Transcript Physics 2211: Lecture 8 Notes

Physics 2211: Lecture 07
A Gallery of Forces
 Newton’s 2nd Law of Motion
 Newton’s 1st Law of Motion

Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 1
Force

A force is a push or a pull.
agent
A force is caused by an agent
and acts on an object. More
precisely, the object and agent
INTERACT.


A force is a vector.
F1
Physics 2211 Spring 2005
Dr. Bill Holm
object
Lecture 07, Page 2
Review: Newton's Laws
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
Law 2: For any object,
Fnet   F  ma .
Law 3: Forces occur in action-reaction pairs: Fab  Fba
where Fab is the force due to object a acting on object b.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 3
Weight
The Earth is the agent (on this planet)
The “weight force” pulls the box down
(toward the center of the Earth)
long-range force
W
Physics 2211 Spring 2005
Dr. Bill Holm
no contact needed!
Lecture 07, Page 4
Gravity

What is the force of gravity exerted by the earth on a typical
physics student?
Typical student mass m = 55kg
g = 9.8 m/s2.
Fg = mg = (55 kg)x(9.8 m/s2 )
Fg = 539 N (weight)
FES  Fg  mg
FSE  mg
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 5
Example
Mass vs. Weight

An astronaut on Earth kicks a bowling ball and hurts his foot.
A year later, the same astronaut kicks a bowling ball on the
moon with the same force.
Ouch!
His foot hurts...
(1)
(2)
(3)

more
less
the same
The masses of both the bowling
ball and the astronaut remain the
same, so his foot will feel the same
resistance and hurt the same as
before.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 6
Example
Mass vs. Weight
Wow!

However the weights of the
bowling ball and the astronaut are
less:
W = mgMoon

That’s light.
gMoon < gEarth
Thus it would be easier for the
astronaut to pick up the bowling
ball on the Moon than on the
Earth.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 7
The Spring Force
The spring is the agent
A compressed spring
pushes on the object.
Physics 2211 Spring 2005
Dr. Bill Holm
A stretched spring pulls
on the object.
Lecture 07, Page 8
Springs

Hooke’s Law: The force exerted by a spring is proportional
to the distance the spring is stretched or compressed from
its relaxed position (linear restoring force).
FX = -k x
Where x is the displacement from the
relaxed position and k is the constant
of proportionality.
relaxed position
FX = 0
x=0
Physics 2211 Spring 2005
Dr. Bill Holm
x
Lecture 07, Page 9
Springs

Hooke’s Law: The force exerted by a spring is proportional
to the distance the spring is stretched or compressed from
its relaxed position (linear restoring force).
FX = -k x
Where x is the displacement from the
relaxed position and k is the constant
of proportionality.
relaxed position
FX = -kx > 0
x0
Physics 2211 Spring 2005
Dr. Bill Holm
x
Lecture 07, Page 10
Springs

Hooke’s Law: The force exerted by a spring is proportional
to the distance the spring is stretched or compressed from
its relaxed position (linear restoring force).
FX = -k x
Where x is the displacement from the
relaxed position and k is the constant
of proportionality.
relaxed position
FX = - kx < 0
x
x>0
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 11
Tension
The rope is the agent
The rope force (tension) is a pull
along the rope, away from the object
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 12
Tools: Ropes & Strings


Ropes & strings can be used to pull from a distance.
Tension (T) at a certain position in a rope is the magnitude
of the force acting across a cross-section of the rope at that
position.
The force you would feel if you cut the rope and
grabbed the ends.
An action-reaction pair.
T
cut
T
T
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 13
Tools: Ropes & Strings

Consider a horizontal segment of rope having mass m:
Draw a free-body diagram (ignore gravity).
m
T1
a
T2

Using Newton’s 2nd law (in x direction):
FNET = T2 - T1 = ma

So if m = 0 (i.e., the rope is light) then T1 = T2
Physics 2211 Spring 2005
Dr. Bill Holm
x
Lecture 07, Page 14
Tools: Ropes & Strings

