Transcript Introduction to 2-Dimensional Motion

Introduction to 2-Dimensional
Motion
2-Dimensional Motion
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Definition: Motion that occurs with
both x and y components.
Example:
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Playing pool .
Throwing a ball to another person.
Each dimension of the motion can
obey different equations of motion.
Solving 2-D Problems
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Resolve all vectors into components
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Work the problem as two one-dimensional
problems.
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x-component
Y-component
Each dimension can obey different equations of
motion.
Re-combine the results for the two
components at the end of the problem.
Sample Problem
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You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a)
b)
How far west have you traveled in 2.5 minutes?
How far south have you traveled in 2.5 minutes?
Sample Problem
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You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a)
b)
How far west have you traveled in 2.5 minutes?
How far south have you traveled in 2.5 minutes?
v = 40 m/s
Sample Problem
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You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a)
b)
How far west have you traveled in 2.5 minutes?
How far south have you traveled in 2.5 minutes?
v = 5 m/s, Q = 40o,
t = 2.5 min = 150 s
vy
vx = v cosQ
vy = v sin Q
v = 5 m/s
vx = 5 cos 40
vy = 5 sin 40
vx =
vy =
x = vx t
y = vyt
x = ( )(150) y = ( )(150)
x=
y=
vx
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
a)
What are the x and y positions at 5.0 seconds?
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
a)
What are the x and y positions at 5.0 seconds?
Vo,y = 6.2 m/s, Vo,x = 0 m/s, t = 5 s,
ax = -4.4 m/s2, ay = 0 m/s2
x=?
y=?
x = vo,x + at
y = vo,y + at
x=
y=
x=
y=
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
b)
What are the x and y components of velocity at this time?
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
b)
What are the x and y components of velocity at this time?
vx = vo,x + axt
vx =
vx =
vy = voy + axt
vy =
vy =
Projectiles
Projectile Motion
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Something is fired, thrown, shot,
or hurled near the earth’s surface.
Horizontal velocity is constant.
Vertical velocity is accelerated.
Air resistance is ignored.
1-Dimensional Projectile
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Definition: A projectile that moves in a
vertical direction only, subject to
acceleration by gravity.
Examples:
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Drop something off a cliff.
Throw something straight up and catch it.
You calculate vertical motion only.
The motion has no horizontal component.
2-Dimensional Projectile
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Definition: A projectile that moves both
horizontally and vertically, subject to
acceleration by gravity in vertical
direction.
Examples:
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Throw a softball to someone else.
Fire a cannon horizontally off a cliff.
Shoot a monkey with a blowgun.
You calculate vertical and horizontal
motion.
Horizontal Component of
Velocity
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Is constant
Not accelerated
Not influence by gravity
Follows equation:
x = Vo,xt
Horizontal Component
of Velocity
Vertical Component of
Velocity
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Undergoes accelerated motion
Accelerated by gravity (9.8 m/s2
down)
Vy = Vo,y - gt
y = yo + Vo,yt - 1/2gt2
Vy2 = Vo,y2 - 2g(y – yo)
Horizontal and Vertical
Horizontal and Vertical
Zero Launch Angle Projectiles
Launch angle
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Definition: The angle at which a
projectile is launched.
The launch angle determines what the
trajectory of the projectile will be.
Launch angles can range from -90o
(throwing something straight down)
to +90o (throwing something straight
up) and everything in between.
Zero Launch angle
vo
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A zero launch angle implies a perfectly
horizontal launch.
Sample Problem
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The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
Sample Problem
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The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = -9.8 m/s2, vo,x = 3.6 m/s
v=?
v  vx  vy
2
2
Sample Problem

The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v=?
v  vx  vy
2
2
Gravity doesn’t change horizontal velocity.
vo,x = vx = 3.6 m/s
Vy2 = Vo,y2 - 2g(y – yo)
Vy2 =
Vy =
Sample Problem

