Transcript Lecture-04-09

Lecture 3:
2-Dimensional Motion
Launch angle: direction of initial velocity with
respect to horizontal
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Zero Launch Angle
In this case, the initial velocity in the y-direction is
zero. Here are the equations of motion, with x0 = 0
and y0 = h:
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Zero Launch Angle
Eliminating t and solving for y as a function of x:
This has the form y = a + bx2, which is the
equation of a parabola.
The landing point can be found by setting y = 0
and solving for x:
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Trajectory of a zero launch-angle projectile
horizontal points equally spaced
vertical points not
equally spaced
parabolic
y = a + bx2
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General Launch Angle
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General Launch Angle
-g
In general,
v0x = v0 cos θ and
v0Sin( )
v0y = v0 sin θ
This gives the equations
of motion:
v0Cos( )
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Range: the horizontal distance a projectile travels
As before, use
(y = 0 at landing)
and
Eliminate t and solve
for x when y=0
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Sin 

Range is maximum at 45o
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Range Gun
If the range gun launches a ball 6
meters with a launch angle of
45 degrees, at which of these
angles should a ball be
launched to land in a bucket
at 3 meters?
a) 10 degrees
b) 22.5 degrees
c) 60 degrees
d) 75 degrees
e) one would also need the launch
velocity of the range gun to know
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Range Gun
If the range gun launches a ball 6
meters with a launch angle of
45 degrees, at which of these
angles should a ball be
launched to land in a bucket
at 3 meters?
a) 10 degrees
b) 22.5 degrees
c) 60 degrees
d) 75 degrees
e) one would also need the launch
velocity of the range gun to know
The range is proportional to sin2θ, so to travel half the distance,
the ball would need to be launched with sin2θ = 0.5.
sin(2*75o) = 0.5, so θ=75o
Note: there are two angles that would work [sin(2*15o) = 0.5 also].
How are the two solutions different?
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Symmetry in projectile motion
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•• On a hot summer day, a young girl swings on a rope above
the local swimming hole. When she lets go of the rope her initial
velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If
she is in flight for 0.616 s, how high above the water was she when
she let go of the rope?
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Time to hit the water: t=0.616
Initial velocity: 2.25 m/s at 35o above the horizontal
y = (v0 sinθ) t - 1/2 g t2
y = 2.25 m/s * sin(35o) * (0.616 s) - 1/2 (9.8 m/s2) (0.616 s)2
= - 1.07 m
At time t= 0.616 s, the girl is 1.07 m below her starting position,
so her initial position was 1.07m above the water.
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Punts I
Which of the
three punts
h
has the
longest hang
time?
a
b
c
d) all have the same hang time
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Punts I
Which of the
three punts
h
has the
longest hang
time?
a
b
c
d) all have the same hang time
The time in the air is determined by the vertical motion!
Because all of the punts reach the same height, they all
stay in the air for the same time.
Follow-up: Which one had the greater initial velocity?
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Dropping the Ball III
A projectile is launched
from the ground at an
angle of 30°. At what point
in its trajectory does this
projectile have the least
speed?
a) just after it is launched
b) at the highest point in its flight
c) just before it hits the ground
d) halfway between the ground and the
highest point
e) speed is always constant
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Dropping the Ball III
A projectile is launched
from the ground at an
angle of 30º. At what point
in its trajectory does this
projectile have the least
speed?
a) just after it is launched
b) at the highest point in its flight
c) just before it hits the ground
d) halfway between the ground and the
highest point
e) speed is always constant
The speed is smallest at the
highest point of its flight
path because the
y-component of the velocity
is zero.
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A battleship simultaneously fires two shells at two enemy
submarines. The shells are launched with the same magnitude
of initial velocity. If the shells follow the trajectories shown,
which submarine gets hit first ?
a
b
c) both at the same time
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A battleship simultaneously fires two shells at two enemy
submarines. The shells are launched with the same magnitude
of initial velocity. If the shells follow the trajectories shown,
which submarine gets hit first ?
The flight time is fixed by the
motion in the y-direction. The
higher an object goes, the longer it
stays in flight. The shell hitting
submarine #2 goes less high,
therefore it stays in flight for less
time than the other shell. Thus,
submarine #2 is hit first.
a
b
c) both at the same time
Follow-up: Did you need to know that they had the same initial speed?
