Physics 201: Lecture 5, Pg 1 Lecture 5 l Goals: (Chapter 4.1-3) Introduce position, displacement,...
-
Upload
whitney-thornton -
Category
Documents
-
view
214 -
download
0
Transcript of Physics 201: Lecture 5, Pg 1 Lecture 5 l Goals: (Chapter 4.1-3) Introduce position, displacement,...
Physics 201: Lecture 5, Pg 1
Lecture 5 Goals: (Chapter 4.1-3)
Introduce position, displacement, velocity and acceleration in 2D Address 2D motion in systems with constant acceleration (i.e. both magnitude and acceleration) Discuss horizontal range (special)
Key point 1: In many case motion occurs 2D with constant acceleration (typically on the surface of a planet)
Key point 2: The “superposition principle” allows us to discuss the x & y motion individually
Physics 201: Lecture 5, Pg 2
Decomposing vectors Any vector can be resolved into components along the x and y axes
tan-1 ( y / x )
22 yxr
y
x
(x,y)
rry
rx
j i
sin
cos
yxr
ryr
rxr
y
x
vy
vx
(vx,vy)
vvy
vx
ay
ax
(ax,ay)
aay
ax
j i
sin
cos
yx
y
x
vvv
vv
vv
tan-1 ( vy / vx )
22yx vvv
j i
sin
cos
yx
y
x
aaa
aa
aa
22yx aaa
tan-1 ( ay / ax )
Physics 201: Lecture 5, Pg 3
Dynamics: Motion along a line but with a twist(2D dimensional motion, magnitude and directions)
Particle motions involve a path or trajectory
In 2D the position of a particle is r = x i + y j
and this vector is dependent on the origin. (i , j unit vectors )
ri’
rf’
O’Displacement, r, is independent of origin
Physics 201: Lecture 5, Pg 4
Motion in 2DPosition
Displacement
Velocity (avg)
ffii trtr , and ,
if rrr
t
rv
avg
Physics 201: Lecture 5, Pg 5
Instantaneous Velocity
As t 0 r shrinks and becomes tangent to the path The direction of the instantaneous velocity is along a line
that is tangent to the path of the particle’s direction of motion.
v
dt
rd
t
rv
t
0lim
Physics 201: Lecture 5, Pg 6
Average Acceleration The average acceleration of particle motion reflects
changes in the instantaneous velocity vector (divided by the time interval during which that change occurs).
Instantaneous
acceleration
t
vv
tv if
avga
avga
t
va
t
0
lim
Physics 201: Lecture 5, Pg 7
Instantaneous Acceleration
The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path
Changes in a particle’s path may produce an acceleration The magnitude of the velocity vector may change The direction of the velocity vector may change Both may change simultaneously (depends: path vs time)
t
va
t
0
lim
Physics 201: Lecture 5, Pg 8
2 D Kinematics Position, velocity, and acceleration of a particle:
r = x i + y j
v = vx i + vy j (i , j unit vectors )
a = ax i + ay j
2
2
dtxd
ax 2
2
dtyd
ay
dt
dxvx
dt
dyvy
)( txx )( tyy
ji
ji
)jˆ(
yx vv
dtdy
dtdx
dtyixd
dtrd
v
Physics 201: Lecture 5, Pg 9
2 D Kinematics (special case) If ax and ay are constant then
tavtv
tatvyty
tavtv
tatvxtx
t
tatyty
tatxtx
yyiy
yyii
xxix
xxii
i
yyii
xxii
)(
)(
)(
)(
0 if and
v)(
v)(
221
221
221
221
In many case one of the a’s is zero
Physics 201: Lecture 5, Pg 10
Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like? x
t = 0
4y
x vs y
Position
Special Case:
ax=0 & ay= -g
vx(t=0) = v0 vy(t=0) = 0
vy(t) = – g t
x(t) = x0 v0 t
y(t) = y0 - ½ g t2
Physics 201: Lecture 5, Pg 11
Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the curve!
x
t = 0
4y
x vs yVelocity
Physics 201: Lecture 5, Pg 12
Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the curve!
Acceleration may or may not be!
x
t = 0
4y
x vs yAcceleration
PositionVelocity
Physics 201: Lecture 5, Pg 13
Another trajectory
x vs yt = 0
t =10
Can you identify the dynamics in this picture?
