PHY 130 FUNDAMENTAL PHYSICS 1
CHAPTER 3 K I N EM ATI C I N T WO D I M ENS IO N
LESSON OUTCOMES • M u l t i p l y o r d i v i d e a v e c t o r q u a n t i t y b y a s c a l a r q u a n t i t y. • Use
the
direction
methods of
the
of
graphical
vector
resultant
analysis in
to
problems
determine involving
the
magnitude
vector
addition
and or
subtraction of two or more vector quantities.
• Use the trigonometric component method to resolve a vector components in the x and y direction and the vector resultant in problems involving vector addition or subt raction of t wo or more vect or quant ities.
• Use the kinematics equations along with the vector component method to solve problems involving two dimensional motion of projectiles.
VECTORS
Vectors • Physical quantities could be categorized to:
ď ą
Scalars – quantities which has only magnitude. (example: mass, volume, etc)
ď ą
Vectors – quantities which have both magnitude and direction. (example: velocity, acceleration, etc)
• Vector are usually written with their magnitude and directions
• Vectors are written in two ways: •
Method 1:
• Method 2:
đ?‘Ž (using an arrow above) đ?’‚ (using bold face print)
Example :
QUANTITIES
SYMBOLS
WRITTEN AS
Force
F
đ?‘ or đ??š
Momentum
p
đ?’‘ or đ?‘?
Vector Representation • Graphically vector can be represented on paper with an arrowed straight line.
• Magnitude represented by the length of the line (scaled) and direction represented by the angle and arrow • Example: N
30
d = 100 m, East E
v = 20 ms-1, 30 north of east
Equality Of Vectors •
Equality of vectors – two vectors A and B are said to be equal if i. magnitude of A = magnitude of B ii. A and B are parallel or in the same direction.
A
A=B
B
A
B
AB Different directions
A
B
AB Different magnitudes
Multiplication Vector With Scalar •
Scalar multiplication – if v is a vector and k is a scalar, then when k is multiplied with v, a new vector, kv, is formed where this new vector has the same direction as v but a magnitude that is k times the initial value.
suppose k = 3 and v = 3 ms-1, North-east
•
Example:
•
Then the multiplication of k and v produces a new vector say w = 3v
•
Where w = 9 ms-1, North-east
w v
Vector Addition/ Subtraction In One Dimension
• For vectors in one dimension, simple addition and subtraction are all that is needed. • If the motion is in two dimensions, the situation is somewhat more complicated.
• Methods of addition (considering only two vectors)
geometric method
analytical method
Vector Addition/ Subtraction In Two Dimension •
Geometrical method: ď ą Triangle method ď ą Parallelogram method
•
Triangle method: i. Sketch vector đ?‘Ž using appropriate scale.
ii.
Sketch vector đ?‘? using the same scale.
iii.
Placed the tail of đ?‘? at the tip of đ?‘Ž đ?‘? đ?‘? đ?‘Ž
đ?‘? =đ?‘Ž+đ?‘?
Vector Addition/ Subtraction In Two Dimension • Parallelogram method: the two vectors are joined at the tails, the resultant is the diagonal of a parallelogram formed with the two vectors.
Vectors Component • Component of vectors: Vectors can be described by its components • Consider horizontal (x) and vertical (y) components • A vector A can be written as A = Ax + Ay
A
Ay – vertical component or y-component
Ax- horizontal component or x - component
If the components are perpendicular, they can be found using trigonometric functions.
