Physics 231 Topic 2: Vectors and two dimensional motion Wade Fisher

Physics 231
Topic 2: Vectors and two dimensional motion
Wade Fisher
September 5-7 2012
MSU Physics 231 Fall 2012
1
House Keeping
Students with RCPD forms MUST give them to me ASAP
• Nominally you have 1 week to turn them in and we are past that
deadline now.
Please go online and register your iClickers!
• I will also do roll calls periodically
• Sections 1 & 3 will NOT get credit for clicker questions in Section 2
(transfers / overrides are special cases)
The TA schedule for the Physics Learning Center (BPS
1248) is now posted:
http://www.pa.msu.edu/~fisherw/Phy231_Fall2012/taSchedule.html
MSU Physics 231 Fall 2012
2
Recall from Last Lecture
Position and coordinate systems
Displacement vs. Distance
Velocity
 Distinguish from speed
 Average vs instantaneous velocity
Acceleration
 Relationship to velocity
 Impact on position in 1D motion
Constant velocity and constant acceleration equations
 How to set up & solve problems
 Graphical representations of distance, velocity, acceleration
MSU Physics 231 Fall 2012
3
Galileo’s Free Fall Experiment
Galileo throws a canon ball from the tower of Pisa. Ignoring
friction what is the distance covered between t=1 and t=3
seconds? The initial velocity was 0 m/s.
MSU Physics 231 Fall 2012
4
MSU Physics 231 Fall 2012
5
Key Concepts: 2D Motion
Vectors and Scalars
Two Dimensional Motion
 Velocity in 2D
 Acceleration in 2D
Projectile motion
 Throwing a ball or cannon fire
Uniform Circular Motion
 Centripetal acceleration
Covers chapter 3 in Rex & Wolfson
MSU Physics 231 Fall 2012
6
Extra Credit Quiz
MSU Physics 231 Fall 2012
7
Vectors and Scalars
• Scalar: A quantity specified by its magnitude only
• Vector: A quantity specified both by its magnitude and
direction.
• To distinguish a vector from a scalar quantity, it is usually
written with an arrow above it, or in bold to distinguish it
from a scalar.
• Scalar: A
• Vector: A or A
MSU Physics 231 Fall 2012
8
Question
• Are these two vectors the same?
• Are the lengths of these two vectors the same?
Two vectors are equal if both their length and direction
are the same!
MSU Physics 231 Fall 2012
9
Vector Addition
A+B
B
A
B+A
B
A+B=B+A
A
MSU Physics 231 Fall 2012
10
Vector Subtraction
-B
B
A
A-B=A+(-B)
MSU Physics 231 Fall 2012
11
Vector Addition
MSU Physics 231 Fall 2012
12
Vector operations in equations
 X a b   X a   X b   X a  X b 

        

 Ya b   Ya   Yb   Ya  Yb 
 X a b   X a   X b   X a  X b 

        

 Ya b   Ya   Yb   Ya  Yb 
(xa+b,ya+b)
y
(xb,yb)
B
A+B
A (xa,ya)
x
Example:
 X a b   5    3   2 

         
 Ya b   2   2   4 
MSU Physics 231 Fall 2012
13
Clicker Question
begin
Which route is
shorter?
a)
b)
c)
d)
Red
Black
The same
Don’t know
end
MSU Physics 231 Fall 2012
14
More on Coordinate Systems
Cartesian Coordinates:
(x,y)=(a,b)
r
b
Also, r = a ∙ î + b ∙ ĵ
or r = aî + bĵ

