kinematics in two dimensions - uml.edu
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Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Lecture 5
Chapter 4
Kinematics in two Dimensions
Physics I
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Today we are going to discuss:
Chapter 4:
Motion in Two Dimensions: Section 4.1 Projectile motion: Section 4.2
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Motion in Two Dimensions(Vector Kinematics)
We use the vector mathematics to consider motion in more than one dimension.
Previously described 1D displacement as Δx, where motion could only be positive or negative.
In more than 1D, displacement are 2D vectors
v
r
r (t) v(t)
r
v v
v
rr
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Displacement in two dimensions
12 rrr
x
y
r1 r2
r
displacement (in unit vector notation):
jyyixxr ˆ)(ˆ)( 1212
1t
2t
In two dimensions, the displacement is a vector:
ij
jyixr ˆˆ222
jyixr ˆˆ
111
Rabbit’s path
jtyitxr ˆ)(ˆ)( Position of an object
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Instantaneous Velocity in two dimensions
x
y
r1 r2
r
tr
ttrrv
12
12
1t2t
Average velocity is the displacementdivided by the elapsed time
As Δt and Δr become smaller and smaller, the average velocity approaches the instantaneous velocity.
v limt 0
r
t dr
dtThe instantaneous velocity indicates how fastthe object moves and the direction of motionat each instant of time
dtrdv
1v
r12r
3r
2v
3v
x
y Tangent
jvivjdtdyi
dtdx
dtrdv yx
ˆˆˆˆ
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Instantaneous acceleration in two dimensions
Average acceleration
The instantaneous acceleration is in the direction of ,and is given by:12 vvv
1v
r12r 2v
x
y 1v
2v v
v
tv2
v1
t2 t1
tv
ta
0lim
2
2
dtrda
jdt
dvi
dtdva yx ˆˆ
jdt
ydidt
xd ˆˆ2
2
2
2
dtvd
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
jtyitxr ˆ)(ˆ)(
jvivjdtdyi
dtdx
dtrdv yx
ˆˆˆˆ
jdt
dvi
dtdva yx ˆˆ
jdt
ydidt
xd ˆˆ2
2
2
2
Instantaneous acceleration
kjiofterminwrittenavr ˆ,ˆ,ˆ,,
Instantaneous velocity
Position of an object
)(tx
)(ty
rt v
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Example A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn. The coordinates as functions of time are given
jtittr ˆ3ˆ)14()( 2
)14()( 2 ttxtty 3)(
jtyitxtr ˆ)(ˆ)()(
The x, y coordinates are components of the rabbit’s position vector
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Projectile motionA projectile is an object moving in two dimensions under the influence of Earth's
gravity; its path is a parabola.
?
r v
)(tx
)(ty
xagay
,
?
0
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
By splitting the equations of motion into component form, we can solve problems one direction at a time
tavvtatvrr
0
221
00
There is only one parameter, which connects X and Y motion: time X and Y motions are completely independent Work the problem as two one-dimensional problems
2
221
00 tatvxx xxf
(3)No time equation
Position equationtavtv xoxx )(
Velocity equation
(2)
(1)
2
221
00 tatvyy yyf
(3)No time equation
Position equationtavtv yoyy )(
Velocity equation
(2)
(1)
0xa gay
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
No forces in x direction (air resistance is neglected),
as a result, no acceleration
Equations in the X - direction
ax = 02
02
xx vv
xx vv 0
00 tvxx x
vx is constant!!!!
2
221
00 tatvxx xxf
(3)No time equation
Position equationtavtv xoxx )(
Velocity equation
(2)
(1)
So, from here, we can get only one equation.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Equations in the Y- direction
y
g
y
g
ay = ‐g
ay = g
if then
if then
)(2
02
02
221
00
0
yygvv
gttvyy
gtvv
yy
y
yy
)(2
02
02
221
00
0
yygvv
gttvyy
gtvv
yy
y
yy
x
x
2
221
00 tatvyy yyf
(3)No time equation
Position equationtavtv yoyy )(
Velocity equation
(2)
(1)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Helicopter flying horizontally at 70m/s wants to drop supplies on mountain top 200m below. How far in advance (horizontal distance) should the package be dropped?
Rescue Helicopter
• Draw diagram, choose coordinates• Knowns and unknowns• Divide equations into x and y• Solve, noting that in the x and y
calculations the common parameter is the time interval t
From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first?
A) the “dropped” ballB) the “fired” ballC) they both hit at the same timeD) it depends on how hard the ball
was firedE) it depends on the initial height
Both of the balls are falling vertically under the influence of
gravity. They both fall from the same height. Therefore, they will
hit the ground at the same time. The fact that one is moving
horizontally is irrelevant—remember that the x and y motions are
completely independent !!
OR
ConcepTest 1 Dropping the Ball I
221
00 gttvyy y
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
A golf ball is hit with initial velocity v0 at an angle θ0above the horizontal.
Draw diagram and choose coordinate systemFill in knowns
x
y
v0
v0x
v0y
Example (Golf Ball)
00 x 00 y
0xa gay 000 cos vv x 000 sin vv y
g
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal.
Find: the time of flight (how long the ball is in the air)This depends only on the y-component equations, as the motion in y direction stops the flight.
0 voyt 12
gt2y yo voyt 12
ayt2
Since both y0 and y are zero at the beginning/end
y=0 when the ball was hit
The second is the time of flight t 2v0 y
g
0 t(voy gt2
)
gv
t y021
2 tand 0
Cont. Example (Golf Ball).
0 0
y=0 when the ball landed
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
gvR oo cossin2 0
2
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal.
Find: Range (how far does ball travel on flat ground)Use constant x-velocity to calculate how far ball travels horizontally during time of flight (Range)
gv
gv
t y sin22 002 time of flight
000 cos vvv xx Constant x velocity
Range
Maximum RangeRmax vo
2
g
Cont. Example (Golf Ball)
200 tvxxR x
4512sin 00 or
gvo 0
2 2sin
when
Range
0
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal.
Find: trajectory (height y as a function of position x)
Since common parameter is time t, eliminate t to get y(x)
x v0 xt y voyt 12
gt2
y v0 y
v0 x
x g
2v0 x2
x2
y Ax Bx2
Cont. Example (Golf Ball)
Equation of parabola
t xv0 x
Which of the three
punts has the
longest hang time?
D) all have the same hang timeA B C
h
The time in the air is determined by the vertical motion!
Because all of the punts reach the same height, they all
stay in the air for the same time.
ConcepTest 2. Punts I
)(2 02
02 yygvv yy 2
21 tatvyy yoyo
A B
C) both at the same time
A battleship simultaneously fires two shells at two enemy
submarines. The shells are launched with the same initial
velocity. If the shells follow the trajectories shown, which
submarine gets hit first ?
The flight time is fixed by the motion in the y-direction. The higher an object goes, the longerit stays in flight. The shell hitting submarine #2 goes less high, therefore it stays in flight for less time than the other shell. Thus, submarine #2 is hit first.
ConcepTest 3 Punts II
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Solving Problems Involving Projectile Motion
1. Read the problem carefully, and choose the object(s) you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.
5. Examine the x and y motions separately.
6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.
7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 5 Danylov
Thank youSee you on Monday