Motion in Two Dimensions:

1
Motion in Two Dimensions

• Projectile Motion
• Circular Motion
• Angular Speed
• Simple Harmonic Motion
• Torque
• Center of Mass

2
Projectile Motion
A red marble is dropped off a cliff at the same
time a black one is shot horizontally. At any
point in time the marbles are at the same height,
i.e., theyre falling down at the same rate, and
they hit the ground at the same time. Gravity
doesnt care that the black ball is moving
sideways it pulls it downward just the same.
Since gravity cant affect horiz. motion, the
black particle continues at a constant rate.
With every unit of time, the marbles vertical
speed increases, but their horiz. speed remains
the same (ignoring air resistance).
continued on next slide
3
Projectile Motion
9.8 m/s2
Gravitys downward pull is independent of horiz.
motion. So, the vertical acceleration of each
marble is -g (for the whole trip), and the
sideways acceleration of each is zero. (Gravity
cant pull sideways). Whatever horiz. velocity
the black one had when shot is a constant
throughout its trip. Only its vertical velocity
changes. (A vertical force like gravity can only
produce vertical acceleration.)
9.8 m/s2
9.8 m/s2
continued on next slide
4
Projectile Motion (cont.)
t 0
vy 0
t 1
vy 1
If after one unit of time the marbles have one
unit of speed downward, then after two units of
time they have two units of speed downward, etc.
This follows directly from vf v0 a t. Since
v0 0, downward speed is proportional to time.
Note The vectors shown are vertical components
of velocity. The shot marble has a horizontal
component too (not shown) the dropped one
doesnt.
t 2
vy 2
t 3
vy 3
vy 4
continued on next slide
t 4
5
Projectile Motion (cont.)
vx v
vx
vy
Since the shot black marble experiences no horiz.
forces (ignoring air), it undergoes no horiz.
acceleration. Therefore, its horiz. velocity,
doesnt change. So, the horiz. vector has a
constant magnitude, but the vertical vector gets
longer. The resultant (the net velocity vector
in blue) gets longer and points more downward
with time. When t 0, v vx for the shot
marble. v vy for the dropped marble for the
whole trip.
v
vx
vy
vy
v
vx
vy
vy
v
continued on next slide
vx
6
Projectile Motion (cont.)
v
The trajectory of any projectile is parabolic.
(Well prove this later.) If its initial
velocity vector is horizontal, as with the black
marble, the launch site is at the vertex of the
parabola.
v
The velocity vector at any point in time is
tangent to the parabolic trajectory. Moreover,
velocity vectors are always tangent to the
trajectory of any moving object, regardless of
its shape.
v
continued on next slide
7
Projectile Motion (cont.)
?y 1
The vertical displacements over consecutive units
of time show the familiar ratio of odd numbers
that weve seen before with uniform acceleration.
Measured from the starting point, the vertical
displacements would be 1, 4, 9, 16, etc.,
(perfect squares), but the horiz. displacements
form a linear sequence since there is no
acceleration in that direction.
x 1
?y 3
x 2
?y 5
x 3
?y 7
x 4
continued on next slide
8
Projectile Example
A rifle is held perfectly horizontally 1.5 m over
level ground. At the instant the trigger is
pulled, a second bullet is dropped from the tip
of the barrel. The muzzle velocity of the gun is
80 m/s. 1. Which bullet hits the ground first?
answer 2. How fast is each bullet moving
after 0.3 s ? answer
They hit at same time.
Use vf v0 a t and use vertical info only
v0 0, a -9.8 m/s2, and t 0.3 s. We get
vy in the pic for each bullet is -2.94 m/s.
Using the Pythagorean theorem for the fired
bullet we get 80.054 m/s in a direction tangent
to its path.
continued on next slide
80 m/s
80 m/s
dropped bullet after 0.3 s
fired bullet after 0.3 s
vy
vy
9
Projectile Example (cont.)
3. How far away does the fired bullet land (its
range)? answer
The first step is to find the its hang time.
This is the same hang time as the dropped bullet.
Use ?y v0 t 0.5 a t 2 with only vertical
data -1.5 (0) t (0.5) (-9.8) t 2. So, t
0.5533 s. The whole time the bullet is
falling its also moving to the left at a
constant 80 m/s. Since horizontally v is
constant, we use d v t with only horiz. info
d (80 m/s) (0.5533 s) 44.26 m. Note When
a 0, ?x v0 t 0.5 a t 2 breaks down to d
v t.
80 m/s
1.5 m
10
Projectiles Fired at an Angle
Now lets find range of a projectile fired with
speed v0 at an angle ?. Step 1 Split the
initial velocity vector into components.
continued on next slide
11
Projectiles Fired at an Angle (cont.)

Step 2 Find hang time. Use ?y v0t ½ a t 2
with only vertical data
?y (v0 sin? ) t ½ (-g)t 2
Over level ground, ?y 0. Divide through by t
0 v0 sin? - 4.9 t, and t (v0 sin? ) / 4.9
Note If we had shot the projectile from a 100
m cliff, ?y would be -100 m.
ALL THINGS VERTICAL
continued on next slide
12
Projectiles Fired at an Angle (cont.)

