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This Week
Moving in circles
Angular momentum
Spinning at the Olympics
Torques
Why is a wrench useful?
Center of Gravity
Useful info for crossing Niagara falls on a wire
The New Hubble Telescope
Views of Saturn
5/2/2017
Physics 214
Fall 2009
1
General case of motion
In general the motion of an object consists of translation and
rotation.
Translation we have dealt with in straight line motion
+
A wheel is an excellent example of rotation
We define an axis and counterclockwise
is positive.
3600
One full circle =
= 2π radians
Circumference = 2πR
Time for one revolution = 2πR/v
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R
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2
Rotational Motion
Consider an object moving in a circular path. It has velocity, acceleration,
kinetic energy and momentum but these are not the simplest variables
Displacement we use the angle θ measured in radians
Angular velocity ω = Δθ/Δt
Angular acceleration α = Δω/Δt
v
1 revolution/sec = 2π radians/sec
Since the time for one revolution is
2πr/v = 2π/ω then v = rω
so Δv = rΔω and Δv/Δt = rΔω/Δt
and
a = rα
R
θ
All parts of a rotating wheel have the same ω but
The further from the center the bigger is v
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Constant angular acceleration
Apart from changing variables the equations are identical to linear
motion with constant acceleration
Displacement
Velocity
Acceleration
Constant
d
v = Δd/Δt
a = Δv/Δt
v = v0 + at
d = v0t + 1/2at2
v2 = v02 + 2ad
d = ½(v + v0)t
θ
ω = Δθ/Δt
α = Δω/Δt
ω = ω0 + αt
θ = ω0t + 1/2αt2
ω2 = ω02 +2αθ
θ = ½(ω + ω0)t
Once again all variables except time can be positive or negative
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Forces and torques
If we apply a force to a bicycle wheel that is free to rotate for a given force
it is easier to rotate the wheel the further you are from the axle.
In the picture shown below each of the single weights on it’s own will
cause the rule to rotate but the two together can be balanced.
A force applied to an object, in general makes that object rotate and the
action of the force we call a TORQUE and Τ = FL where L is the
perpendicular distance to the line of action of the force.
Once again + is counterclockwise and – is clockwise and the net torque is
sum of all torques.
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Using a wrench
In our everyday life we are limited in
the force we can apply but if we
increase the lever arm we can
increase the torque. We can turn a
very tight nut by applying a force F at
a large radius R
If the radius of the nut is r then to just
move the nut
FR = Ffr so if
R/r = 30 then Fs/F = 30
Ff
F
R
The work done is the same because
in one turn
2πRF = 2πrFs
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Center Gravity
There is a point associated with any body that allows it to be
balanced by a single force upward without any rotation.
This point is called the Center of Mass
It is like the whole mass of the body
is concentrated at that point.
If we hold an object up on a piece of string
the center of mass will lie on the line of the string.
And when forces act Fnet = Macm
and in addition the object might rotate
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Walking the plank
If we have an object like a plank
or a teeter totter that pivots about
a single point then it will be
exactly balanced if the clockwise
torques equal the
counterclockwise torque.
As the boy walks out on the
plank the clockwise torque
increases until it becomes larger
than the counterclockwise torque
and then the plank rotates so at
the point where the plank tips
Mplank gd = Mchildgx
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x
8
Newtons second Law
Let us look at the rotation of a simple
Object with all the mass at a radius R
For the object shown F = ma
v = rω and Δv = Δω x r and a = αr
F = ma
Fr = T = mar = mr2α = Iα
I plays the role of mass for rotation
Kinetic energy = 1/2mv2 = 1/2 Iω2
1/2 Iω2 is the general expression for any
rotating body and each body has a moment
of inertia which depends on the distribution
of mass about the axis of rotation.
The work done by the force W = Fd. For one
revolution W = F2πr = T2π
and the general expression is W = Tθ
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Moment of Inertia
The Moment of Inertia is
always of the form
I = mass times a length
squared
and it depends on the
distribution of mass
about the axis of rotation
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Conservation of angular momentum
Linear momentum for a particle P = mv
For a rotating object L = Iω is the angular
momentum
Angular momentum is conserved in a closed system.
That is one in which no external torques are acting.
A simple closed system is a skater and if I is changed
ω will to keep L at the same value.