Spike it or launch it?
You are standing in a large
parking lot with a large walnut
in your hand. Ignoring air
resistance, at what angle
should you throw the walnut
to maximize your chance of
breaking it on the pavement?
(The initial speed and altitude of
release are independent of angle.)
a) throw it upwards, so that it falls farthest
b) throw it straight down
c) 45 degrees, so that it flies the farthest
d) this depends on how tall you are
e) up OR down, whatever. It’s all the same.
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Spike it or launch it?
You are standing in a large
parking lot with a large walnut
in your hand. Ignoring air
resistance, at what angle
should you throw the walnut
to maximize your chance of
breaking it on the pavement?
(The initial speed and altitude of
release are independent of angle.)
a) throw it upwards, so that it falls farthest
b) throw it straight down
c) 45 degrees, so that it flies the farthest
d) this depends on how tall you are
e) up OR down, whatever. It’s all the same.
If you throw it up, eventually it will come
back to its launch altitude, and when it
does it will have the same speed as you
threw it with!
You know this due to symmetry: it takes
the same amount of time to reach the
maximum altitude as it does to fall from the
maximum to the launch point, so the net
acceleration on the way down must equal
that on the way up.
(Of course, you do want to throw it up
OR down, so the acceleration is
maximized on contact with the
pavement. The horizontal velocity
doesn’t help break the nut.)
Monkey and the hunter: A monkey hangs from a
tree at a height h from the ground. A hunter lying
on the ground a distance D from the tree wants to
shoot the monkey, firing just as the monkey drops.
At what angle should the hunter shoot if the muzzle
speed of his bullet is v0?
Solution:
Monkey:
x0  D
y0  h
v x0  0
v y0  0
ax  0
ay   g
 xm  t   D  0  0  D
1
1
ym  t   h  0  gt 2  h  gt 2
2
2
Bullet:
x0  0
y0  0
v x0  v0 cos
v y0  v0 sin
ax  0
ay   g
 x b  t   0  v0 cos t  0
1
yb  t   0  v0 sin t  gt 2
2
1
 v0 sin t  gt 2
2
 v0 cos t
For the bullet to hit the monkey, the bullet and the
monkey must at some time tc be at the same place.
That is,
xm  t c   x b  t c  and ym  t c   yb  t c 
Thus,
x : D  v0 cos t c  t c 
D
v0 cos
1 2
1 2
h
D
h
y : h  gt c  v0 sin t c  gt c  h  v0 sin t c  t c 

 tan 
2
2
v0 sin v0 cos
D
Relative Motion
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You are riding on a Jet Ski at an angle of 35° upstream on a river
flowing with a speed of 2.8 m/s. If your velocity relative to the
ground is 9.5 m/s at an angle of 20.0° upstream, what is the speed
of the Jet Ski relative to the water? (Note: Angles are measured
relative to the x axis shown.)
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Now suppose the Jet Ski is moving at a speed of 12 m/s relative
to the water. (a) At what angle must you point the Jet Ski if your
velocity relative to the ground is to be perpendicular to the shore
of the river? (b) If you increase the speed of the Jet Ski relative
to the water, does the angle in part (a) increase, decrease, or stay
the same? Explain. (Note: Angles are measured relative to the x
axis shown.)
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Relativity Car
A small cart is rolling at
constant 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?
a) it depends on how fast the cart is
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
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Relativity Car
A small cart is rolling at
constant 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?
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.
a) it depends on how fast the cart is
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
when viewed
from train
when viewed
from ground
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Relativity Cart
As you stand on the platform, a train moves past you at a speed V. A man
standing on the train throws a ball straight up with an initial upward speed of v0.
Describe what each of you sees. Are the two views compatible?
Train view:
v0  v0 yˆ ; r0  0
1 2
y  t   v0t  gt
2
x t   0
Ball goes straight up, and
then straight down
Platform view:
v0  Vxˆ  v0 yˆ ; r0  0
1 2
y  t   v0t  gt
2
x t 
x  t   Vt  t 
V
x t  1  x t  
 v0
 g

V
2  V 
2

v 
 g 
y t    0  x t    2  x2 t 
 2V 
V 
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Relativity Car
+
=
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2-Dimensional Motion
&
Newton Laws of Motion
(sections 5.1-5.4)
Kinematics: Assumptions,
Definitions and Logical Conclusions
What have we done so far?
