Transcript Two-Dimensional Motion and Vectors

Chapter 3:
Two-Dimensional Motion
and
Vectors
By: Amber Jubie
Introduction
Magnitude: a
measurement
represented by a
number.
Direction: an
indication of
orientation.
Scalar: has
magnitude only.
Examples include
temperature, speed,
mass, and volume.
Vector: has
magnitude and
direction.
Displacement,
acceleration, and
velocity are
examples.
Vectors can be
represented with
bold type with an
arrow above the
symbol.
Vectors are represented with arrows.
Length and direction
should be drawn to
scale.
10km ---------->
20km -------------------->
Vector Properties
Conventions
Compass Rose
North (0)
South (180)
West (270)
East (90)
Normal (Mr. Unit Circle)
not told N, S, W, E
“east” (0) “north” (90)
“west” (180) “south” (270)
Example Problems
Draw each vector on a separate sheet of paper. Answers on next
slide.
1. Given the SCALE: 1 cm = 10 m, represent the vector 50 m, 30-degrees
by a scaled vector diagram.
2. Given the SCALE: 1 cm = 10 m, represent the vector 60 m, 150-degrees
by a scaled vector diagram.
3. Given the SCALE: 1 cm = 15 m/s, represent the vector 120 m/s, 240degrees by a scaled vector diagram.
1.
Answers
2.
3.
Vector Addition
• Must have the same units.
• A negative vector has the
same magnitude, but
opposite in direction.
• Always add a negative
vector.
• Resultant: sum of vectors
• Always draw head to tail.
Example: Eric leaves the base camp and hikes 11 km, north and then
hikes 11 km east. Determine Eric's resulting displacement.
Finding Direction
Once you find
displacement, next you
must find the direction by
using Trig. Equations.
Now you can finish
the Hiker problem!
One Thing To Remember
The measure of an angle as determined through use of SOH CAH TOA is
not always the direction of the vector. The following vector addition
diagram is an example of such a situation. Observe that the angle within
the triangle is determined to be 26.6 degrees using SOH CAH TOA. This
angle is the southward angle of rotation which the vector R makes with
respect to West. In conclusion, the angle is actually 206.6 degrees.
Some review problems involve representing a vector as a
tailwind or river currents. Here’s some examples to help you on
those problems.
Tips!
Non-Perpendicular Motion
Find the components of each vector. Each vector has two
components, one parallel to the x-axis and the other
parallel to the y-axis.
Learning to solve non perpendicular vectors
Solving these at first is a little confusing, but once you practice you’ll think they are
easy. First look at each vector separately and solve the components for each. Right
down which components are going in the “y” direction and add them together
(make sure you remember what direction they are going!), then add the
components going in the “x” direction. Once you have the two numbers put them
into a new right triangle and solve for the other side and you will have the answer!
Example Problem
Jimmy walked 50.0km due East and then 40.0 km at 30.00
North of East. Calculate Jimmy's displacement.
X1= 50km
40km
X1= 50 km
Y2
X2
Y1= 0 km
X2= (40) (cos 30)= 34.6 km
Y2= (40) (sin 30)= 20 km
Example Problem Continued
Horizontal Component
Vertical Component
X1 + X2 = 50km + 34.6km
Y1 + Y2 = 0km + 20km
X= 84.6 km
Y= 20 km
R2 = A2 + B2
Tan 0 = opposite/adjacent
R2 = 84.62 + 20.02
R = (84.62 + 20.02)1/2
R = 87.0 km
Tan 0 = 20km/84.6km
0 = tan-1(20.0 / 84.6)
20 km
84.6 km
0
0 = 13.30
Projectile Motion
Three types of projectiles
include: objects dropped
from rest, objects thrown
vertically in the air, and
objects thrown at an
angle.
Regardless of whether a projectile is
moving downwards, upwards,
upwards and rightwards, or
downwards and leftwards, the only
force acting upon the object is
gravity.
Motion of a Projectile
The motion of a projectile in the air is actually twodimensional motion. Its velocity can be broken down
into perpendicular components that are independent of
each other.
If two balls are
at the same
height, one is
dropped from
rest and the
other is
launched with
an initial
horizontal
velocity they
will fall at the
same rate.
Basically they
will be in the air
the same
amount of time.
Objects Falling
at the same rate
Understanding of Projectiles
•A projectile is any object upon which the only force is gravity.
•Projectiles travel with a parabolic trajectory due to the influence of
gravity.
•There are no horizontal forces acting upon projectiles and thus no
horizontal acceleration.
•The horizontal velocity of a projectile is constant (a never changing in
value).
•There is a vertical acceleration caused by gravity; its value is 9.8
m/s/s, down.
•The vertical velocity of a projectile changes by 9.8 m/s each second.
•The horizontal motion of a projectile is independent of its vertical
motion.
Two key principles of projectile
motion
The two key principles of projectile motion - there is a horizontal
velocity which is constant and a vertical velocity which changes by -9.8
m/s each second. As the projectile rises towards its peak, it’s slowing
down, and as it falls from its peak, it’s speeding up.
Notice, the vertical speed one second before reaching its peak is the same
as the vertical speed one second after falling from its peak.
