chapter 3 kinematics in two dimensions;...

Chapter 3

Kinematics in Two Dimensions; Vectors

• Vectors and Scalars

• Addition of Vectors – Graphical Methods (One and Two-Dimension)

• Multiplication of a Vector by a Scalar

• Subtraction of Vectors – Graphical Methods

• Adding Vectors by Components

• Projectile Motion

• Projectile Motion Is Parabolic

• Relative Velocity

Adding Vectors by ComponentsAdding Vectors by Components

Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicularto each other.

In the figure above, the vector points at an angle of 300 with respect to the +xdirection (in fact, it can point at any direction θ).The component is called the projection of on the x axis with magnitude Vxgiven by the line 0C, while the component is the projection of on the y axis with magnitude Vy given by the line 0B.

Using the parallelogram method for vector addition:

(3.4)

Vy

Vx


From now on, we will denote the magnitude of a vector by the following expression:

Where the symbol | | represents the module of the vector , or, in other words, its magnitude.

We can also write , and

Note that the eq. 3.6 followsfrom the Pythagorean theorem.

Vy

Vx

(3.6)

(3.5)

Vy

Vx


Example 3.4: Using the figure below, show that if and are two vectors in a plane defined by two perpendicular axes x and y, and is the resultant of the addition of these two vectors, then:

Equation 3.7 states that the resultant of the sum of two vectors in a two dimensional plane is equivalent to the sum of their components on each of the x and y axes.

(3.7) x

y

Solution developed on the blackboard


Equation 3.7 also applies when both vectors start away from the origin of the coordinate system:


If the components are perpendicular, they can be found using trigonometric functions.

In the figure, the vector makes and angleθ with the x axis.

Note: We define (by convention) θ with respect to the +x direction:

sin θ = sine of theta =

cos θ = cosine of theta =

tan θ = tangent of theta =

θ-θ

(3.8)

(3.9)

(3.10)


You can use your calculator to compute sin, cos and tan. If you want to find the angle, you can use the functions sin-1 , cos-1 and tan-1 for equations 3.8, 3.9 and 3.10,respectively. We write:

If you calculator does not havethese functions, it should have an key called“invert” or “inv” or something similar.

Trigonometric functions can be added,multiplied, etc when needed. You can findseveral functions and identities in Appendix A of the text book.

Three very important trigonometric identities are the following:

(3.8)

(3.9)

(3.10)


Using equations 3.8 – 3.10, we can write the components of the vector as:

(3.8)

(3.9)

(3.10)

(3.11)

(3.12)


Problem 3.10 (textbook): Three vectors are shown in Fig. 3–32. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the x axis.

Fig. 3-32

o o44.0 cos 28.0 38.85 44.0 sin 28.0 20.66x yA A= = = =o o

o o

26.5 cos 56.0 14.82 26.5sin 56.0 21.97

31.0 cos 270 0.0 31.0 sin 270 31.0

x y

x y

B B

C C

= − = − = =

= = = = −

( ) ( )38.85 14.82 0.0 24.03 24.0x

= + − + = =A + B + Cr rr

( ) ( )20.66 21.97 31.0 11.63 11.6y

= + + − = =A + B + Cr rr

( ) ( )2 224.03 11.63 26.7= + =A + B + Cr rr

1 o11.63tan 25.8

24.03θ −= =

(a)

(b)

10. Three vectors are shown in Fig. 3–32. Their magnitudes are given in arbitrary units.

Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b)

magnitude and angle with the x axis.


Problem 3.13 (textbook): For the vectors given in Fig. 3–32, determine

(a)

(b)

(c)

Fig. 3-32

,CBArrr

+−

,CBArrr

−+

.BACrrr

−−

13.For the vectors given in Fig. 3–32, determine (a) (b) and

(c)

,CBArrr

+− ,CBArrr

−+

.BACrrr

−−

o o44.0 cos 28.0 38.85 44.0 sin 28.0 20.66x yA A= = = =o o

o o

26.5 cos 56.0 14.82 26.5sin 56.0 21.97

31.0 cos 270 0.0 31.0 sin 270 31.0x y

x y

B B

C C

= − = − = =

= = = = −

( ) ( )38.85 14.82 0.0 53.67x

− + = − − + =A B Cr rr

( ) ( )20.66 21.97 31.0 32.31y

− + = − + − = −A B Cr rr

( ) ( )2 2 1 o32.3153.67 32.31 62.6 tan 31.0 below axis

53.67xθ − −

− + = + − = = = +A B Cr rr

(a)

Note that since the x component is positive and the ycomponent is negative, the vector is in the 4th quadrant.

(b)

( ) ( )38.85 14.82 0.0 24.03x

+ − = + − − =A B Cr rr

( ) ( )20.66 21.97 31.0 73.63y

+ − = + − − =A B Cr rr

( ) ( )2 2 1 o73.6324.03 73.63 77.5 tan 71.9

24.03θ −+ − = + = = =A B C

r rr

(c)

( ) ( )0.0 38.85 14.82 24.03x

− − = − − − = −C A Br r r

( ) 31.0 20.66 21.97 73.63y

− − = − − − = −C A Br r r

( ) ( )2 2 1 o73.6324.03 73.63 77.5 tan 71.9 below axis

24.03xθ − −

− − = − + − = = = −−

C A Br r r

Note that since both components are negative, the vector is in the 3rd quadrant.

Note that the answer to (c) is the exact opposite of the answer to (b).