Plane Polar Coordinates
(r,)
a
Frame
Transformation
r  a2  b2
tan( )  b / a
MSU Physics 231 Fall 2012
15
Trigonometry
SOH-CAH-TOA:
sin =opposite/hypotenuse =a/c
cos =adjacent/hypotenuse =b/c
tan =opposite/adjacent
=a/b
c
a
Pythagorean theorem:
c  a b
2

b
2
2
Note that sin,cos,tan are
dimensionless.
2 radians corresponds to 360o
MSU Physics 231 Fall 2012
16
Vector length and its components
Y
(xa,ya)

x
Length of vector (use pythagorean theorem):
x a  l cos 
l  xa2  y a2
y a  l sin 
tan   y a / x a
MSU Physics 231 Fall 2012
17
Be Careful….
h

y

y = h sin() = h cos()
Always check carefully which angle is given
MSU Physics 231 Fall 2012
18
Question
A man walks 5 km/h. He travels 12 minutes to the east,
30 minutes to the south-east and 36 minutes to the north.
A) What is the displacement of the man when he’s done?
B) What is the total distance he walked?
3 km
1 km
=315o
2.5 km
Y2=2.5sin()=-1.77
x2=2.5cos()=1.77
MSU Physics 231 Fall 2012
19
Relative motion
Motion is relative to a viewing frame!
A woman in a train moving 50 m/s throws a ball straight up
with a velocity of 5 m/s. A second person watches the
train pass by and sees the woman through a window.
What is the motion of the ball seen from the point of
view from the man outside the train?
Motion of the ball in rest-frame of train
Resulting motion
Motion of the train
MSU Physics 231 Fall 2012
20
Boat crossing the river
MSU Physics 231 Fall 2012
21
Question
A boat is trying to cross a 1-km wide river in
the shortest way (straight across).
Its maximum speed (in still water) is 10 km/h.
The river is flowing with 5 km/h.

1) At what angle  does the captain
have to steer the boat the go
straight across?
2) How long does it take for the boat
to cross the river?
3) If it doesn’t matter at what point
the boat reaches the other side,
at what angle should the captain steer
to cross in the fastest way?
MSU Physics 231 Fall 2012
22
Answer
Counter balance
flow=5km/h
A boat is trying to cross a 1-km wide river in the
shortest way (straight across).
Its maximum speed (in still water) is 10 km/h. The
river is flowing with 5 km/h.
1) At what angle  does the captain have to steer
the boat the go straight across?

Flow=5km/h
2) How long does it take the boat to cross the river?
MSU Physics 231 Fall 2012
23
Answer
Counter balance
flow=5km/h