An ideal (massless) rope has constant tension along the
rope.
T


T
If a rope has mass, the tension can vary along the rope
 For example, a heavy rope
hanging from the ceiling...
T = Tg
T=0
We will deal mostly with ideal massless ropes.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 15
Tools: Ropes & Strings

The direction of the force provided by a rope is along the
direction of the rope:
T
Since ay = 0 (box not moving),
m
T = mg
mg
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 16
The Surface Contact Force
(one force — two components)
Component perpendicular
Component parallel
(normal) to surface
to surface
n
f
NORMAL FORCE—the surface
pushes outward against the object.
Physics 2211 Spring 2005
Dr. Bill Holm
FRICTION FORCE pulls in a
direction to (try to) prevent
slipping of the object
Lecture 07, Page 17
Surface Contact Force
(microscopic view)
FRICTION is caused by
the making and
breaking of molecular
bonds between the two
surfaces.
The NORMAL force is caused
by the molecular-scale
repulsion between the
object and the surface.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 18
There are two types of Friction
Kinetic Friction
Static Friction
v
fs
fk
FRICTION doesn’t
prevent slipping of
object along surface.
Physics 2211 Spring 2005
Dr. Bill Holm
FRICTION
successfully
prevents
slipping of the
object along
surface
Lecture
07, Page 19
Rolling friction
| f r |  r | N |
coefficient of rolling friction
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 20
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 21
Drag
A force experienced by a body that moves through air.
Experiment shows that to a good approximation,
1 2
D   Av vˆ
4
Unit vector that points in the
direction of the velocity of the body
Cross sectional area of the body measured in m2
Square of the body's velocity measured in m2 /s2 .
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 22
Drag Force
DRAG occurs when an object
moves in a gas or liquid. Like
friction, the force of drag
always points opposite to the
direction of motion
The fluid is the agent.
Skin Drag (like friction)
Form Drag (rowboat)
Drag is very small for most of the
objects we will discuss. We will always
neglect drag in our models unless we
explicitly state otherwise.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 23
Throw a ball upward vertically
Drag decreases as
the ball slows down
Drag adds to the
weight
Physics
2211 Springas
2005it rises
Dr. Bill Holm
Drag increases
as the ball speeds
up slows
Drag opposes the
weight as it falls
Lecture 07, Page 24
Terminal velocity
At the terminal velocity,
the drag force balances
the force of gravity.
F
y
 Dw0
1
4
Av  mg  0
2
T
4mg
vT 
A
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 25
Drop a particle from rest (v = 0)
As the speed (and thus drag) increases,
the slope decreases
vT
Slope approaches zero as
v approaches terminal velocity
Without
drag, v = - at
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 26
The dimensions of a 1500 kg car, as seen from the front, are 1.6 m wide by 1.4m high.
At what speed does the drag force equal the force of rolling friction?
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 27
Empirical (Contact) Forces
(examples)

Linear Restoring Force:

Friction Force:

Fluid Force:
F  kx
F  N Fs  sN; Fk  k N
F  bv n n  1 (v small);
n  2, (v large)
Characterized by variable, experimentally
determined “constants”: k, , b, etc.
These forces are all electromagnetic in origin.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 28
Thrust
THRUST is a contact force exerted on rockets
and jets (and leaky balloons) by exhaust
gases (the agent).
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 29
Electric and Magnetic Forces (Non-Contact)
E
E
E
q
E
E
-q
E
E
E
Electric Field
Physics 2211 Spring 2005
Dr. Bill Holm
Magnetic Field
Lecture 07, Page 30
Identifying Forces---The Skier
Tension T
1. Identify SYSTEM
Normal force N
3. Find contact points
between SYSTEM &
ENVIRONMENT—
Name and label the
CONTACT FORCES
Friction force f
2. Draw PICTURE with a
closed curve around the
Weight W
SYSTEM. (Everything
outside the curve is the
environment.
Physics 2211 Spring 2005
Dr. Bill Holm
4. Name and label
LONG-RANGE FORCES
Lecture 07, Page 31
Newton’s 2nd Law of Motion
Newton’s 1st Law of Motion
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 32
Newton’s 2nd Law
the connection between force and motion
Fnet
a
m