We can invert the bicycle wheel and ω will change to
keep L constant
L=Iω
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Rotating object
To describe the properties of a rotating object
I is the moment of inertia which depends on the
distribution of mass about the axis of rotation.
The kinetic energy = ½ Iω2
Angular momentum L = Iω is a vector perpendicular to the rotation plane
Rotation is changed by torque T = Iα (+ counter clockwise)
Closed system angular momentum
is conserved L = I ω.
What about energy and work
Kinetic energy = ½ Iω2
Suppose I changes by a factor of 3 smaller
Then since L is conserved the ωnew = 3 ωold
KEnew = 3 KEold this energy comes from the work done
internally to change I. The skater needs to exert a force
to pull her arms in to her body.
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Summary of Chapter 8
Rotational motion
Displacement
Velocity
Acceleration
Constant
5/2/2017
Angular velocity ω = Δθ/Δt
Angular acceleration α = Δω/Δt
One full circle = 3600 = 2π radians
Circumference = 2πR
Time for one revolution = 2πR/v
2πr/v = 2π/ω and v = rω
d
v = Δd/Δt
a = Δv/Δt
v = v0 + at
d = v0t + 1/2at2
v2 = v02 + 2ad
d = ½(v + v0)t
Physics 214
Fall 2009
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v
R
57.30
θ
ω = Δθ/Δt
α = Δω/Δt
ω = ω0 + αt
θ = ω0t + 1/2αt2
ω2 = ω02 +2αθ
ω = ½(ω + ω0
13
Torques
Torque = FL where
where L is the perpendicular distance
to the line of action of the force
+ is counterclockwise and – is clockwise
net torque is sum of all torques
For the boy on the plank he will fall
when Wpdp > Wc dc
Torque = FL = Iα
I plays the role of mass for rotation
Kinetic energy = 1/2Iω2
Work = Tθ ( full circle T2π = Fr2π
Angular momentum L = Iω
There is a point in the geometry of a
body at which all the mass appears to
act and one can balance the body with a
single force (center of mass/gravity)
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Conservation of angular momentum
In a closed system L = Iω
we can change I and ω will change like a
spinning skater.
We can invert the bicycle wheel and ω will
change to keep L constant
The kinetic energy changes because the
person does work and has to exert force in
order to change I
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1J-16 Walk the Plank
What happens when a mass is placed at the end of a massive plank?
Sum of Torque about Pivot
XMg–xmg=0
m=MX/x
Can you
safely walk
to the end of
the plank ?
One can solve for either M or m, if
the other quantity is known
EVEN WITH A MASS AT THE END OF THE PLANK, THE SYSTEM CAN
STILL BE IN EQUILIBRIUM
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1C-05 Velocity of Rifle Bullet
How can we
MEASURING SPEEDS OF OBJECTS MOVING VERYmeasure
FAST MAY
theNOT BE
DIFFICULT.
speed of a
THIS TECHNIQUE IS THE SAME ONE USED TO MEASURE
OF
bulletSPEEDS
?
MOLECULES.
We know that the distance between two disks is L. If the second disk rotates
an angle Δθ before the bullet arrives, the time taken by this rotation is
t = Δθ / 2πn , where n is the angular frequency of the shaft.
Therefore we come up with v = L / t = 2πn L / Δθ .
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1J-21 Center of Gravity of an Irregular Lamina
How can we find the Center of Gravity of Irregular shapes?
Why does the
mass on the
string hang
straight down ?
Where is the
Center of Gravity
of the cut board ?
Does the
CG have
to lie on
the object
?
THE CENTER OF GRAVITY IS THE CENTER OF MASS AND IN
EQUILIBRIUM THE CENTER OF GRAVITY IS ALWAYS LOCATED
DIRECTLY BELOW THE SUSPENSION POINT. IF IT IS HUNG FROM
SEVERAL DIFFERENT POINTS THE VERTICAL LINES WILL ALL
INTERSECT AT ONE POINT.
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1J-23 Corks & Forks
Can the Center of Gravity lie at a point not on the object?
How difficult is it to
balance this system
on a sharp point ?
Where is the
C of G ?
