Tài liệu Chapter 4: Motion in Two and Three Dimensions docx

Chapter 4 Motion in Two and Three Dimensions In this chapter we will continue to study the motion of objects without the restriction we put in chapter 2 to move along a straight line.. I

Chapter 4 Motion in Two and Three Dimensions

In this chapter we will continue to study the motion of objects without the restriction we put in chapter 2 to move along a straight line Instead we will consider motion in a plane (two dimensional motion) and motion in space

The following vectors will be defined for two- and three- dimensional

motion:

Average and instantaneous acceleration

We will consider in detail projectile motion and uniform circular motion as examples of motion in two dimensions

Finally we will consider relative motion, i.e the transformation of velocities between two reference systems which move with respect to each other with

Position Vector

The position vector of a particle is defined as a vector whose tail is at

a reference point (usually the origin O) and its tip is at the particle at

point P

The position vecto

Example: r in the f

r

igure is:

ˆ

r   xi  yj zk 

 3 ˆ 2 ˆ 5 ˆ 

P

t 2

t 1

Displacement Vector

For a particle that changes postion vector from to we define the displacement vector as follows:

r





    

The position vectors and are written in terms of components as:r r

r x i  y j z k r2 x i2ˆ  y j z k2 ˆ  2 ˆ

(4 -3)

   The displacement r can then be written as:

t + Δt

Average and Instantaneous Velocity

Following the same approach as in chapter 2 we define the average velocity as: displacement

average velocity =

time interval

avg

v





We define as the instantaneous velocity (or more simply the velocity) as the limit:

lim 0

r dr v

t dt t





 



t + Δt

If we allow the time interval t to shrink to zero, the following things happen:

1 Vector moves towards vector and 0

2 The direction of the ratio (and thus )approaches

r

v



 









the direction

of the tangent to the path at position 1

3 vavg  v



(4 - 5)

x

dx v

dt



y

dy v

dt



z

dz v

dt



The three velocity components are given by the equations:

dr v

dt







Average and Instantaneous Acceleration

The average acceleration is defined as:

change in velocity average acceleration =

time interval

avg

a



We define as the instantaneous acceleration as the limit:

lim

0

y

dv dv

t





 



The three acceleration components are given by the equations:

x x

dv a

dt

a

dt

a

dt



Note: Unlike velocity, the acceleration vector does not have any specific relationship with the path

dv a

dt







Projectile Motion

The motion of an object in a vertical plane under the influence of

gravitational force is known as “projectile motion”

The projectile is launched with an initial velocity

The horizontal and vertical velocity components are:

o

v 

cos

ox o o

v  v  voy  vo sin o

Projectile motion will be analyzed in

a horizontal and a vertical motion along the x- and y-axes,

respectively These two motions are independent of each other Motion along the x-axis has zero

acceleration Motion along the

y-axis has uniform acceleration ay = -g

(4-7)

g

(4 - 7)

 

0 The velocity along the x-axis does not change (eqs.1)

Horizontal Motion:

Vertical Mot

(eqs.2) Along the y-axis

x

y

a

v

g

t

a







2

2

0

2

is in free fall (eqs.3) (eqs.4)

If we eliminate between equations 3 and 4 we get

sin

2

2

sin

gt

v v







Here and are the coordinates

of the launching point For many problems the launching point is taken at the origin In this case

0 and 0

In this analysis of projectile moti

Note:

on we

neglect the effects of air resistance

g

(4-8)

   

2

2 2

cos (eqs.2) sin (eqs.4)

2

If we eliminate between equations 2 and 4 we get:

t

The equation of the p

an

2 cos This equatio

a

e

:

d

th

n

o

o

o

o

o

g

gt

x

t

v

t







2

scribes the path of the motion The path equations has the form: y ax bx  This is the equation of a parabola

The equation of the path seems too complicated to be useful Appearances can deceive: Complicated as it is, this equation can be used as a short cut in many projectile motion prob

Note:

lems

(4 - 9)

O A

R

t

2

v cos (eqs.1) cos (eqs.2)

sin (eqs.3) sin (eqs.4)

2 The distance OA is defined as the horizantal range

Horizontal Rang

At point A

e:

y

gt

R

2

we have: 0 From equation 4 we have:

Solution 1 0 This solution correspond to point O and is of no interest

Solution 2

y

t





sin 0 This solution correspond to point A

2

2 sin From solution 2 we get: If we substitute in eqs.2 we get:

gt v

v

g







o 2

2 sin cos sin 2 has its maximum value when 45

o

R

R

v R



 

/2

3/2

 sin

O

t

H

2

v H

g





2 2

The y-component of the projectile velocity is: sin

sin

At point A: 0 sin

sin

2

y

y

v

g

v

H

g









 

2

We can calculate the maximum height using the third equation of kinematics for motion along the y-axis: 2

In our problem: 0 , , sin , 0 , and

2

yo



sin



2

v H

g





Maximum height H (encore)

A

t

H

g

(4 -12)

Uniform circular Motion:

A particles is in uniform circular motion it moves on a circular path of

radius r with constant speed v Even though the speed is constant, the

velocity is not The reason is that the direction of the velocity vector

changes from point to point along the path The fact that the velocity

changes means that the acceleration is not zero The acceleration in uniform circular motion has the following characteristics:

1 Its vector points towards the center C of the circular path, thus the name

“centripetal”

2 Its magnitude a is given by the equation:

2

v a

r



R

Q

r

r

r

The time T it takes to complete a full revolution is

known as the “period” It is given by the equation:

2 r

T

v





(4 -13)

P

ˆ ˆ sin ˆ cos ˆ sin cos Here and are the coordinates of the rotating particle





We note that: P cos and P sin



2 2

/ sin

/ cos

y x

v r

a

a











sin

x

cos

y

Relative Motion in One Dimension:

The velocity of a particle P determined by two different observers A and B varies

from observer to observer Below we derive what is known as the “transformation equation” of velocities This equation gives us the exact relationship between the

velocities each observer perceives Here we assume that observer B moves with a

known constant velocity v BA with respect to observer A Observer A and B determine

the coordinates of particle P to be x PA and x PB , respectively

Here is the coordinate of B with respect to A

We take derivatives of the above equation: d x PA d x PB d x BA

PA PB BA

v  v  v If we take derivatives of the last equation and take

into account that dv BA 0

Even though observers A and B measure different velocities for P,

they measure the same acceler

Note:

ation

(4 -15)

Relative Motion in Two Dimensions:

Here we assume that observer B moves with a known constant velocity v BA with

respect to observer A in the xy-plane

Observers A and B determine the position vector of particle P to be

and , respectively

rPA rPB rBA We take the time derivative of both sides of the equation

PA PB BA

v   v   v 

If we take the time derivative of both sides of the last equation we have:

If we take into account that BA 0



As in the one dimensional case, even though observers A and B measure different velocities for P, they measure the same acceler

Note:

ation