Fnet  ma
The acceleration of an object a is proportional to the net
force F net that acts upon it.
The constant of proportionality is called the "mass".
The unit of mass m is the kilogram (kg).
Mass is an intrinsic property of matter.
The unit of force is [F] = [m][a] = kg-m s2 = N (Newton)
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 33
Newton’s First Law
An object moves with constant velocity v if and only if
the total force F net acting on the object is zero.
dynamics:
a  F net m  a  0
kinematics:
v  v0  at  v  v0
Rest is just a special case where v0  0.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 34
Mechanical Equilibrium: (Fnet  0)
Dynamic Equilibrium
(object moving with constant velocity)
Static Equilibrium
(object at rest)
N
W
Physics 2211 Spring 2005
Dr. Bill Holm
An object moving in a straight line at constant
Lecture 07, Page 35
velocity is in dynamic equilibrium
Problem: Accelerometer

A weight of mass m is hung from the ceiling of a car with a
massless string. The car travels on a horizontal road, and
has an acceleration a in the x direction. The string makes
an angle  with respect to the vertical (y) axis. Solve for 
in terms of a and g.
a
iˆ
Physics 2211 Spring 2005
Dr. Bill Holm

Lecture 07, Page 36
Accelerometer

Draw a free body diagram (FBD) for the mass:
 What are all of the forces acting?

a
T (string tension)
y
m
x
mg (gravitational force)
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 37
Accelerometer

Resolve forces into components:

Sum forces in each dimension separately:
x : SFx = Tx = T sin  = ma
y:

SFy = Ty - mg
T
= T cos - mg = 0

Tx iˆ
Eliminate T :
T sin  = ma
T cos = mg
Physics 2211 Spring 2005
Dr. Bill Holm
=>
a
tan  
g

Ty ˆj
y
x
a
mg  mg ˆj
Lecture 07, Page 38
Accelerometer
tan  
a
g

Let’s put in some numbers:

Say the car goes from 0 to 60 mph uniformly in 10 seconds:
 60 mph = (60 x 0.45) m/s = 27 m/s.
Acceleration a = Δv/Δt = 2.7 m/s2.
 So a/g = 2.7 / 9.8 = 0.28 .
  = arctan (a/g) = 15.6 deg
a
Physics 2211 Spring 2005
Dr. Bill Holm

Lecture 07, Page 39
Problem: Inclined plane

A block of mass m slides down a frictionless ramp that
makes angle  with respect to the horizontal. What is its
acceleration a ?
m
a

Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 40
Inclined plane

Define convenient axes parallel and perpendicular to plane:
 Acceleration a is in x direction only.
y
m
a
x

Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 41
Inclined plane


Draw a FBD.
Resolve forces into components & sum forces in x and y
directions separately:
F
F
y
: N - mg cos  = may= 0
x
: mg sin  = max = ma
Assume forces
are acting at
“center of mass”
of block.
- mg cos 
a = g sin 
N
a
ˆj

mg
Physics 2211 Spring 2005
Dr. Bill Holm
N = mg cos 
y
mg sin 
iˆ
x

Lecture 07, Page 42
Angles of an Inclined plane
Lines are perpendicular, so the angles are the same!
- mg cos 
ˆj

mg

Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 43
Vocabulary

A Reference Frame is the (x,y,z) coordinate system you
choose for making measurements.

An Inertial Reference Frame is a frame where Newton’s Laws
are valid.
USE NEWTON’S LAWS TO TEST
FOR INERTIAL REFERENCE FRAMES
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 44
Test for Inertial Reference Frame
EXAMPLE: Airplane parked on runway
A ball is placed on the floor of the plane; no net forces act on the ball.
Observation : The ball's motion doesn't change (no acceleration)
Conclusion : The plane is an inertial reference frame.
Less obvious: a plane cruising at constant velocity is also an inertial
reference frame!
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 45
Test for Inertial Reference Frame


EXAMPLE: Airplane taking off.
Ball placed on floor of plane; no net forces act on ball.
Observation: The ball rolls back (it accelerates)
Conclusion : Newton's 1st Law is violated.
The accelerating plane is NOT an inertial frame.
.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 46
We live in a World of Approximations . . .
Strictly speaking, an Inertial Reference Frame has zero
acceleration with respect to the “distant stars”.
The Earth accelerates a little (compared to the distant
stars) due to its daily rotation and its yearly revolution
around the Sun. Nevertheless, to a good approximation,
the Earth is an Inertial Reference Frame.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 47
Is Atlanta a good IRF?