ALL THINGS HORIZONTAL
Step 3 Now that we know how long its in the
air, we know how long it travels horizontally.
(The projectiles vertical and horizontal
movements are completely independent.) Use ?x
v0t ½ a t 2 again, this time with only
horizontal data
?x (v0 cos? ) t ½ (0) t 2 (v0 cos? ) t
This is the same as sayinghoriz. distance
horiz. speed ? time In other words, d v t
v0
v0 sin?
?
continued on next slide
v0 cos?
13
Picklemobile Example
A stuntman drives a picklemobile off a 350 m
cliff going 70 mph. The angle of elevation of
the cliff is 21?. Hes hoping to make it across
a 261 m wide river and land on a ledge 82 m high.
Does he make it ?
70 mph
Well, the first thing we have to do is convert
the initial velocity into m/s
21?
70 mi
1609 m
h
31.2861 m/s



h
mi
3600 s
350 m
continued on next slide
261 m
82 m
14
Picklemobile Example (cont.)
We resolve the initial velocity into components.
Then we find the picklemobiles hang time (which
is the same as if it had been shot straight up at
about 11.2 m/s), with ?y 82 m - 350 m -268
m.
31.2861 m/s
-268 11.2119 t - 4.9 t 2 4.9
t 2 - 11.2119 t - 268 0 t -6.3394 s
or 8.6276 s (using quadratic formula or
computer)
350 m
continued on next slide
261 m
82 m
continued on next slide
15
Picklemobile Example (cont.)
We want the positive answer for t. The
interpretation of the negative answer is that if
the pickle car had been launched from the
height of the ledge, it would have taken about
6.3 s to reach the edge of the cliff.
Anyway, for 8.62757 s the pickle mobile is in
the air and traveling to the right at about 29
m/s. Therefore, its range is (29.2081 m/s)
(8.6276 s) ? 252 m lt 261 m. Alas, the poor
picklemobile doesnt make it.
29.2081 m/s
82 m
continued on next slide
16
Picklemobile Example (cont.)
What max height does the pickle mobile attain?
11.2119 m/s
It attains the same max height as if it had been
shot up at about 11.2 m/s. Since its vertical
velocity is zero at its high pt., we have 02 -
(11.2119)2 2(-9.8) ?y. So, ?y 6.41 m. Add
350 m and the max height is 356.41 m.
parabolic trajectory
350 m
continued on next slide
82 m
17
Picklemobile Example (cont.)
What is the impact velocity of the pickle mobile
(the velocity upon splash down)?
11.2119 m/s
The horiz. component is the same at landing as it
was on liftoff. We must find the final vertical
velocity vf2 - (11.2119)2 2(-9.8) (-350). So,
vf -83.5805 m/s.
29.2081 m/s
29.2081 m/s
?
The Pythag. theorem gives us the magnitude of the
resultant. ? tan-1 (83.5805 / 88.5371)
70.74?. Thus the impact velocity is about 88.5
m/s at 71? below the horizontal.
83.5805 m/s
350 m
88.5371 m/s
18
Parabolic Proof
A projectile is shot with speed v0 at an angle
?. Its vertical position is given by y (v0
sin? ) t ½ (-g) t 2. Here y is the dependent
quantity, and t is the independent quantity.
Everything else is a constant.
The projectiles horizontal position is given by
x (v0 cos? ) t. Only x and t are
variables, and t x / (v0 cos? ). Lets
substitute this for t in the equation for y
y (v0 sin? ) t ½ (-g) t 2 y (v0 sin? )
x / (v0 cos? ) - ½ g x / (v0 cos? )2
g
x 2
y (tan? ) x -
2 v02 cos2?
The coefficients of x and x 2 are constants.
Since the leading coef. is negative, this is the
equation of a parabola opening down.
19
Symmetry and Velocity
The projectiles speed is the same at points
directly across the parabola (at the same
vertical position). The angle is the same too,
but with opposite orientation.
Horizontal speeds are the same throughout the
trajectory.
Vertical speeds are
the same only at points of
equal height.
?
?
The vert. comp. shrinks then grows in opposite
direction at a const. rate (-g). The resultant
velocity vectors orientation and magni-tude
changes, but is always tangent.
The horiz. comp. doesnt change. At the peak,
the horiz. comp. equals the resultant velocity
vector.
?
20
Symmetry and Time
Over level ground, the time at the peak is half
the hang time. Notice the symmetry of times at
equal heights relative to the 10 unit mark. The
projectile has covered half its range when it has
peaked, but only over level ground. Note near
the peak the object moves more slowly than
when lower to the ground. It rises
3/4 of its max height in only 1/2 of
its rising time. (See