THE CENTER OF GRAVITY IS NOT ON THE OBJECT. IT ACTUALLY LIES
ALONG THE VERTICAL BELOW THE SHARP POINT.
WHEN THE FORK IS MOVED THE CM RISES AND THIS MEANS THE
SYSTEM IS IN STABLE EQUILIBRIUM .
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1J - 24 Climbing double cone
An object appears to run uphill
What happens to the
center of mass ?
The force causing the object to move is gravity and we know that by
energy conservation that if the object gains kinetic energy it must lose
potential energy. Therefore the center of mass must be falling and the
kinetic energy = mgh where h is the distance the CM falls.
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1J-28 Wine Bottle Holder
Balance a Bottle and a Wooden Holder by Eliminating Net Torque
M
How does this
system
Balance?
m1
x1
x2
m2
Sum of Torque about Pivot
m1x1g - m2x2g = 0
THE CENTER OF GRAVITY OF THE BOTTLE PLUS THE WOOD MUST LIE
DIRECTLY OVER AND WITHIN THE BOUNDARY OF THE SUPPORT (PIVOT).
FOR BALANCE THERE CAN BE NO NET TORQUE ON SYSTEM.
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1J-29 Overhanging Blocks
How far out from the table can a stack of bricks be balanced ?
Length = L, CG = L/2
Length = 3L/2, CG = 3L/4
Length = 7L/4, CG = 25L/24
How do the blocks
stay balanced
when the top block
extends beyond
the bottom block ?
Δx
Blocks can over hang but the Center of Gravity
of a block must be inside the block below
For 6 blocks max extension Δx :
Δx = L/2 + L/4 + L/6 + L/8 + L/10 + L/12 = 1.22(L)
IN REALITY EACH BLOCK HAS TO BE MOVED SLIGHTLY BACK TO
AVOID TIPPING, SO THE TOTAL EXTENSION WILL BE A LITTLE LESS.
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1Q-20 Conservation of angular momentum
Changing the moment of inertia for a closed system
What happens when
we pull the cord so
that the two spheres
come closer together?
The dominant physical law is conservation of angular momentum. There
is no net torque (the forces are internal)so L is a constant but the
moment of inertia changes and since L = Iω as I decreases ω increases.
So if I decreases by a factor of 2 then ω increases by a factor of 2.
The kinetic energy is = Iω2 so if I decreases the kinetic energy increases
by the amount of work Fd where F is the force applied to the cord and d
is the distance moved. So in the above example the kinetic energy
increases by a factor of 2. It requires significant force for a skater to pull
in his arms.
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1Q-21 Conservation of angular momentum
Conservation of angular momentum using a spinning wheel
What
happens
when the
wheel is
inverted ?
The dominant physical law is conservation of angular momentum. To
change the angular momentum of the wheel requires an external torque.
So although we can change the direction of the angular momentum of
the wheel the force we use is internal to the wheel/stool system so the
the stool rotates to keep the net angular momentum the same
To turn the wheel requires significant force and work is needed. The
energy of the final system is greater than the initial energy by the
amount of work that is done.
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1Q-30 Bicycle Wheel Gyroscope
Gyroscopic action and precession
L
What
happens to
the wheel,
does it fall
down?
F
F = mg
mg
The counterclockwise torque adds to L and
produces a precession, providing L is large
and the torque is small
The torque causes the vector L to precess and changes the
direction of the angular momentum vector which is
perpendicular to the plane of rotation. This is a very large top.
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1Q-32 Stability Under Rotation
Example of Gyroscopic Stability: Swinging a spinning Record
Why does the Record
not “flop around”
once it is set
spinning ?
L
L
The dominant physical law is conservation of angular momentum. With
no torque the vector L, perpendicular to the plane of rotation always
points in the same direction.
SINCE THERE IS NO TORQUE ABOUT THE CENTER OF ROTATION OF
THE RECORD, THE ANGULAR MOMENTUM VECTOR CANNOT CHANGE.
THIS IS GYROSCOPIC STABILITY. THIS IS A VERY SIMPLE
GYROSCOPE AND SOPHISTICATED GYROSCOPES ARE USED TO
STEER AIRCRAFT AND ORIENT THE HUBBLE TELESCOPE
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1Q- 23 Conservation of angular momentum
Changing the moment of inertia of a skater
How does
conservation of
angular momentum
manifest itself ?