• Defined displacement, velocity, acceleration (also
position, distance, speed)...
• Defined scalers (like speed) and vectors (like velocity)
• Laid out assumptions about free-fall
• noticed that 2-dimensional motion is really just two,
simultaneous, 1-dimensional motions.
Used this to shoot a monkey, range
out a small cannon, etc.
This wasn’t physics. This was preparing the language
needed to talk about physics.
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Newton’s Laws
How can we consistently and generally describe
the way objects move and interact?
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Isaac Newton
1643-1727
Newton in a 1702
1689 portrait by
Godfrey Kneller
Nature and nature's laws lay hid in night;
God said "Let Newton be" and all was light.
Newton’s epitaph, Alexander Pope
I do not know what I may appear to the world, but to myself I
seem to have been only like a boy playing on the sea-shore,
and diverting myself in now and then finding a smoother
pebble or a prettier shell than ordinary, whilst the great ocean
of truth lay all undiscovered before me.
from a memoir by Newton
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Force
Force: push or pull
Force is a vector – it has magnitude and direction
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Mass
Mass is the measure of
how hard it is to change
an object’s velocity.
Mass can also be thought
of as a measure of the
quantity of matter in an
object.
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Newton’s First Law of Motion
If you stop pushing an object, does it stop
moving?
Only if there is friction!
In the absence of any net external force, an object at rest
will remain at rest.
In the absence of any net external force a moving object
will keep moving at a constant speed in a straight line.
This is also known as the Law of Inertia.
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Inertia
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Newton’s First Law
A hockey puck
slides on ice at
constant velocity.
What is the net
force acting on the
puck?
a) more than its weight
b) equal to its weight
c) less than its weight but more than zero
d) depends on the speed of the puck
e) zero
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Newton’s First Law
A hockey puck
slides on ice at
constant velocity.
What is the net
force acting on
the puck?
a) more than its weight
b) equal to its weight
c) less than its weight but more than zero
d) depends on the speed of the puck
e) zero
The puck is moving at a constant velocity, and
therefore it is not accelerating. Thus, there must be
no net force acting on the puck.
Follow-up: Are there any forces acting on the puck? What are they?
Newton’s First Law
You put your book on
a) a net force acted on it
the bus seat next to
b) no, or insufficient, net force acted on it
you. When the bus
c) it remained at rest
stops suddenly, the
book slides forward off
the seat. Why?
d) it did not move, but only seemed to
e) gravity briefly stopped acting on it
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Newton’s First Law
You put your book on
a) a net force acted on it
the bus seat next to
b) no, or insufficient, net force acted on it
you. When the bus
c) it remained at rest
stops suddenly, the
book slides forward off
the seat. Why?
d) it did not move, but only seemed to
e) gravity briefly stopped acting on it
The book was initially moving forward (because it was on
a moving bus). When the bus stopped, the book
continued moving forward, which was its initial state of
motion, and therefore it slid forward off the seat.
Calibrating force
Two equal weights exert twice the force of one; this
can be used for calibration of a spring:
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Experiment: Acceleration vs Force
Now that we have a calibrated spring, we can do
more experiments.
Acceleration is proportional to force:
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Experiment: Acceleration vs Mass
Acceleration is inversely proportional to mass:
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Newton’s Second Law of Motion
Acceleration is proportional to force:
Acceleration is inversely proportional to mass:
Combining these two observations gives
Or, more familiarly,
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Newton’s Second Law of Motion
An object may have several forces acting on it; the
acceleration is due to the net force:
SI unit for force Newton is defined using this equation as:
1 N is the force required to give a mass of 1 kg an
acceleration of 1 m/s2
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Units of force: Newtons
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Force of Gravity
The weight of an object is the force acting on it due to gravity
Weight: W = Fg = ma = mg vertically downwards
Weight is not mass!
Since
, the weight of an object in Newtons is
approximately 10 x its mass in kg
adult human
70 kg
700 N ~ 160 lbs.
There is no “conversion” from kg to pounds!
(Unless you specify what planet you are assuming)
Newton’s First and Second Laws
(I)
In order to change the velocity of an object –
magnitude or direction – a net force is required.