if you can prove this for

an arbitrary launch

velocity.)
t 10
t 15
t 5
t 3
t 17
t 0
t 20
21
Max height hang time depend only on initial
vertical velocity
Each initial velocity vector below has the a
different magnitude (speed) but each object will
spend the same time in the air and reach the same
max height. This is because each vector has the
same vertical component. The projectiles will
have different ranges, however. The greater the
horizontal component of initial velocity, the
greater the range.
22
Max Range at 45?
Over level ground at a constant launch speed,
what angle maximizes the range, R ? First
consider some extremes When ? 0, R 0,
since the object is on the ground from the moment
its launched. When ? 90?, the object goes
straight up and lands right on the launch site,
so R 0 again. The best angle is 45?, smack dab
between the extremes.
proof on next slide
Here all launch speeds are the same only the
angle varies.
76?
45?
38?
23
Range Formula Max Range at 45?
First find the time. Note that ?y 0, since
the projectile starts and stops at ground level
(no change). ?y v0 t ½ at 2. So, 0
(v0 sin? ) t - ½ g t 2 Since the ground is
level we divide through by t giving us t 2
v0 sin? / g. Then, R (v0 cos? ) t (v0 cos?
) (2 v0 sin? / g) 2 v02 sin? cos? / g. By the
trig identity sin 2? 2 sin? cos?, we get R
v02 sin 2? / g. Since v0 and g are fixed, R
is at a max when sin 2? is at a max. When the
angle, 2?, is 90?, the sine function is at its
maximum of 1. Therefore, ? 45?.
v0
v0 sin?
?
v0 cos?
24
Max Range when ?y ? 0
When fired from a cliff, or from below ground, a
projectile doesnt attain its max range at 45?.
45? is only the best angle when a projectile is
fired over level ground. When fired from a cliff,
a projectile attains max range with a launch
angle less than 45 ? (see next slide). When
fired from below ground, a projectile attains max
range with a launch angle greater than 45 ?.
25
Range when fired from cliff
If ground were level, the 45? launch would win.
45?
lt 45?
Launch speeds are the same.
Because the lt 45? parabola is flatter it
eventually overtakes 45 ? parabola.
26
Ranges at complementary launch angles
An object fired at angle ? will have the same
range as when its fired at the same speed at an
angle 90? - ?. Reason R 2v02 sin? cos? / g,
and the sine of an angle is the cosine of its
complement (and vice versa). For example, R at
40? is 2v02 sin 40? cos 40? / g
2v02 cos 50? sin 50? / g
R at 50?.
75?
15?
50?
40?
27
Monkey in a Tree
Heres a classic physics problem You want to
shoot a banana at a monkey up in a tree. Knowing
that the monkey will get scared and let go of the
branch the instant he hears the sound of the
banana gun, how should you aim a little above, a
little below, or right at him?
Monkey in a tree web site
28
Monkey in a tree explanation
The reason you should aim right at the monkey
even though the monkey lets go right when you
pull the trigger is because both the monkey and
the banana are in the air for the same amount of
time before the collision. So, with respect to
where they would have been with no gravity, they
fall the same distance.
gravity-free monkey
banana path without gravity
monkey w/ gravity
path with gravity
29
Homerun Example
From home plate to the center field wall at a
ball park is 130 m. When a batter hits a long
drive the ball leaves his bat 1 m off the ground
with a velocity of 40 m/s at 28? above the
horizontal. The center field wall is 2.6 m high.
Does he hit a homerun?
40 m/s
2.6 m
28?
1 m

130 m
Lets first check the range to see if it even has
a chanceR v02 sin 2? / g 402 sin 56 / 9.8
135.35 m. So, if it can clear the wall, it
will make it. We need to determine its vertical
position when its horizontal position is 130 m.
If its 1.6 m or more, its a homer. Lets first
find the time when the ball is 130 m away
(horizontally) from the point where it was
walloped.
continued on next slide
30
Homerun (cont.)
t d / v (130 m) / (40 cos 28?) 3.68085 s.
Lets see how high up it is at this time ?y
(40 sin 28?) (3.68085) - 4.9 (3.68085)2 2.73
m, which is 3.73 m above the ground, out of the
reach of a leaping outfielder. Therefore, its a
homer!
In real life the batter wouldnt be so fortunate.
His hit would only have gone a fraction of the
distance with air resistance. What is barely a
homer in a vacuum is a mere blooper in air.
Homerun balls in the major league would go about
700 or 800 ft in a vacuum! This means balls
would routinely be hit out over the stadium into
the parking lot.
31
Trebuchet Example
A trebuchet launches a 180 kg lead sphere at the
wall of a medieval castle 120 m away. The
projectile impacts the wall 23 m up. Its high
point in the air occurs 2/3 of the way to the
wall. The trebuchet takes 0.6 s to launch the
sphere and releases it at a height of 5 m. Find
the launch velocity of the sphere and the average
force the trebuchet exerts on it.
continued on next slide
lead sphere
tree trunk
rope
axle
swinging counterweight
wheeled wooden frame that rolls backwards during
launch
Trebuchet
32
Trebuchet (cont.)
The high pt. occurs horizontally at a distance of
half of what its range would be over level
ground, which is 2/3 of 120 m.
(80, h )
(120, 23) impact
So, 80 R / 2 v 2 sin 2? / 2g ? v 2
1568 / sin 2?
v
v sin?
?
continued on next slide
v cos?
(0, 5)
33
Trebuchet (cont.)
Let t time at impact. Horizontally 120 (v
cos? ) t. Vertically 18 (v sin? ) t - 4.9 t
2. Now substitute for t 18 (v sin? )
120 / (v cos? ) - 4.9 120 / (v cos? ) 2
14400
? 18 120 tan? - 4.9
v2 cos2?
From the last slide, v 2 1568 / sin 2?.
Substitute for v 2 into the equation above
14400
18 120 tan? - 4.9
1568
cos2?
sin 2?
15 sin 2?
45 sin 2?
? 6 40 tan ? -
? 18 120 tan? -
cos2?
cos2?
continued on next slide
34
Trebuchet (cont.)
Now we have an equation with just one variable,
but its not an easy one to solve. We can solve
for ? by graphing the equation below and
looking for a y-intercept (in radian mode).
15 sin 2?
y 40 tan? -
- 6
cos2?
Domain 0, ? / 2
Domain 0.52, 0.55
180?
? 0.5404 radians
30.9626?
? radians
continued on next slide
35
Trebuchet (cont.)
Now lets substitute this value for ? into v 2
1568 / sin 2?. This gives us v 42.1557 m/s.
Using the fact that the trebuchet pushes on the
lead sphere for 0.6 s, we can find the spheres
acceleration, (since it starts from rest and we
now know vf for the launch phase).
vf v0 a t aavg (42.1557 - 0) / (0.6)
70.2595 m/s2 So, Favg m aavg 180 (70.2595)
12 646.71 N
Because the force (and therefore the
acceleration) is not constant, what weve
calculated is the average force and acceleration.
36
Circular Motion
v
v
v
Suppose you drive a go cart in a perfect circle
at a constant speed. Even though your speed
isnt changing, you are accelerating. This is
because acceleration is the rate of change of
velocity (not speed), and your velocity is
changing because your direction is changing!
Remember, a velocity vector is always tangent to
the path of motion.
37
Tangential vs. Centripetal Acceleration
Suppose now you drive your go cart faster and
faster in a circle. Now your velocity vector
changes in both magnitude and direction. If you
go from start to finish in 4 s, your average
tangential acceleration is at (18 m/s - 10
m/s) / 4 s 2 m/s2
15 m/s
18 m/s
finish