Is Atlanta accelerating?
YES!
Atlanta is on the Earth.
The Earth is rotating.
What is the acceleration (centripetal) of Atlanta?
2
T = 1 day = 8.64 x 104 sec,
2
v  2 R  1
aatl   
R ~ RE = 6.4 x 106 meters .

R



 T  R
Plug this in: aatl = 0.034 m/s2 ( ~ 1/300 g)
Close enough to zero that we will ignore it.
Atlanta is a pretty good IRF.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 48
Physics 2211: Lecture 12

2-D, 3-D Kinematics and Projectile Motion
Independence of x and y components
*Georgia Tech track and field example
*Football example
*Shoot the monkey
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 49
3-D Kinematics

The position, velocity, and acceleration of a particle in 3
dimensions can be expressed as:
r  xiˆ  yjˆ  zkˆ
v  vxiˆ  v y ˆj  vz kˆ
a  axiˆ  ay ˆj  az kˆ

We have already seen the 1-D kinematics equations:
x  x(t )
Physics 2211 Spring 2005
Dr. Bill Holm
dx
v
dt
dv d 2 x
a
 2
dt dt
Lecture 07, Page 50
3-D Kinematics

For 3-D, we simply apply the 1-D equations to each of the
component equations.
x  x(t )
vx 
ax 

dx
dt
d2x
dt
2
y  y( t )
vy 
ay 
dy
dt
vz 
d2y
dt
z  z( t )
2
az 
dz
dt
d2z
dt 2
Which can be combined into the vector equations:
r  r t 
Physics 2211 Spring 2005
Dr. Bill Holm
dr
v
dt
dv d 2 r
a
 2
dt dt
Lecture 07, Page 51
3-D Kinematics

So for constant acceleration we can integrate to get:
a  constant
v  v0  at
r  r0  v0t  12 at 2

Aside: the “4th” kinematics equation can be written as
v  v0  2a  r
2

Physics 2211 Spring 2005
Dr. Bill Holm
2
(more on this later)
Lecture 07, Page 52
2-D Kinematics

Most 3-D problems can be reduced to 2-D problems when
acceleration is constant:
 Choose y axis to be along direction of acceleration
 Choose x axis to be along the “other” direction of
motion

Example: Throwing a baseball (neglecting air resistance)
 Acceleration is constant (gravity)
 Choose y axis up: ay = -g
 Choose x axis along the ground in the direction of the
throw
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 53
Uniform Circular Motion
Physics 2211 Spring 2005
Dr. Bill Holm

What does it mean?

How do we describe it?

What can we learn about it?
Lecture 07, Page 54
What is Uniform Circular Motion?
v is constant

Motion with
Constant Radius R
(Circular)
y
v
(x,y)
R

 Constant Speed
(Uniform)
Physics 2211 Spring 2005
Dr. Bill Holm
x
Lecture 07, Page 55
How can we describe
Uniform Circular Motion?


In general, one coordinate system is as good as any other:
 Cartesian:
y
» (x,y) [position]
» (vx ,vy)
[velocity]
v
 Polar:
(x,y)
R
» (R, )
[position]

» (vR , ) [velocity]
In uniform circular motion:
 R is constant (hence vR = 0).
  (angular velocity) is constant.
 Polar coordinates are a natural way to describe
Uniform Circular Motion!
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 56
x
Polar Coordinates
Conversion from polar to
Cartesian coordinates:
y
And back:
R
x  R cos
R  x2  y 2

y  R sin
1
(x,y)
x
sin
0
/2
  arctan  y x 
cos


3/2
2
-1
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 57
Angular Motion

The arc length s (distance along the circumference) is
related to the angle in a simple way:
s = R, where  is the angular displacement.
 units of  are called radians.
y