(II)
What about the bus... From the perspective of
someone who didn’t know they were on the bus?
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Inertial Reference Frames
(I)
In order to change the velocity of an object –
magnitude or direction – a net force is required.
(II)
Newton’s First and Second Laws do not
work in an accelerating frame of reference
An inertial reference frame is one in which the
first and second laws are true. Accelerating
reference frames are not inertial.
Was the bus an inertial reference frame?
Is the earth an inertial reference frame?
No, but acceleration due to earth’s rotation around
Its axis (0.034 m/s2), and due to earth’s rotation around sun
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(smaller) are negligible compared to g; so approximately yes.
Analyzing the forces in a system
Free-body diagrams:
A free-body diagram shows every force acting on
an object.
• Sketch the forces
• Isolate the object of interest
• Choose a convenient coordinate system
• Resolve the forces into components
• Apply Newton’s second law to each coordinate
direction
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Free-body Diagram
Example of a free-body diagram:
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Newton’s First Law
A book is lying at
rest on a table.
The book will
remain there at
rest because:
a) there is a net force but the book has too
much inertia
b) there are no forces acting on it at all
c) it does move, but too slowly to be seen
d) there is no net force on the book
e) there is a net force, but the book is too
heavy to move
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Newton’s First Law
A book is lying at
rest on a table.
The book will
remain there at
rest because:
a) there is a net force but the book has too
much inertia
b) there are no forces acting on it at all
c) it does move, but too slowly to be seen
d) there is no net force on the book
e) there is a net force, but the book is too
heavy to move
There are forces acting on the book, but the only
forces acting are in the y-direction. Gravity acts
downward, but the table exerts an upward force
that is equally strong, so the two forces cancel,
leaving no net force.
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Newton’s Third Law of Motion
Forces always come in pairs, acting on different
objects:
If object 1 exerts a force
object 2 exerts a force –
on object 2, then
on object 1.
These forces are called action-reaction pairs.
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Some action-reaction pairs
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Action-reaction pair?
a) Yes
b) No
Newton’s 3rd: F12 = - F21
action-reaction pairs are equal and
opposite, but they act on different bodies
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Newton’s Third Law of Motion
Although the forces are the same, the
accelerations will not be unless the objects
have the same mass.
Q: When skydiving, do you
exert a force on the
earth? Does the earth
accelerate towards you?
Is the magnitude of the
acceleration of the earth the
same as the magnitude of
your acceleration?
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Newton’s Third Law of Motion
Contact forces:
The force exerted by one
box on the other is
different depending on
which one you push.
Assume the mass of the two objects
scales with size, and the forces
pictured are the same. In which case
is the magnitude of the force of box 1
on box 2 larger?
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Truck on Frozen Lake
A very large truck sits on a
frozen lake. Assume there
is no friction between the
tires and the ice. A fly
suddenly smashes against
the front window. What
will happen to the truck?
a) it is too heavy, so it just sits there
b) it moves backward at constant speed
c) it accelerates backward
d) it moves forward at constant speed
e) it accelerates forward
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Truck on Frozen Lake
A very large truck sits on a
frozen lake. Assume there
is no friction between the
tires and the ice. A fly
suddenly smashes against
the front window. What
will happen to the truck?
a) it is too heavy, so it just sits there
b) it moves backward at constant speed
c) it continuously accelerates backward
d) it moves forward at constant speed
e) it continuously accelerates forward
When the fly hit the truck, it exerted a force on the truck
(only for a fraction of a second). So, in this time period, the
truck accelerated (backward) up to some speed. After the
fly was squashed, it no longer exerted a force, and the truck
simply continued moving at constant speed.
Follow-up: What if the fly takes off, with the same speed in the
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direction from whence it came?
A 71-kg parent and a 19-kg child meet at the center of an ice rink. They
place their hands together and push.
(a) Is the force experienced by the child more than, less than, or the
same as the force experienced by the parent?
(b) Is the acceleration of the child more than, less than, or the same as
the acceleration of the parent? Explain.
(c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the
magnitude of the parent’s acceleration?
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On vacation, your 1300-kg car pulls a 540-kg trailer away from a
stoplight with an acceleration of 1.9 m/s2
(a) What is the net force exerted by the car on the trailer?
(b) What force does the trailer exert on the car?
(c) What is the net force acting on the car?
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