10 m/s
start
So youre speeding up at a rate of 2 m/s per
second. This is the rate at which your velocity
changes tangentially. But what about the rate at
which your velocity changes radially, due to its
changing direction? This is your centripetal (or
radial) acceleration.
So how do we calculate the centripetal
acceleration ? ? ? Stay tuned!
38
Centripetal Acceleration
Lets find a formula for centripetal acceleration
by considering uniform circular motion. By the
definition of acceleration, a (vf - v0) / t.
We are subtracting vectors here, not speeds,
otherwise a would be zero. (v0 and vf have the
same magnitudes.) The smaller t is, the
smaller ? will be, and the more the blue sector
will approximate a triangle. The blue triangle
has sides r, r, and v t (from d v t ). The
vector triangle has sides v, v, and vf - v0 .
The two triangles are similar (side-angle-side
similarity).

vf v
r
v0 v
?
r
vf - v0
v t
v0
r
?
?
r
vf
continued on next slide
39
Centripetal Acceleration (cont.)

vf v
By similar triangles,
vf - v0
v
r

v0 v
r
v t
?
r
So, multiplying both sides above by v, we have
vf - v0
v 2
ac

r
t
v t
vf - v0
m2 / s2
(m/s)2
r
v
Unit check

r
m
m
?
v
?
m

s2
40
Centripetal acceleration vector always points
toward center of circle.
v

ac

at
moving counterclockwise speeding up
moving counterclockwise slowing down
Centripetal means center-seeking. The
magnitude of ac depends on both v and r.
However, regardless of speed or tangential
acceleration, ac always points toward the center.
That is, ac is always radial (along the
radius).
41
Resultant Acceleration
The overall acceleration is the vector sum of the
centripetal acceleration and the tangential
acceleration. That is, a ac at This is
true regardless of the direction of motion. It
holds true even when an object is not moving in a
perfect circle. Note The equation above does
not include v. Vectors of different quantities
cannot be added!
a
ac
at

moving counterclockwise while speeding up or
moving clockwise while slowing down
42
Non-circular paths
Here we have an object moving along the brown
path at a constant speed (at 0). ac changes,
though, since the radius of curvature changes.
At P1 the path is approximated by the large
green circle, at P2 by the smaller orange one.
The smaller r is, the bigger ac is.
v 2
ac

r
R2
v
R1
P2
ac
ac
ac
P1
v
43
Friction and ac
Youre cruising at a constant 20 m/s on a winding
highway. The radius of curvature where you are
is 60 m. Your centripetal acceleration is ac
(20 m/s)2 / (60 m) 6.67 m/s2 The force that
causes this acceleration is friction, which is
why its hard to turn on ice. Friction, in this
case, is the centripetal force. The sharper
the turn or the greater your speed, the greater
the frictional force must be.
60 m
ac
20 m/s

overhead view
continued on next slide
44
Friction and ac (cont.)
Since youre not speeding up, f is the net
force, so Fnet f ?s N ?s mg ma. We
use ?s because youre not sliding (or even
moving) radially. Thus, ?s mg ma. Mass
cancels out, showing that your centripetal
acceleration doesnt depend on how heavy your
vehicle is. Solving for a we have a ?s
g. In the diagram N is pointing out of the slide.
Also, a ac in this case since at 0.
r
f
ac
v
m

overhead view
45
Centripetal Force, Fc
From F ma, we get Fc mac mv2 / r.
If a body is turning, look at all forces acting
on it, and find the net force. The component of
the net force that acts toward the center of
curvature (perpendicular to the bodys motion) is
the centripetal force. The component that acts
parallel to its motion (forward or backwards) is
the tangential component of the net force.
46
Forces that can provide a centripetal force

• Friction, as in the turning car example
• Tension, as in a rock whirling around while
  attached to a string, or the tension in the
  chains on a swing at the park.
• Normal Force, as in a round-up ride at an
  amusement park (that spins the floor drops
  out), or the component of normal force on a car
  on a banked track that acts toward the center.
• Gravity The force of gravity between the Earth
  and sun keeps the Earth moving in a nearly
  circular orbit.
• Any force directed toward your center of
  curvature, such as an applied force.

Picture on upcoming slides
47
Simple Pendulum
Two forces act on a swinging pendulum, tension
and weight. Tension acts radially. We break the
weight vector into a radial component (green) and
tangential one (violet). Blue is bigger than
green, otherwise there would be no net
centripetal force, and the mass couldnt turn.
Fc T - mg cos? mac mv2 / L. Ft mg
sin? mat Ft is the force that speeds the
mass up or slows it down. The weight is a
constant, but since ? changes, so do the
weights components. mg cos? is greatest when
the mass is at its low point. This is also where
its moving its fastest. For these two reasons
T is the greatest when the mass is at the low
point.
string of length L
?
T
m
mg sin?
mg cos?
?
mg
48
Facts about the Simple Pendulum

• A pendulums period is the same for big arcs as
  it is for little arcs, so long as the angle
  through which it swings isnt real large. (The
  average speed, therefore, is greater when the arc
  is bigger, because it must cover a bigger
  distance in the same time.)
• The period is independent of the mass.
• The period depends only on the pendulums
  length.
• The period T 2? (proven in advanced
  physics)
• This formula gives us a way to measure the
  acceleration due to gravity (which varies
  slightly with location) by measuring the period
  and length of a pendulum.
• Dont confuse the symbol T, which is used for
  both period and tension.