For one complete revolution:
2R = Rc
 c = 2
 has a period 2.
v
(x,y)
R

s
x
1 revolution = 2radians
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 58
Angular Motion



In Cartesian coordinates, we say velocity vx = dx/dt.
 x = vx t
(vx constant)
In polar coordinates, angular velocity d/dt = .
=t
( constant)
y
 has units of radians/second.
Displacement s = v t.
but s = R = R t, so:
v=R
Physics 2211 Spring 2005
Dr. Bill Holm
v
(x,y)
R
  t
s
x x
Lecture 07, Page 59
Aside: Period and Frequency



Recall that 1 revolution = 2 radians
(a) frequency ( f ) = revolutions / second
(b) angular velocity (  ) = radians / second
By combining (a) and (b)
 (rad/s) = [2 rad/rev] x f (rev/s)
 = 2f
Realize that:
period (T) = seconds / revolution
So T = 1 / f = 2/
y
v
(x,y)
R

s
x x
 = 2 / T = 2f
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 60
Summary

Relationship between Cartesian and Polar coordinates:
x  R cos
R  x2  y 2
y  R sin
  arctan  y x 
y

Angular motion:
displacement:
velocity:
constant velocity:
Linear
Angular
Links
s

s  R
ds
v
dt
s  vt
d

dt
  t
Physics 2211 Spring 2005
Dr. Bill Holm
v  R
v
(x,y)
R
  t
s
x x
Lecture 07, Page 61
Polar Unit Vectors


We are familiar with the Cartesian unit vectors: iˆ, ˆj , kˆ
Now introduce “polar unit-vectors” rˆ and ˆ :
rˆ points in radial direction
ˆ points in tangential direction (counterclockwise)
ˆ
rˆ
y
ˆj
R

iˆ
Physics 2211 Spring 2005
Dr. Bill Holm
x
Lecture 07, Page 62
Acceleration in Uniform Circular Motion


Even though the speed is constant, velocity is not constant
since the direction is changing: acceleration is not zero!
Consider average acceleration in time t
v
v2
R
t
Physics 2211 Spring 2005
Dr. Bill Holm
v2
v1
v1
aav  v / t
Notice that v (hence v / t )
points at the origin!
In the limit t  0 ,
a  dv / dt
and a points in the
rˆ direction.
Lecture 07, Page 63
Acceleration in Uniform Circular Motion


This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
v
v2
R
R
v2
v1
Similar triangles:
v1
v R

v
R
But R = v t for small t
v vt

So:
v
R
v v 2

t
R
R
v2
a
R
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 64
Centripetal Acceleration

Uniform Circular Motion results in acceleration:
Magnitude:
a = v2 / R
Direction:
-^
r (toward center of circle)
a
R

Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 65
Centripetal Acceration
(in terms of )
We know that
v2
a
R
v = R
and
Substituting for v we find that:
R 

a
2
R
a = 2R
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 66
Example:
Uniform Circular Motion

A fighter pilot flying in a circular turn will pass out if the
centripetal acceleration he experiences is more than about
9 times the acceleration of gravity g. If his F18 is moving
with a speed of 300 m/s, what is the approximate diameter
of the tightest turn this pilot can make and survive to tell
about it ?
(1) 500 m
(2) 1000 m
(3) 2000 m
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 67
Example:
Solution
v2
a
 9g
R
R
10000
R
m  1000 m
9.81
2
2
90000 ms2
v

9g  9  9.81 sm2
D  2R  2000m
2km
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 68
Example: Propeller Tip

The propeller on a stunt plane spins with frequency
f = 3500 rpm. The length of each propeller blade is L = 80 cm.
What centripetal acceleration does a point at the tip of a
propeller blade feel?
what is a here?
f
L
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 69
Example
First calculate the angular velocity of the propeller:

1 rpm  1
rev
x
min
 1 min 
 60 s  x


rad
 rad 
-1
2


0.105

0.105
s
 rev 
s


so 3500 rpm means  = 367 s-1

Now calculate the acceleration.
a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g
 direction of acceleration points towards the propeller hub.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 70
Example: Newton & the Moon