A boat is trying to cross a 1-km wide river in the
shortest way (straight across).
Its maximum speed (in still water) is 10 km/h. The
river is flowing with 5 km/h.
3) If it doesn’t matter at what point the boat
reaches the other side, at what angle should the
captain steer to cross in the fastest way?
Flow=5km/h
MSU Physics 231 Fall 2012
24
plane in the wind
MSU Physics 231 Fall 2012
25
Key Concepts: 2D Motion
Vectors and Scalars
Two Dimensional Motion
 Velocity in 2D
 Acceleration in 2D
Projectile motion
 Throwing a ball or cannon fire
Uniform Circular Motion
 Centripetal acceleration
Covers chapter 3 in Rex & Wolfson
MSU Physics 231 Fall 2012
26
Clicker Quiz
Galileo and Newton stand on top of the tower of Pisa.
Galileo drops a stone of mass 2 kg straight down (no initial
velocity).
Newton throws a 2 kg stone with an initial horizontal
velocity of 3 m/s.
Which stone will hit the ground first? (ignore
effects of friction)
a) The stone thrown by Galileo
b) The stone thrown by Newton
c) Both stones arrive at the same time
d) Not enough information to say.
MSU Physics 231 Fall 2012
27
v(t) = vo + at
x(t) = xo + vt + ½at2
Time (s)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A (m/s2)
0
2
3
0
1
-1
1
1
-2
-1
0
0
0
0
n/a
V (m/s)
0
0
2
5
5
6
5
6
7
5
4
4
4
4
4
X (m)
0
0 1
4.5
9.5
15
20.5 26 32.5 38.5 43 47
51
55
59
X(3s) = X(2s) + V(2s) t + ½A(2s) (t)2 = 1 + 21 + ½312 = 1+2+1.5 = 4.5 m
MSU Physics 231 Fall 2012
28
Time (s)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A (m/s2)
0
2
3
0
1
-1
1
1
-2
-1
0
0
0
0
n/a
V (m/s)
0
0
2
5
5
6
5
6
7
5
4
4
4
4
4
X (m)
0
0
1
4.5
9.5
15
20.5 26 32.5 38.5 43
47
51
55
59
Calculate the speed of the car at t = 3 s.
Calculate the distance traveled during the first 5 s.
Calculate the distance traveled from t=10 s to t=14 s.
Calculate the car's average speed from t = 6 s to t=9 s.
MSU Physics 231 Fall 2012
Answer = 5 m/s
Answer = 15 m
Answer = 59-43 = 16 m
Answer = 18/3 = 6 m/s
29
Displacement in 2D
y
Often, we replace motion
in 2D into horizontal and
vertical components.
r
In vector notation:
r= x+y (r,x,y: vectors)
x
We can then work on
horizontal and vertical
components separately.
MSU Physics 231 Fall 2012
30
1d motion
x(t) = x0 + v0t + ½at2
v(t) = v0 + at
decomposition for 2D
2D motion; decompose into
horizontal and vertical components
x(t) = x0 +v0xt + ½axt2
vx(t) = v0x + axt
y(t) = y0 + v0yt + ½ayt2
vy (t) = v0y + ayt
MSU Physics 231 Fall 2012
31
MSU Physics 231 Fall 2012
32
Clicker Quiz Firing Balls I
A small cart is rolling at constant a) it depends on how fast the cart is
velocity on a flat track. It fires a
ball straight up into the air as it
moves. After it is fired, what
happens to the ball?
moving
b) it falls behind the cart
c) it falls in front of the cart
d) it falls right back into the cart
e) it remains at rest
In the frame of reference of
the cart, the ball only has a
vertical component of velocity.
So it goes up and comes back
down. To a ground observer,
both the cart and the ball
have the same horizontal
velocity, so the ball still
returns into the cart.
MSU Physics 231 Fall 2012
33
MSU Physics 231 Fall 2012
34
MSU Physics 231 Fall 2012
35
Pop and Drop
MSU Physics 231 Fall 2012
36
A
B
Pop and Drop
For A: Vy = -½gt2
Vx=0
For B: Vy = -½gt2
Vx=V0
For A: Y = Y0- ½gt2
X= 0
For B: Y = Y0 – ½gt2
X =V0t
MSU Physics 231 Fall 2012
37
While studying motion in 2D one almost
always makes a decomposition into
horizontal and vertical components of the
motion, which are both described in 1D
• Remember that the object can accelerate in one direction, but
remain at the same speed in the other direction.
• Remember that after decomposition of 2D motion into
horizontal and vertical components, you should should
investigate both components to understand the complete motion
of a particle.
• After decomposition into horizontal and vertical directions,
treat the two directions independently.
MSU Physics 231 Fall 2012
38
Parabolic motion
MSU Physics 231 Fall 2012
39
Parabolic motion
vx=v0cos
vy=v0sin-1g
vx=v0cos
vy=v0sin-2g=0
vx=v0cos
vy=v0sin-3g
V=v0
vx=v0cos
vy=v0sin
Vx remains constant
throughout the flight!

t=0
vx=v0cos
vy=v0sin-4g
t=1
t=2
MSU Physics 231 Fall 2012
t=3
t=4
40
Parabolic motion
X(t)=X0 + V0cost
Y(t)=Y0 + V0sint - ½gt2
X=X0
Y=Y0
Where is the speed…
1) …highest?
2) …lowest ?
Assume height of catapult is
negligible to the maximum height
of the stone.