49
Conical Pendulum
Like a tether ball, the mass hangs from a rope
and sweeps out a circular path, and the rope a
cone. ? is a constant. The vertical component
of T balances mg. The horiz. comp. of T is
the centripetal force.
?
?
T
T
ac
ac
v
m
m
v
mg
mg
continued on next slide
50
Conical Pendulum (cont.)
All vectors shown are forces. v points into or
out of the slide, depending on the direction in
which the mass moves. ac is parallel to T
sin?, which serves as the centripetal force.
?
T cos?
T
?
T cos? mg T sin? mv2 / r
T sin?
m
mg
Dividing equations tan? v2 / rg
51
Loop-the-loop in a Plane
v
A plane flies in a vertical circle, so its
upside down at the high point. Its speed is
constant, but because of its nonlinear motion,
the pilot must experience centripetal
acceleration. This ac is provided by a
combination of her weight and her normal force.
mg is constant N is not. N is the force her
pilots chair exerts on her body.
Ntop
mg
Nbot
v
mg
continued on next slide
52
Loop-the-loop (cont.)
Top Normal force and weight team up to provide
centripetal force Ntop mg mv2 / r. If the
pilot were sitting on a scale, it would say shes
very light.
v
Ntop
mg
r
Bottom Weight works against normal force, so N
must be bigger down here to provide the same
centripetal force Nbot - mg mv2 / r. (Fc
has a constant magnitude since m, v, and r
are constants.) Here a scale would say that the
pilot is very heavy.
Nbot
v
mg
continued on next slide
53
Loop-the-loop (cont.)
The normal force (force on pilot due to seat)
changes throughout the loop. This case is
similar to the simple pendulum (the only
difference being that speed is constant here).
Part of the weight opposes N, and the net radial
force is the centripetal force N - mg cos? mac
mv2 / r
r
?
N
?
mg sin?
mg cos?
Weve been discussing the pilot, but what force
causes the plane to turn? Answer
mg
The air provides the centripetal force on plane
54
Angular Speed, ?
B
Linear speed is how fast you move, measured as
distance per unit time. Angular speed is how
fast you turn, measured as an angle per unit
time. The symbol for angular speed is the small
Greek letter omega, ?, which looks like a curvy
w. Units for angular speed include degrees
per second radians per second and rpm
(revolutions per minute).
110?
A
3 m
Suppose an object moves steadily from A to B
along the circle in 5 s. Then ? 110? / 5 s
22? / s. The distance it covers is (110? /
360?)(2?)(3) 5.7596 m. So its linear speed is
v (5.7596 m) / (5 s) 1.1519 m / s.
55
Arc length s r ?
s
If ? is in radians, then the arc length, s, is
? times r. This follows directly from the
definition of a radian. One radian is the angle
made when the radius of a circle is wrapped along
the circle.
?
r
r
When the arc length is as long as the radius, the
angle subtended is one radian. (A radian is
really dimensionless, since its found by
dividing a length by a length.)
1 radian
r
56
v r ?
M
The Three Stooges go to the park. Moe and Larry
are on a merry-go-round ride of radius r that
Larry is pushing counterclockwise, running at a
speed v. In a time t, Moe goes from M to M
and Curly goes from C to C. Moe is twice as
far from the center as Curly. The distance Moe
travels is r ?, where ? is in radians.
Curlys distance is ½ r ?. Both stooges sweep
out the same angle in the same time, so each has
the same angular speed. However, since Moe
travels twice as far, his linear speed in twice
as great.
C
?
v / 2
C
v
M
s r ?
r ? t
? t
st
r