What is the acceleration of the Moon due to its motion
around the Earth?
What we know (Newton knew this also):
T = 27.3 days = 2.36 x 106 s
(period ~ 1 month)
R = 3.84 x 108 m
(distance to moon)
RE = 6.35 x 106 m
(radius of earth)
R
Physics 2211 Spring 2005
Dr. Bill Holm
RE
Lecture 07, Page 71
Moon

Calculate angular velocity:
1 rev
1 day
rad
x
x 2
 2.66 x106 s -1
27.3 day 86400 s
rev

So  = 2.66 x 10-6 s-1.

Now calculate the acceleration.
 a = 2R = 0.00272 m/s2 = 0.000278 g
direction of acceleration points is towards the
center of the Earth.
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 72
Moon


So we find that amoon / g = 0.000278
Newton noticed that RE2 / R2 = 0.000273
amoon
g
R

RE
This inspired him to propose that FMm  1 / R2
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 73
Example
Centripetal Acceleration

The Space Shuttle is in Low Earth Orbit (LEO) about 300 km
above the surface. The period of the orbit is about 91 min.
What is the acceleration of an astronaut in the Shuttle in the
reference frame of the Earth?
(The radius of the Earth is 6.4 x 106 m.)
(1)
(2)
(3)
0 m/s2
8.9 m/s2
9.8 m/s2
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 74
Example
Centripetal Acceleration

First calculate the angular frequency :
1 rev
1 min
rad

x
x 2
 0.00115 s -1
91min 60 s
rev

Realize that:
RO
RO = RE + 300 km
= 6.4 x 106 m + 0.3 x 106 m
= 6.7 x 106 m
300 km
RE
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 75
Example
Centripetal Acceleration

Now calculate the acceleration:
a = 2R
a = (0.00115 s-1)2 (6.7 x 106 m)
RO
a = 8.9 m/s2
300 km
RE
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 76
Physics 2211: Lecture 16
Physics 2211 Spring 2005
Dr. Bill Holm

Circular Orbits

Fictitious forces
Lecture 07, Page 77
SEEM to be very different,
BUT they have the same free body diagram . . .
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 78
Orbital Motion is Projectile Motion!
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 79
Orbiting projectile is in free fall
w  mg (downward )
w  mg (center )
2
vorbit
ar 
g
r
Fnet
a
 g (center )
m
Flat earth approximation
Spherical Earth, near Earth
h
vorbit  rg  RE g
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 80
Fictitious forces in Non-Inertial Frames of Reference
A car and driver move a constant speed. Suddenly, the driver brakes
Outside observer: if the seat is frictionless,
the driver continues forward at constant
speed and collides with the front window.
F
Physics 2211 Spring 2005
Dr. Bill Holm
Inside observer: a force F “throws” the driver
against the front window.
F is fictitious; the observer is in a noninertial (accelerated) frame where
Newton’s laws do not apply. There is no
true “push” or “pull” from Lecture
anything.
07, Page 81
Fictitious forces in Non-Inertial Frames of Reference
path of book
(Newton’s 1st Law)
● Car turns
● Book continues on straight line
● In driver’s reference frame,
apparent (fictitious) force moves book
across the bench seat
book
v
car
frictionless bench seat
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 82
A car is passing over the top of a hill at (non-zero)
speed v. At this instant,
1
2
3
4
n>w
n=w
n<w
Can’t tell without knowing v
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 83
A ball on a string swings in a vertical circle. The
3
2
string breaks when the string is horizontal and
the ball is moving straight up. Which trajectory
1
does the ball follow thereafter?
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 84
4
Roller-Coaster
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 85
Bottom:
mv 2
 Fr  n  w  mar  r
mv 2
n  w
 wapp
r
Top:
mv 2
 Fr  n  w  mar  r
mv 2
n
 w  wapp
r
Lose contact : n  0  vcritical 
Physics 2211 Spring 2005
Dr. Bill Holm
rw
 rg
m
Lecture 07, Page 86
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 87
Full Disclosure
vtop  vbottom and force is
purely radial (centripetal).
mg
Except at the top and bottom,
N
N
mg
aT  0 since there is a non-zero
N
a
component of the weight in the
N
a
mg
tangential direction. Since v is
entirely tangential, the speed v
must speed up or slow down
on the sidewalls
mg
Physics 2211 Spring 2005
Dr. Bill Holm
acceleration is not purely
07, tangential
Page 88
radial orLecture
purely
Example
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 89
Example

