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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


Today we are going to discuss:

Chapter 4:

Motion in Two Dimensions: Section 4.1 Projectile motion: Section 4.2


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


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


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


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


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


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


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


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


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.


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)


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


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


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


gvR oo cossin2 0

2


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



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


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.


Thank youSee you on Monday

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