v r ?
57
Ferris Wheel Problem
Schmedrick is working as a miniature ferris wheel
operator (radius 2.1 m). He gets a little
overzealous and cranks it up to 75 rpm. His
little brother Poindexter flies out at point P,
when he is 35? from the low point. At the low
point the wheel is 1 m off the ground. A 1.5 m
high wall is 27 m from the low point of the
wheel. Does Poindexter clear the wall?
P
Strategy outlined on next slide
58
Ferris Wheel Problem-Solving Strategy
1. Based on his angular speed and the radius,
calculate Poindexters linear speed. 2.
Break his launch velocity down into vertical
and horizontal components. 3. Use trig to
find the height of his launch. 4. Use trig to
find the horizontal distance from the
launch site to the wall. 5. Calculate the
time it takes him to go that far
horizontally. 6. Calculate height at that
time. 7. Draw your conclusion.
16.4934 m/s
vx 13.5106 m/s vy0 9.4602 m/s
1.3798 m
25.7955 m
1.9093 s
1.5798 m
He just makes it by about 8 cm !
59
Springs
m mass of weight hanger and weight. (Were
assuming the mass of the spring itself is
negligible.) x the amount the spring stretches
due to some force, in this case the weight mg.
x
m
The amount of stretch (x) depends on how much
force is applied to the spring (mg) and the
stiffness of the spring.
continued on next slide
60
Hookes Law F - k x
Hookes Law says that the stretch is proportional
to the force the spring exerts, F, which is the
reaction pair to the force causing the stretch.
Here F is mg up, since the spring is pulling up
on the weight hanger with a force of mg. The
negative sign is there because as the spring is
stretched downward, the force it exerts on the
weight is upward. The constant of
proportionality is the spring constant, k. The
bigger k is, the stiffer the spring, and the
harder it is to stretch it.
x
m
continued on next slide
61
Spring Constant, k
Suppose that to stretch a certain spring 5 cm,
you grab it and pull with 100 N of force. From
Hookes Law, we get k (100 N) / (5 cm) 20 N
/ cm. This means for every cm you want to
stretch it, you must apply another 20 N.
100 N (by hand)
5 cm
Note k is always positive. In the formula F
- k x, F is to the right (the direction the
spring pulls on your hand), and x is to the
left (the direction in which the spring was
elongated). The minus sign accounts for this
difference in direction.
If you applied different forces and plotted the
spring forces against the corresponding stretches
they produced (an F vs. x graph), what would
graph look like?
a line with a slope of -k
answer
62
Simple Harmonic Motion
A mass bobbing up and down on a spring undergoes
simple harmonic motion (SHM). SHM occurs
whenever a bodys position as a function of time
is of the form y A cos b (t - c) d.
Youll see this in trig class too.
y vertical position (our dependent variable)
t time (our independent variable)A
amplitude (maximum distance from equilibrium) d
vertical displacement of the equilibrium from
your reference point (point from which
you measure) c phase shift (time from start of
experiment until the mass reaches
its first maximum displacement). c 0 if
experiment starts with spring fully
stretched or compressed. The period of the mass
is the amount of time it takes to bob up and
down once. b is used to determine the period.
The period, T, is given by T 2? / b. 2?
is the normal period of the sine cosine
functions (see next slide).
m
equilibrium pt.
63
Sine Cosine Graphs y A b (t - c)
d
y cos t
period 2?
y 1 cos 1(t - 0) 0
amp 1
-Pi
Pi
A 1b 1c 0d 0T 2? / 1 2?
y cos t begins at a max y sin t begins at
equilibrium. Their periods and amplitudes are
the same. Their graphs are congruent, differing
only by a 90? horizontal shift.
y 1 sin 1(t - 0) 0
period 2?
y sin t
A 1b 1c 0d 0T 2? / 1 2?
amp 1
64
Cosine graph Amplitude y A cos b (t - c)
d
The leading coefficient determines the amplitude,
which equals A. If A is negative, the graph
is reflected about the time axis. For the red
graph, our mass is pushed up, compressing the
spring 2 m and is released at time zero. In the
green graph, its pulled down 1.5 m and released
at time zero. Note all graphs have the same
period or 2?, and none has a phase (horiz.) shift
or a vertical shift.
y cos t
A 1 m
-Pi
Pi
y 2 cos t
A 2 m
y -1.5 cos t
A 1.5 m
65
Cosine graph Period y A cos b (t - c)
d
T 2? / b
y cos t
2?
The bigger b is, our scrunch factor, the more
scrunched up the graph gets. The red graph
completes 3 cycles in same time the blue graph
does one, and reds period is 3 times shorter.
So, the mass in the red graph goes up and down 3
times more often. The green graph completes half
as many cycles as blue and its mass has twice the
period. Note bs units cancel ts.
-Pi
Pi
b 1 s-1
y cos 3 t
2? / 3
b 3 s-1
y cos 0.5 t
b ½ s-1
4?
66
Cosine graph vertical displacement
y A cos b (t - c) d
y cos t
Adding or subtracting outside of the cosine
function shifts the graph vertically. The dotted
lines are lines of symmetry. For our mass on a
spring, this is where its in equilibrium (where
the spring is neither stretched nor compressed).
For the red graph, the mass is in equilibrium 1.5
meters above the point from which we chose to
make our measurements.
-Pi
Pi
d 0
y cos t 1.5
d 1.5
d -0.5
y cos t - 0.5
67
Cosine graph Phase Shift y A cos b (t -
c) d
Adding or subtracting inside the cosine function
shifts the graph horizontally. (Adding shifts it
left, subtracting right.) Like the mass in the
blue graph, the red mass was compressed 1 m and
then released, but the clock was started 1.2 s
after it had reached max compression. The green
mass was also compressed 1 m and released, but
the clock was started 0.9 s before the mass
reached max compression.
y cos t
c 0
c -1.2 s

