use constant -elocity to calculate how far ball traels horizontally during time of flight (Range) Constant = cos time of flight t = 2 y g = 2 sin g Range R = 2 2 o sin cos o g = 2 o sin 2 g Maimum Range R ma = 2 o occurs when sin(2θ )=1 or θ =45 g

20 Eample (Golf Ball) A golf ball is hit with initial elocity at an angle θ aboe the horizontal. Trajectory (height y as a function of position ) = t y = oy t 1 2 gt 2 Since common parameter is time t, eliminate t to get y() t = y = $ y # & '# " % " g 2 2 $ & 2 % y = A B 2 Equation of parabola

21 The Speed Bus In the moie, a bus tries to jump a gap in the highway.

22 So we know: The Speed Bus o, 1) DRAW DIAGRAM 2) Determine knowns 3) Pick equations o t = y = = 31.3m/s Δ = 5m Is this een close to possible? Δy 1 2 gt 2 = 15.24m 31.3m / s.5s = 1 2 (9.8)*(.5)2 m 1.2m Bad news

23 The Speed Bus But.

24 The Speed Bus But. 3º It looks like the bus is magically launched at an angle of 3º Now can the bus make it? CHOWDHURY PHYSICS I SPRING 213 LECTURE 5

25 Speed Bus with Magic Launch o 3º Range R = 2 o sin 2 g = (31.3)2 sin m

26 The Speed Bus Let s check CHOWDHURY PHYSICS I SPRING 213 LECTURE 5

27 Summary Relatie Velocity Vector kinematics Projectile motion