How many distinct regimes are there?
0 < t < 3 3 < t < 7 7 < t < 10
I. vx = constant = v0 ; vy = 0
II. vx = vy = v0
III. vx = 0 ; vy = constant < v0
y
x
What can you say about the acceleration?
Physics 201: Lecture 5, Pg 14
Exercises 1 & 2 Trajectories with acceleration
A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces.
Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C Sketch the shape of the path
from B to C. At point C the engine is turned off.
Sketch the shape of the path
after point C (Note: a = 0)
Physics 201: Lecture 5, Pg 15
Exercise 1Trajectories with acceleration
A. A
B. B
C. C
D. D
E. None of these
B
C
B
C
B
C
B
C
A
C
B
D
From B to C ?
Physics 201: Lecture 5, Pg 16
Exercise 2Trajectories with acceleration
A. A
B. B
C. C
D. D
E. None of these
C
C
C
C
A
C
B
D
After C ?
Physics 201: Lecture 5, Pg 17
Kinematics in 2 D; Projectile Motion The position, velocity, and acceleration of a particle
moving in 2-dimensions can be expressed as:
r = x i + y j v = vx i + vy j a = ax i + ay j
Special Case: ax=0 & ay= -g
vx(t) = v0 cos
vy(t) = v0 sin – g t
x(t) = x0 v0 cos t
y(t) = y0 v0 sin t - ½ g t2
cos00 vvx
0yv0v
Physics 201: Lecture 5, Pg 18
Kinematics in 2 D; Horizontal Range
Given v0 and how far will and object travel horizontally?Let y0 = 0 = yinitial = yfinal x0 =0
Again: ax=0 & ay= -g
1. vx(t) = v0 cos
2. vy(t) = v0 sin – g t
3. x(t) = 0 v0 cos t = R (range)
4. y(t) = 0 = 0v0 sin t - ½ g t2
4 gives: 0 = t (v0 sin - ½ g t) t = 0, 2 v0 sin g
R = v0 cos 2 v0 sin g = v02 sin 2 / g
Maximum when dR/d = 0 cos 2 = 0 or 45°
cos00 vvx
0yv0v
Physics 201: Lecture 5, Pg 19
Parabolic trajectories ( v=10 m/s , g = - 10 m/s2)
90° R:0.0m H:5.0m t=2.00s
75° R:5.0m H:4.7m t=1.93s
60° R:8.7m H:3.7m t=1.73s
45° R:10.0m H:2.5m t=1.41s
30° R:8.7m H:1.2m t=1.00s
15° R:5.0m H:0.3m t=0.52s
Physics 201: Lecture 5, Pg 20
Example Problem
A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1st one he fires from the cannon and it lands 200 m away. Simultaneously the 2nd cannon ball is dislodged and falls directly down. The 2nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s2 & no air resistance)?
Physics 201: Lecture 5, Pg 21
Example Problem
A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1st one he fires from the cannon and it lands 200 m away. Simultaneously the 2nd cannon ball is dislodged and falls directly down. The 2nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s2 & no air resistance)?
Find height of wall from 2nd cannon ball
0 = h + 0 – ½ 10 t2 h = 5(4) m = 20 m
Find angle
0 = h + v sin (t+2) – ½ 10 (t+2)2 0 = 20 m + 4v sin - 80 m
R = 200 m = v cos (t+2) 4v = 200 / cos
Combining 60 m = 200 m tan 16.7 degrees
Physics 201: Lecture 5, Pg 22
Example Problem
Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel be for it hit the level ground (if g= -10 m/s2 & no air resistance)?
Physics 201: Lecture 5, Pg 23
Example Problem
If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel before it hits the level ground
(assume g= -10 m/s2 & no air resistance)? at t=0 the knife is at (0.0 m, 1.0 m) with vy=0 after t the kinfe is at (x m, 0.0 m) x = x0 + vx t and y = y0 – ½ g t2
So x = 10 m/s t and
0.0 m = 1.25 m – ½ g t2 2.5/10 s2 = t2 t = 0.50 sx = 10 m/s 0.50 s = 5 .0 m
Physics 201: Lecture 5, Pg 24
Fini
For Thursday, Read all of Chapter 4