Projectile MotionProjectile Motion

Projectile motion is a generalization of purely vertical motion (falling objects) we studied in chapter 2. Now we will consider the motion in two dimensions (can be easily generalized to three dimensions).

An example of projectile motion is that of a tennis ball, speeding bullets, etc..

In this chapter, we will not consider how an object is set in motion, but only its motion after it start moving and before it lands or is caught.


Projectile motions can be approached using the method of vector components as described before.

We can analyze the motion by studying its vertical and horizontal componentsseparately (Galileo was the first to describe projectile motion using this method).

In this figure, a ball rolls off the endof a table with an initial velocity Vx0at an instant of time t = 0in the horizontal direction. Furthermore, its velocity points in the direction of the motion at any instant of time t, and is tangent to its path.


We can then analyze the motion in the x and y direction separately using the equations of motion we obtained in chapter 2:

(2.12)(2.13)

(2.10)horizontalmotion

(2.17)

(2.18)

(2.16)verticalmotion


y direction

In the vertical direction, we have the acceleration, g, of gravity. The initial velocity is zero (Vy0 = 0). The coordinate y is chosen to be positive upward and zero at the origin selected as where the point the ball start experiencing vertical motion as depicted in the figure.

Then, at t = 0 we have:

y0 = 0Vy0 = 0a = –g (a points downward)

Using equation 2.17 we have:

Using eq. 2.16:

(3.13)

(3.14)


x direction

In the horizontal direction, there is no acceleration. The initial velocity is Vx0. The coordinate x is chosen to be eastward and zero at the origin selected as the point where the ball start experiencing vertical motion as depicted in the figure.

Then, at t = 0 we have:

x0 = 0initial velocity = Vx0a = 0

Using equation 2.12 we have:

Using eq. 2.10:

The velocity is constant in the horizontal direction.

(3.15)

(3.16)


Example 3.5:Show that an object projected horizontally will reach the ground in the same time as an object dropped vertically from the same height (use the figure below).

The only motion that matters here is the verticalone since (displacement in y direction).

Equation 3.13 gives the y displacementfor both balls since they are subject to similarinitial conditions, namely:

y0 = 0Vy0 = 0a = –g (a points downward)

Therefore

applies in both cases at any instant of time.

Thus, the two ball reach the ground together.


If an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

You can use equations 3.11 and 3.12 and then develop the problem in the very same way as on the previous slides.


Problem 3.21 (textbook) : A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base. What was the ball’s initial speed?


0 0yv =29.80 m sya =

0 0y =

( ) ( )2 2 21 10 0 2 2 2

2 45.0 m 45.0 m 9.80 m s 3.03 sec

9.80 m sy yy y v t a t t t= + + → = → = =

24.0 m 3.03 s 7.92 m sx xx v t v x t∆ = → = ∆ = =

Choose downward to be the positive y direction. The origin will be at the point where the ball is thrown from the roof of the building. In the vertical direction,

and the displacement is 45.0 m. The time of flight is found from applying Eq. 2.17 to the vertical motion.

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity:

.

Problem 3.21 (textbook) : A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base. What was the ball’s initial speed?


Problem 3.31 (textbook) : A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37.0º with the horizontal, as shown in Fig. 3–35. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the range X of the projectile as measured from the base of the cliff. At

the instant just before the projectile hits point P(c) Find the horizontal and the vertical components of its velocity(d) Find the magnitude of the velocity(e) Find the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile.


0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=

Choose the origin to be at ground level, under the place where the projectile is launched, and upwards to be the positive y direction. For the projectile,

The time taken to reach the ground is found from Eq. 2.17, with a final height of 0.

( )( )( )

2 21 10 0 0 02 2

2 2 10 0 0 0 2

12

0 125 sin

sin sin 4 125 39.1 63.110.4 s , 2.45 s 10.4 s

2 9.8

y yy y v t a t v t gt

v v gt

g

θ

θ θ

= + + → = + − →

− ± − − − ±= = = − =

− −

Choose the positive sign since the projectile was launched at time t = 0.

(a)

0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=


(b)

The horizontal range is found from the horizontal motion at constant velocity.

( ) ( ) ( )o0 0cos 65.0 m s cos 37.0 10.4 s 541 mxx v t v tθ∆ = = = =

0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=


(c)

( ) o0 0cos 65.0 m s cos 37.0 51.9 m sxv v θ= = =

( ) ( ) ( )o 20 0 0sin 65.0 m s sin 37.0 9.80 m s 10.4 s

63.1m s

y yv v at v gtθ= + = − = −

= −

At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant

The vertical component is found from Eq. 2.16:

0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=


(d)

The magnitude of the velocity is found from the x and y components calculated in part c) above.

( ) ( )2 22 2 51.9 m s 63.1m s 81.7 m sx yv v v= + = + − =

0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=


(e)

1 1 o63.1tan tan 50.6

51.9y

x

vv

θ − − −= = = −

o50.6 below the horizon

The direction of the velocity is (see slide number 7 of this lecture)

and so the object is moving .

0 65.0 m sv =o

0 37.0θ =

ya g= −

0 125y =

0 0 0sinyv v θ=


(f)

The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from Eq. 2.18.

( )( )

( )

2 2 2 20 0 0 0 m ax

2 2 o2 20 0

m ax 2

2 0 sin 2

65.0 m s sin 37 .0sin78.1 m

2 2 9.80 m s

y y yv v a y y v gy

vy

g

θ

θ

= + − → = −

= = =

chapter 3 kinematics in two dimensions;...

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