y cos (t 1.2)
c 0.9 s

y cos (t - 0.9)
68
SHM / Spring Problem part 1
Schmedrick wants to go bungee jumping but cant
afford it, so he takes a dive off a bridge 60 ft
over a ranging river with a giant slinky
connecting him to the bridge. After the leap he
begins bobbing up and down. While bobbing, hes
moving the fastest when hes 27 ft above the
river. He gets as high as 39 ft above the
river. It takes him 5 s to go from his high
point to his low point. Schmedrick weighs 105
lb, and the natural length of the slinky is 20
ft. 1. Whats the spring constant of the
slinky? The point at which Schmedrick is moving
the fastest while bobbing is the equilibrium
point. (v 0 at the high and low points and v
is a max half way in between.) This happens at
27 ft above the river, which is 60 - 27 33 ft
from the bridge. So, if Schmed were just
hanging from the slinky, his weight would stretch
it 33 - 20 13 ft. Thus, k (105 lb) / (13
ft) 8.08 lb / ft. This means every foot of
stretch requires about an additional 8 pounds of
force.
69
SHM / Spring Problem part 2
2. If the clock starts when he jumps off the
bridge, write Schmedricks position as a function
of time using the river as a reference
point. Ignoring air resistance, a mass on a
spring undergoes SHM (derived in advanced
physics). Hence we can use y A cos b (t -
c) d. Our task, then, is to find A, b, c,
and d. d 27 ft since the equilibrium pt,
Eq, is 27 ft above our reference position. The
high pt, H, is given to be 39 ft. This is 12 ft
above Eq, so A 12 ft, and the low pt, L, is 12
ft below Eq. Since it takes 5 s for him to go
from H to L, his period is 10 s. So, T 2?
/ b ? b 2? / (10 s) 0.628 s-1. If
Schmedrick had been at H or L at t 0, c
would be zero. But the clock starts early (when
y 60 ft), so lets figure out how long it
takes him to fall to H (assume free fall).
continued on next slide
70
SHM / Spring Problem part 2 (cont.)
bridge
y 60 ft
If wed like to continue working in feet, we
cant use 9.8 m/s2 for g. Instead we use its
equivalent 32 ft / s2. He falls 21 ft to H.
Thus, -21 (0) t 0.5 (-32) t 2 ? t
1.146 s, and this is our c value. (This is
only an approximation, since the spring begins
stretching after he falls 20 ft.) Putting it all
together, we have
y 39 ft
H
y(t) 12 cos 0.628 (t - 1.146) 27
where t is the time in seconds from the instant
he jumps, and y is his height above the water
in feet. Note This formula only works for t gt
1.146 s, since thats when SHM begins.
y 27 ft
Eq
y 15 ft
L
Check by plugging in some values into the
function (using radian mode in the calculator)
y(1.146) 12 cos 0 27 12(1) 27 39 ft.
(H) y(6.146) 12 cos 0.628 (5) 27 12
(-1) 27 15 ft. (L) So hes at L
5 s after hes at H.
y 0
river
71
SHM / Spring Problem parts 3 4
3. How high up is Schmedrick 14 s into the
experiment?
y (14) 12 cos 0.628 (14 - 1.146) 27 24.4
ft. (When we calculated b, we used 2? rather
than 360? for the normal period of the cosine
function, so we must put our calculators in
radian mode.)
4. At what times is he 30 ft above the river?
He can only be at one position at a particular
time, but he can be at a particular position at
many different times. (This is the nature of a
function.) Thus, 30 12 cos 0.628 (t -
1.146) 27, and we must solve for t. First
isolate the cosine function by subtracting 27 and
dividing by 12 0.25 cos 0.628 (t - 1.146)
. Think of the quantity in the brackets as an
angle. We want to know what angles have a cosine
of 0.25. One angle is cos-1 (0.25) 1.318
(radians). But this is only one of an infinite
number of angles whose cosine is 0.25.
continued on next slide
72
SHM / Spring Problem part 4 (cont.)
y
P1
On a unit circle (radius 1, center at origin)
?
x
One angle in which cos? 0.25 is 1.318 radians ?
75.5? (in Q I). Another occurs at 4.965 ? 284.5?
(in Q IV). To get to P1 or P2 from the origin,
you must go 0.25 units to the right. Thus, both
P2
angles have the same cosine of 0.25. Negative
angles and angles over 2? (or 360?) that
terminate at P1 or P2 will also have a cosine of
0.25. These angles are 1.318 2n? and 4. 965
2n?, where n 0, ?1, ?2, ?3, . Ex when n
5, cos1.318 2(5)? 0.25. Here weve gone
around the circle 5 times and stopped at P1.
continued on next slide
73
SHM / Spring Problem part 4 (cont.)
We are in the process of solving 0.25 cos
0.628 (t - 1.146) . Our angle is 0.628 (t -
1.146). Therefore,
t 1.146 (1.318 2n?) / 0.628or t 1.146
(4.965 2n?) / 0.628
0.628 (t - 1.146) 1.318 2n? or0.628 (t -
1.146) 4. 965 2n?
We only want t values gt 1.146 s (when SHM
begins). Substituting in for n, we get
t 3.245 s, 13.250 s, 23.255 s, ort
9.052 s, 19.057 s, 29.062 s,
These are all the times when Schmedrick is 30 ft
above the water. The top row lists times when
hes on his way down. The bottom row lists times
when hes bouncing back up. The difference
between consecutive times on a given row is 10 s
(neglecting some slight compounded rounding
error), which is the period of his motion. That
is, every 10 s he is at the same position moving
in the same direction.
74
SHM / Spring Problem part 5
Weve been neglecting air resistance throughout
this problem. How would Schmedricks position
vary with time in real life?
Its not exactly SHM any more, but its similar.
He will still bob up and down relative to the
same equilibrium point. It would take him a
little longer to fall to the high point for two
reason. One, drag from the air slows him down.
Two, after falling 20 ft (natural length of the
slinky) the slinky begins to stretch out and
exerts an upward force on him, slowing him down.
This would change our c value. The most
interesting change, however, is the amplitude.
In real life, the amplitude itself is a function
of time, diminishing with each bounce due to air
resistance and friction in the slinky. This is
damped periodic motion. Lets checked out some
graphs of Schmedricks motion, damped (real life)
and undamped (idealized simple harmonic motion).

continued on next slide
75
SHM / Spring Problem part 5 (cont.)
In real life drag forces would decay his
amplitude until hes hanging at the equilibrium
point 27 ft above the water, barely moving. The
amplitude decays exponentially, meaning that a
certain percentage of it is lost with each
bounce. The percentage lost depends on the
viscosity of the fluid in which the mass is
moving. The damping would be more severe in
water than in air.
undamped
damped
76
Things that undergo SHM

• Mass bobbing vertically from a spring w/ no air
  resistance
• Mass on a frictionless floor attached to a spring
  on the wall
• Simple pendulum or person on a swing where max
  displacement is not excessive
• The shadow on a wall of an object moving in a
  vertical circle at constant speed with light
  shining on it from the side
• A person who jumps into a tunnel that goes clear
  through the center of the Earth all the way to
  the opposite side

All of these situations have common features

• periodic motion with constant period
• an equilibrium point half way between its high
  and low (or leftmost and rightmost) points where
  its moving its fastest
• a constant amplitude that is half of its total
  span (from high to low)
• a restoring force that always acts in the
  direction of equilibrium and is a max at the
  extremes (max distance from equil.)