t=0
A
t=1
B
t=2
C
t=3
D
MSU Physics 231 Fall 2012
E
t=5
41
What does the motion
of the object look like
according to the pilot?
MSU Physics 231 Fall 2012
42
Shoot the monkey
The hunter
aims his gun
exactly at
the monkey
At the moment the hunter fires, the monkey
drops off the branch. What happens?
At the moment he fires, the monkey
drops off the branch. What happens?
a)
b)
c)
d)
monkey gets hit
bullet goes over the monkey
bullet goes under the monley
no idea
PHY 231
MSU Physics 231 Fall 2012
43
43
44
MSU Physics 231 Fall 2012
44
MSU Physics 231 Fall 2012
45
Another example
A football player throws a ball with initial
velocity of 30 m/s at an angle of 30o
degrees w.r.t. the ground.
1) How far will the ball fly before hitting
the ground?
2) What about an angle of 60o?
3) At which angle is the distance thrown
maximum?
MSU Physics 231 Fall 2012
46
Uniform Circular Motion
Consider a car at constant speed,
constrained to a circular track
• Velocity is always determined
by the direction the car is
facing at a given time
Velocity is constantly changing, so
this implies an acceleration
• But it’s just the direction of the
velocity that’s changing!
MSU Physics 231 Fall 2012
47
Uniform Circular Motion
𝑣𝑓 − 𝑣𝑖 Δ𝑣
𝑎=
=
𝑡𝑓 − 𝑡𝑖
Δ𝑡
Sin( /2 ) = ( s/2 ) / r
Sin( /2 ) = ( v/2 ) / v
MSU Physics 231 Fall 2012
48
Uniform Circular Motion
𝑣𝑓 − 𝑣𝑖 Δ𝑣
𝑎=
=
𝑡𝑓 − 𝑡𝑖
Δ𝑡
( s/2 ) / r = sin( /2 )
sin( /2 ) = ( v/2 ) / v
( s/2 ) / r = ( v/2 ) / v
( s/t ) v/r = ( v/t )
𝑎𝑐 =
𝑣
𝑣
𝑟
=
MSU Physics 231 Fall 2012
𝑣2
𝑟
( s/t )  v
49
For Next Week
Rex & Wolfson Chapter 4:
Forces & Laws of Motion
Homework Set 2: Covers Chapters 2 & 3
Due Wednesday 9/16 @ 11PM
MSU Physics 231 Fall 2012
50
Solving Quadratic Equations
ax 2  bx  c  0
a0
 b  b 2  4ac
x
2a
In general there are 2 solutions. In physics problems,
One of them is often not realistic and is thrown out.
MSU Physics 231 Fall 2012
51
Calculate the length of the shorter of two sides of a
rectangle, which has an area of 24 m2 and a perimeter
(circumference) of 22 m.
A=LxH
C = 2L + 2H
H
24 = L x H
so L = 24/H
22 = 2L+2H = 2x24/H+2H = 48/H+2H
-2H2+22H-48 = 0
 22  22  4(2)(48)  22  10
H

2(2)
4
H 8
H 3
OR
L3
L8
2
MSU Physics 231 Fall 2012
L
ax 2  bx  c  0
a0
 b  b 2  4ac
x
2a
52
Question
A hunter aims at a bird that is some distance away and flying
very high (i.e. consider the vertical position of the hunter to be
0), but he misses.
If the bullet leaves the gun with a speed of v0 and friction by air
is negligible, with what speed vf does the bullet hit the ground
after completing its parabolic path?
v0
vf
MSU Physics 231 Fall 2012
53
Answer
First consider the horizontal direction:
V0x = V0cos()
Since there is no friction, there is no change in the
horizontal component: Vfx = V0cos() = V0x
Next the vertical direction:
V0y = V0sin()
Vy(t) = V0y - gt
xy(t) = Voyt - ½ gt2 (g = 9.81 m/s2)
Boundary condition: bullet hits the ground:
0 = Voyt - ½gt2
t = 0 or t = 2V0y/g
So, Vfy(t) = V0y - (2V0y/g) g = -V0y
Total speed = (V0x2+(-V0y)2) = V0!!!!
The speed of the bullet has not changed, but the vertical
component of the velocity has changed sign.
MSU Physics 231 Fall 2012
54
And another example…
Calculate how far the ball goes.
Vertical direction
y(t) = y(0) + v0sin()t - ½gt2
11 = 37 + 10.1 sin(59)t - ½9.8t2
4.9t2 - 8.66t - 26=0
t = 3.35 (solve quadratic
equation)
This time correspond to how
long it takes for the ball to
land on the second building.
Horizontal direction
x(t) = x(0) + vocos()t
Use time derived from
vertical direction
X(3.35)=0+10.1cos(59)3.35
X(3.35)=17.4 m
MSU Physics 231 Fall 2012
55