Treat horizontal motion and vertical motion separately

Then just add the results: Principle of Superposition

Horizontal Motion:

Velocity at highest point:
ax 
dv x
0
dt
v x  constant
v x  v 0 x  10 m/s in +x direction
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 90
Example
ay  g
v y  v 0 y  gt
y  y 0  v 0 y t  21 gt 2

Vertical Motion:

Need to determine magnitude of take-off velocity: v 0  v 0  v 02x  v 02y

v 0y  ? We know when y  y max  h, v y  0
v y  v 0 y  gt
0  v 0 y  gth

th  v 0 y g
y  y 0  v 0 y t  21 gt 2
h  0  v 0 y th  21 gth2
v
h  0  v 0 y goy  21 g
 
OR, use v y2  v 02y  2g y . When y  h, v y  0
Physics 2211 Spring 2005
Dr. Bill Holm
 
v oy
g
2
v 02y  2gh
v 02y  2gh
Lecture 07, Page 91
Example

Magnitude of take-off velocity:
v 0  v 0  v 02x  v 02y
 v 02x  2gh

 10.2
Physics 2211 Spring 2005
Dr. Bill Holm


10 ms   2 9.81 sm2  0.25 m
2
m
s
Lecture 07, Page 92
Example
Two
footballs are thrown from the same point on a flat field.
Both are thrown at an angle of 30o above the horizontal. Ball 2
has twice the initial speed of ball 1. If ball 1 is caught a distance
D1 from the thrower, how far away from the thrower D2 will the
receiver of ball 2 be when he catches it?
(1)
D2 = 2D1
Physics 2211 Spring 2005
Dr. Bill Holm
(2) D2 = 4D1
(3) D2 = 8D1
Lecture 07, Page 93
Example

The horizontal distance a ball will go is simply
x = (horizontal speed) x (time in air) = v0x t

To figure out “time in air”, consider the
y  y 0  v 0 y t  21 gt 2
equation for the height of the ball:

When the ball is caught, y = y0
v
0y
t
 gt  t  0
1
2
two
solutions
Physics 2211 Spring 2005
Dr. Bill Holm
v 0 y t  21 gt 2  0
2v 0 y
g
t 0
(time of catch)
(time of throw)
Lecture 07, Page 94
Example


So the time spent in the air is: tR  2v 0 y g
The range, R, is thus : R  x  tR   v 0 x tR 
v0
2v 0 xv 0 y
g
2v 02 sin  cos v 02 sin 2


g
g
v oy  v 0 sin 

v ox  v 0 cos

Ball 2 will go 4 times as far as ball 1!

Notice: For maximum range, sin 2  1    45o for max. range
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 95
Shooting the Monkey
(tranquilizer gun)
Where does the zookeeper
aim if he wants to hit the monkey?
( He knows the monkey will
let go as soon as he shoots ! )
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 96
Shooting the Monkey

If there were no gravity, simply aim
at the monkey
r  r0
r  v0t
Physics 2211 Spring 2005
Dr. Bill Holm
Lecture 07, Page 97
Shooting the Monkey

With gravity, still aim at the monkey!
1 2
r  v0t  gt
2
Physics 2211 Spring 2005
Dr. Bill Holm
1 2
r  r0  gt
2
Dart hits the
monkey!
Lecture 07, Page 98
Shooting the Monkey
x = v0 t
y = -1/2 g t2

This may be easier to think about.
It’s exactly the same idea!!
Physics 2211 Spring 2005
Dr. Bill Holm
x = x0
y = -1/2 g t2
Lecture 07, Page 99