Projectiles Launched at a
Horizontal Velocity
Suppose that the
cannonball is launched
horizontally with no
upward angle whatsoever
and with an initial speed of
20 m/s. The important
concept depicted at the
right vector diagram is that
the horizontal velocity
remains constant during the
course of the trajectory and
the vertical velocity
changes by -9.8 m/s every
second
Describing Projectile Motion
with numbers- Vertical Motion
y = ½ a (t)^2
The above equation pertains to a projectile with no initial vertical
velocity and as such predicts the vertical distance which a projectile
falls if dropped from rest.
Horizontal Motion
X = (v)(t)
The horizontal displacement of a projectile is only
influenced by the speed at which it moves horizontally (vix)
and the amount of time (t) which it has been moving
horizontally.
Practice Problem
Given: ∆y = -321m ∆x = 45m g = 9.81m/s2
Determine: Initial velocity (vi)
•Remember that velocity equals displacement divided by
time or
v = ∆x/∆t
•We know ∆x = 45 but we don’t know ∆t. Using the
equations above we can determine ∆t.
•The only equation above that we can solve for ∆t is the
3rd one
∆y = -½g(Δt)2
•If we substitute and solve for Δt we get
-321 = (-½)(9.81)(Δt)2
65.44 = (Δt)2
Δt = 8.1s
•Now we simply divide ∆x by ∆t to get vi = 5.5m/s
Projectiles launched at an angle
vx = vi (cos θ) =
constant*
∆x = vi (cos θ) ∆t
vy,f = vi (sin θ) – gΔt
vy,f2 = vi2 (sin θ)2 – 2gΔy
Δy = vi (sin θ) Δt ½g(Δt)2
Practice Problem
A baseball is thrown at an angle of 25o relative to the ground at a speed of 23m/s. If
the ball is caught 42m from the thrower, how long was it in the air? How high was the
tallest spot in the ball’s path?
Given: θ = 25o
Δx = 42m
vi = 23m/s
SOLVE FOR Δt
Use this equation: Δx = vi (cos θ) Δt
42 = (23)(cos 25)(Δt)
42 = (23)(.91)(Δt)
2.01s = Δt
SOLVE FOR Δy
Remember: Δy = vy/Δt
Solve for vy: vy = vi (sin θ)
vy = (23)(.422)
vy = 9.72m/s
Solve for Δy: Δy = 9.72/2.01
Δy = 4.83m
~Conceptual Question~ Suppose a zookeeper must shoot a banana
from a banana cannon to a monkey who hangs from the limb of a
tree. This particular monkey has a habit of dropping from the tree
the moment that the banana leaves the muzzle of the cannon. The
zookeeper is faced with the dilemma of where to aim the banana
cannon in order to hit the monkey. If the monkey lets go of the tree
the moment that the banana is fired, then where should she aim the
banana cannon?
Answer
The banana moves in a parabolic path in
the presence of gravity. In the presence
of gravity, the monkey also accelerates
downward once he lets go of the limb.
Both banana and monkey experience the
same acceleration since gravity causes
all objects to accelerate at the same rate
regardless of their mass. Since both
banana and monkey experience the same
acceleration each will fall equal amounts
below their gravity-free path. The
banana misses the monkey, moving over
his head. The banana passes as far above
the monkey's head as it was originally
aimed.
Relative Motion
All motion is relative to the observer or to some fixed object. When you
see a car drive by, it is moving with respect to you. If you are in a car that
is going at the same speed, the other car will not be moving with respect
to you. But both cars are moving with respect to the ground.
Frame of Reference
In talking about motion, it is important to indicate your point of
reference. All motion is relative to what you define as a fixed point. For
an example, the sun appears to move across the sky, when the earth is
actually spinning and causing that apparent motion. If you were standing
on the sun you would know you weren’t moving, but on earth it looks
like the sun is moving. It is all about where your frame of reference is!
Relative Velocity
One must take into account relative velocities to describe the motion
of an airplane in the wind or a boat in a current. Assessing velocities
involves vector addition and a useful approach to such relative
velocity problems is to think of one reference frame as an
"intermediate" reference frame in the form:
This approach is useful for airplane or boat problems.
Practice Problem
Two cars, standing a distance apart, start moving towards each
other with speeds 1 m/s and 2 m/s along a straight road. What is
the speed with which they approach each other ?
Solution
Let us consider that "A" denotes Earth, "B" denotes first car and "C" denotes
second car. The equation of relative velocity for this case is :
⇒vCA=vBA+vCB
vBA=1m/s and vCA=−2m/s.
Now :
vCA=vBA+vCB
⇒−2=1+vCB
⇒vCB=−2−1=−3m/s
This means that the car "C" is approaching "B" with a speed of -3 m/s along the straight
road. Equivalently, it means that the car "B" is approaching "C" with a speed of 3 m/s
along the straight road. We, therefore, say that the two cars approach each other with a
relative speed of 3 m/s.
Review
When vectors are added together they should be drawn head to tail
to determine the resultant or sum vector.
The resultant vector can also be found mathematically
in right triangles using the Pythagorean Theorem.
A single vector can be broken down into two or more
component to find its x and y coordinates.
The vertical component of velocity is affected by the
acceleration of gravity and the horizontal component of
velocity does not change.
The relative velocity of an object depends on
its frame of reference.