The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v=?
v  vx  vy
2
v  (3.6)  (
2
v=
Gravity doesn’t change horizontal velocity.
2
vo,x = vx = 3.6 m/s
)
2
Vy2 = Vo,y2 - 2g(y – yo)
Vy2 =
Vy =
Sample Problem
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An astronaut on the planet Zircon tosses a rock horizontally with a
speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a
horizontal distance of 8.95 m from the astronaut. What is the
acceleration due to gravity on Zircon?
Sample Problem
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An astronaut on the planet Zircon tosses a rock horizontally with a
speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a
horizontal distance of 8.95 m from the astronaut. What is the
acceleration due to gravity on Zircon?
vo,x = 6.75 m/s, x = 8.95 m, y = 0 m, yo = 1.2, Vo,y = 0
g=?
x = vo,xt
y = yo + Vo,yt - 1/2gt2
t = x/vo,x
g = -2(y - yo - Vo,yt)/t2
t=
g=
t=
g=
m/s
Sample Problem
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Playing shortstop, you throw a ball horizontally to the second
baseman with a speed of 22 m/s. The ball is caught by the second
baseman 0.45 s later.
a)
b)
How far were you from the second baseman?
What is the distance of the vertical drop?
Should be able to do this on your own!
General Launch Angle
Projectiles
General launch angle
vo
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Projectile motion is more complicated when the
launch angle is not straight up or down (90o or –
90o), or perfectly horizontal (0o).
General launch angle
vo
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You must begin problems like this by resolving
the velocity vector into its components.
Resolving the velocity
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Use speed and the launch angle to find
horizontal and vertical velocity components
Vo
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Vo,x = Vo cos 
Vo,y = Vo sin 
Resolving the velocity
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Then proceed to work problems just like
you did with the zero launch angle
problems.
Vo
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Vo,x = Vo cos 
Vo,y = Vo sin 
Sample problem
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A soccer ball is kicked with a speed of 9.50 m/s at an angle
of 25o above the horizontal. If the ball lands at the same
level from which is was kicked, how long was it in the air?
Sample problem
A soccer ball is kicked with a speed of 9.50 m/s at an angle
of 25o above the horizontal. If the ball lands at the same
level from which is was kicked, how long was it in the air?
vo = 9.5 m/s,  = 25o, g = 10 m/s2, Remember: because it lands
at the same height: Dy = y – yo = 0 m and vy =- vo,y
Find: Vo,y = Vo sin  and
Vo,x = Vo cos 
Vo,y =
Vo,x =
Vo,y =
Vo,x =
t=?
Vy = Vo,y - gt
t = (Vy - Vo,y )/-g
t=
don’t forget vy =- vo,y
t=
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Sample problem
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Snowballs are thrown with a speed of 13 m/s from a roof
7.0 m above the ground. Snowball A is thrown straight
downward; snowball B is thrown in a direction 25o above the
horizontal. When the snowballs land, is the speed of A
greater than, less than, or the same speed of B? Verify your
answer by calculation of the landing speed of both
snowballs.
We’ll do this in class.
Projectiles launched over
level ground
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These projectiles have highly
symmetric characteristics of motion.
It is handy to know these
characteristics, since a knowledge of
the symmetry can help in working
problems and predicting the motion.
Lets take a look at projectiles
launched over level ground.
Trajectory of a 2-D
Projectile
y
x
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Definition: The trajectory is the path
traveled by any projectile. It is
plotted on an x-y graph.
Trajectory of a 2-D
Projectile
y
x
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Mathematically, the path is defined by
a parabola.
Trajectory of a 2-D
Projectile
y
x
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For a projectile launched over level
ground, the symmetry is apparent.
Range of a 2-D Projectile
y
Range
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x
Definition: The RANGE of the projectile
is how far it travels horizontally.
Maximum height of a
projectile
y
Maximum
Height
Range
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x
The MAXIMUM HEIGHT of the projectile
occurs when it stops moving upward.
Maximum height of a
projectile
y
Maximum
Height
Range
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x
The vertical velocity component is zero at
maximum height.
Maximum height of a
projectile
y
Maximum
Height
Range
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x
For a projectile launched over level ground, the
maximum height occurs halfway through the flight
of the projectile.
Acceleration of a projectile
y
g
g
g
g
g
x
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Acceleration points down at 9.8 m/s2 for
the entire trajectory of all projectiles.
Velocity of a projectile
y
v
v
vo
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v
vf x
Velocity is tangent to the path for the
entire trajectory.
Velocity of a projectile
y
vx
vy
vx
vy
vx
vy
vx
vx

vy
x
The velocity can be resolved into
components all along its path.
Velocity of a projectile
y
vx
vy
vx
vy
vx
vy
vx
vx

vy
x
Notice how the vertical velocity changes
while the horizontal velocity remains
constant.
Velocity of a projectile
y
vx
vy
vx
vy
vx
vy
vx
vx

vy
x
Maximum speed is attained at the beginning,
and again at the end, of the trajectory if the
projectile is launched over level ground.
Velocity of a projectile
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-
vo
vo
Launch angle is symmetric with landing angle
for a projectile launched over level ground.
Time of flight for a
projectile
t
to = 0
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The projectile spends half its time
traveling upward…
Time of flight for a
projectile
t
to = 0
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… and the other half traveling down.
2t
Position graphs for 2-D
projectiles
y
y
x
x
t
t
Velocity graphs for 2-D
projectiles
Vy
Vx
t
t
Acceleration graphs for 2-D
projectiles
ay
ax
t
t
More on Projectile Motion
The Range Equation
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Derivation is an important part of
physics.
Your book has many more equations
than your formula sheet.
The Range Equation is in your
textbook, but not on your formula
sheet. You can use it if you can
memorize it or derive it!
The Range Equation
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R = vo2sin(2)/g.
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R: range of projectile fired over level
ground
vo: initial velocity
g: acceleration due to gravity
: launch angle
Deriving the Range Equation
Sample problem
A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
How far from the golfer does the ball land?
Vo = 42m/s,  = 35o, g = 9.8 m/s2
R=?
Sample problem
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A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
b)
The next golfer hits a ball with the same initial speed, but at a
greater angle than 45o. The ball travels the same horizontal
distance. What was the initial direction of motion?