Vectors • The magnitude of the horizontal and vertical component can be obtained using Law of right-angled triangle, that is
F
Fy= Fsin θ θ Fx= Fcos θ
• Hence vector F can be written as
F F cos , F sin ,
• Where the arrows indicate the direction of the component vector
Fig. 3.14, p.65
Unit Vectors • Unit vector of a vector: • Written as
vector
• Defined as
v̂ vector magnitude of vector v vˆ v
Unit vector of a vector
• Example: given a vector • Therefore, unit vector,
v = 10 ms-1, East
10 ms -1 , East vˆ 10 1 ms -1 , East
• A unit vector is a vector with magnitude equal to 1
Unit Vectors • Writing vectors in the i , j , k form • Let i be the unit vector in the x direction, j be the unit vector in the y direction, and k be the unit vector in the z direction. y
1 j k
1 z
i 1
x
Unit Vectors • A vector can be written in the unit vector notation, for example: • Suppose A is a velocity vector = 20 ms-1, in the x-direction
• Hence A can be written as A = 20 i ms-1 • Let B be force vector, B = 40 N, 30o from the horizontal
30o
Unit Vectors • The vector B = 40 N, 30o from the horizontal can be written in i and j form. • Calculate the x-component and y-component of B • x-component = 40 cos 30 = 34.64 N
• y-component = 40 sin 30 = 20 N • Hence B = (34.64 i + 20 j ) N
B 20 j
34.64 i
Example 1 Component -x
Component -y
F1
6
0
F2
7 cos 120ď‚° = − 3.5
7 sin 120ď‚° = 6.062
F3
4 cos 220ď‚° = − 3.064
4 sin 220ď‚° = − 2.571
R
-0.564
3.491
The Figure shows three forces F1, F2 and F3 acting on a point O. Calculate the resultant force đ?‘… = −0.564đ?‘– + 3.491đ?‘— đ?‘ đ?‘…=
đ?‘…đ?‘Ľ2 + đ?‘…đ?‘Ś2 =
đ?œƒ = đ?‘Ąđ?‘Žđ?‘›âˆ’1
đ?‘…đ?‘Ś đ?‘…đ?‘Ľ
=−
(−0.564)2 +3.4912 = 3.536 đ?‘ 3.491 0.564
= 88.82° above negative-x axis
Example 2 An airplane trip involves three legs, with two stopovers. The first leg is due east for 620 km; the second leg is southeast (45°) for 440 km; and the third leg is at 53° south of west, for 550 km, as shown. What is the plane’s total displacement?
Component -x
Component -y
D1
620
0
D2
440 cos 45ď‚° = 311
− 440 sin 45ď‚° = − 311
D3
− 550 cos 53ď‚° = − 331
− 550 sin 53ď‚° = − 439
DR
600
− 750
đ??ˇđ?‘… = 600đ?‘– − 750đ?‘— đ?‘˜đ?‘š
đ??ˇđ?‘… =
đ??ˇđ?‘Ľ2 + đ??ˇđ?‘Ś2 =
đ?œƒ = đ?‘Ąđ?‘Žđ?‘›âˆ’1
đ??ˇđ?‘Ś đ??ˇđ?‘Ľ
=−
750 600
(600)2 +(−750)2 = 960 km
= 51° south of east
Projectile Motion • A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. • It can be understood by analyzing the horizontal and vertical motions separately
• Example: motion of ball kicked, golf ball hit by a club • Object given some initial velocity, then object moves under free fall or gravitational pull or free fall acceleration ( g = 9.8 ms-2) 21
Projectile Motion • In projectile motion the horizontal component and vertical component are independent of each other. y
v = vx ; vy = 0 vy
v0y
v
Vx
v0 θ v0x
x R = range (maximum horizontal distance)
22
Projectile Motion Generalizing the one-dimensional equations for constant acceleration:
Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.
Projectile Motion 2
vo R sin 2 g
y
• Range equation:
θ x
R/2
R
Projectile Motion y
• Rmax when θ = 45 • H increases as θ increases θ = 45 0
H x Rmax
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. 5.
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. 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.
Projectile Motion – Example 1 y 20
ms-1
α vx
vy
40 R a) Find the velocity at t = 1 s: • vx = vocos θ = 20 cos 40 = 15.32 ms-1
• vy = vosin θ – gt = 20sin 40 – 9.8(1) = 3.06 ms-1 • v = vx i + vy j • v = √(15.32)2 + (3.06)2 = 15.62 ms-1 • α = tan-1 (3.06/15.32) = 11.30
o
x
Projectile Motion – Example 1 y 7.96 hmax
20 ms-1 40
15.32
R
x
b) Find the position of object at t = 1 s: x = (v0cos θ)t = (20 cos 40 )(1) = 15.32 m y = (v0sin θ)t – ½gt2 = (20sin 40)(1) - ½(9.8)12 = 7.96 m
c) Find the maximum height, hmax Max. height when vy = 0 , vy = vosin θ – gt = 20sin 40 – 9.8t = 0, t = 1.31 s
When t = 1.31 s , y = (v0sin θ)t – ½gt2 = (20sin 40)(1.31) - ½(9.8)(1.31)2 = 8.43 m
Projectile Motion – Example 2 A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. 1 đ?‘Ś = đ?‘Łđ?‘Ś0 + đ?‘Žđ?‘Ś đ?‘Ą 2 2 1 −50 = 0 + − 9.8đ?‘Ą 2 2 đ?‘Ą = 3.19 đ?‘ đ?‘Ľ = đ?‘Łđ?‘Ľ0 đ?‘Ą
90= đ?‘Łđ?‘Ľ0(3.19) đ?‘Łđ?‘Ľ0 = 28.2 đ?‘š/đ?‘
known
unkown
x0 =y0 = 0
vx0
x= 90 m
t
y = - 50 m ax = 0 ay = - 9.8 m/s2
vy0 = 0