77
Torque
F
r
Informally speaking, torque is the measure of a
forces turning power. Normally its impossible
to loosen a nut or bolt by hand. A wrench
doesnt make you any stronger, but it increases
your turning power, because the farther a force
is applied from the axis of rotation, the greater
the torque it produces. So long as F and r are
perpendicular, the magnitude of the torque is
given by ? r F, where r is the magnitude of
the displacement vector from the center of
rotation to the point of application of the
force. Technically, torque is a vector quantity,
but well mainly deal with its magnitude.
78
Torque when force is at an angle
F
F sin?
r
?
F cos?
When F is not perpendicular to r, we can split
F into components. The component parallel to
r does nothing in the way of turning the bolt,
so it produces no torque. The perpendicular
component does produce torque. Therefore, our
general formula for the magnitude of torque is
? r F sin?
79
Torque as a vector
Torque is a vector quantity that can be defined
relative to any point of interest by the cross
product
? r ? F
where r is the displacement vector from the
point of interest to the point of application of
the force. As you learned in the last unit, the
magnitude of r ? F is r F sin?, where ? is
the angle between r and F. Use the
right-hand rule to find the direction of ?. In
the wrench example, ? points out toward you,
perpendicular to both r and F. The torque
vector does not point in the direction of motion,
as a velocity vector does. Just as a net force
produces acceleration (a change in velocity), a
net torque produces angular acceleration (a
change in angular speed).
F
r
?
80
See-saw Example part 1
How much do I weigh?
175 lb
35 lb
7 ft
5 ft
6 ft
fulcrum
A man, a turtle, and a bike rider are positioned
on a giant see-saw as shown. If the see-saw is
in equilibrium, how much must the man weigh?
Answer The torque that would produce clockwise
rotation (from the weight of the bike and rider)
must cancel out the torques that would produce
counterclockwise rotation (from the man and
turtle). If the mans weight is W, then (7
ft) W (5 ft) (35 lb) (6 ft) (175 lb)
(7 ft) W 875 ftlb W 125 lb
81
See-saw Example part 2
175 lb
35 lb
7 ft
5 ft
6 ft
The see-saw is in equilibrium at this instant,
but which side will go down after a short time?
Answer Because the biker is heavier and,
presumably, moving faster than the turtle, the
torque he produces will diminish much more
quickly than the torque the turtle produces.
Therefore, the turtles side of the see-saw will
go down. Note A foot-pound is a unit of torque
(a force times a distance). The SI unit for
torque is the Newton-meter (N m). Also note that
the angle ? in the formula ? r F sin? is
90 for each person/animal since each weight
vector is ? to its corresponding displacement
vector from the fulcrum.
82
Equilibrium
N 335 lb
35 lb
175 lb
125 lb
N is the normal force--the force on the see-saw
due to the fulcrum. Its magnitude is the sum of
the weights of the 3 see-saw participants. If
this were not the case, the see-saw would be
accelerating either up or down. So, equilibrium
means no net force and no net torque. Note
that weights on each side dont have to be equal.
The distance each weight lies from the fulcrum
must be taken into account.
83
Center of Mass
c. m. for beam
weight
Schmedrick decides to build his own see-saw, but
of course he screws up, and instead of putting
the fulcrum in the middle, he puts it only 1/ 3
of the way from the right end. The beam of the
see-saw has a mass of 50 kg. If the beam is
uniform (equal density throughout), then we
pretend all of the weight of the beam acts right
at the center of the beam. This point is called
the center of mass because its the point where
the beam would balance if the fulcrum were
beneath it. Schmed figures, rather than cutting
some of the left end off, he could attach a
weight at the right end. In order to attain a
balance, the center of mass of the weight-beam
system must be shifted to the right, so that it
is right above the fulcrum. How much weight
should he add?
continued on next slide
84
Center or Mass (cont.)
m
c. m. for beam
x 0
x L
x L / 2
x 2 L / 3
To calculate the center of mass, we set up a
coordinate system. (The reference point doesnt
matter.) For our purposes, all 50 kg of the the
beam is right at the its center of mass (at L /
2). Mathematically, the center of mass for a
system of masses is given by adding up all the
products of mass and position and then dividing
by the total mass
50 (L / 2) m (L )
xcm
2 L / 3
50 m
The Ls cancel out and we have
25 m
2
75 3 m 100 2 m

50 m
3
So, m 25 kg, and its
weight is (25 kg) (9.8 m/s2) 245 N
85
Center of mass is independent of coordinate system
m
c. m. for beam
x L
x 0
x -2L / 3
x L / 2
x L / 3
x L / 3
x -L / 6
x 0
Lets show we get the same answer using different
coordinate systems one with zero at the right
end where positive is to the left (red) and one
with zero at the fulcrum where positive is to the
right (black). In each equation we set the
center of mass equal to the fulcrums position in
that coordinate system
In each case m comes out to be 25 kg, as it was
on the last slide. (Work the algebra out for
yourself.) This would be true no matter what
coordinate system we chose.
86
Torque Method
L / 6
L / 3
m
c. m. for beam
m g
50 g
Same problem, another method This time well
find m using torque. In order for the beam to
balance where the fulcrum is, the torque (force
? distance) must cancel out on each side of the
fulcrum. The entire weight of the beam acting at
its center of mass is equivalent to its weight
being spread out along its length.
50 g (L / 6) m g (L / 3) Both g and L
cancel out and m 25 kg, just as it did on
previous slides. (Since the weight of the beam
acts at half the distance as the weight of the
green mass, the green mass must have half the
weight.)