This is two examples of Conservation of angular momentum
The first changes the Moment of Inertia (like a skater)
The second shows what happens when you swing a bat.
All forces are internal to the system so L is conserved.
Case 1 the moment of inertia changes and since l = Iω the speed of
rotation changes to keep L constant.
Case 2 since L = 0 swinging the bat causes the person to rotate in the
opposite direction.
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Questions Chapter 8
Q6 Is the linear speed of a child sitting near the center of a
rotating merry-go-round the same as that of another child sitting
near the edge of the same merry-go-round? Explain.
The angular velocity is the same but v = ωr, so speed is greatest at the edge
Q11 The two forces in the diagram have the same magnitude.
Which orientation will produce the greater torque on the wheel?
F1
Explain.
F
2
F1 because it is the tangential component that produces the torque
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Q13 Is it possible for the net force acting on an object to be zero,
but the net torque to be greater than zero? Explain. (Hint: The
forces contributing to the net force may not lie along the same
line.)
F
F
Q20 Two objects have the same total mass, but object A has its
mass concentrated closer to the axis of rotation than object B.
Which object will be easier to set into rotational motion? Explain.
A has a smaller moment of inertia and torque = Iα so will accelerate faster
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Q26 A child on a freely rotating merry-go-round moves from near
the center to the edge. Will the rotational velocity of the merrygo-round increase, decrease, or not change at all? Explain.
L =Iω so I increases and ω will decrease. This requires work
Q29 Suppose you are rotating a ball attached to a string in a
circle. If you allow the string to wrap around your finger, does
the rotational velocity of the ball change as the string shortens?
Explain.
L =Iω so I decreases and ω will increase. This requires work
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Ch 8 E 2
Rate of rotation = 45 RPM (rev/min)
a) What is this in rev/s?
b) How many revolutions in 5 seconds?
a) 45 RPM = 45 rev/min | 1 min/60sec = ¾ rev/sec
b) ¾ rev/s (5s) = 15/4 rev
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Ch 8 E 6
Rotational velocity decreases from 6 rev/s to
3 rev/s in 12 s.
What is rotational acceleration?
w = wf – wi = 3rev/s – 6rev/s = - 3rev/s
= w/t = (-3rev/s)/12s = -1/4 rev/s2
 = -1/4 rev/s | 2rad/rev = -/2 rad/s2
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Ch 8 CP 10
5N placed 10 cm from fulcrum of balance beam,
what weight should be put 4 cm from fulcrum on
other side to balance
 = Fl = 0 = -(5N)(10cm) + x(4cm)
x = 50Ncm/4cm = 12.5 N
?
4cm
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5N
10cm
33
Ch 8 E 16
0.8 kg mass rotating with light, rigid 50 cm rod
with w = 3 rad/s.
a) What is rotational inertia?
b) What is angular momentum?
a) I = mr2 = (0.8 kg)(0.50m)2 = 0.2kg m2
W = 3 rad/s
M
b) L = Iw = (0.2 kg m2)(3 rad/s) = 0.6kg m2/s
50cm
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Ch 8 CP 2
4m plank, weight = 80N.
Pivot = 1m from far end.
150N weight on pivot and moving slowly outward (toward far end).
a) What torque is exerted by plank’s weight about pivot?
b) How far can 150N weight move?
c) How can you test (b) without flipping the plank?
4m
2m
1m
1m
Far
End
W = 80N
a) 1 = Fl = (80N)(1m) = +80Nm
W = 150N
b)  = 0 = 1 – (150N)x , x = 80Nm/150N = 0.53m
c) Think about it!
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Ch 8 CP 4
See Fig 8.23. A student sits on a stool with wheels with a bike
wheel with Ib = 2kgm2 and wb = 5 rev/s. Bike wheel is spinning,
student is not. The student on stool with bike wheel: Is = 6 kgm2.
a) Initially, what is L?
b) Student flips bike wheel. What is student’s L?
c) Where does the torque come from that accelerates student?
a) Lb = Ib wb (just the bike wheel spinning).
wb = 5 rev/s 2/rev = 10 rad/s
Lb = 20 kg m2/s , upwards
b) L = Lb + Ls (flipped bike wheel plus student).
-20 kg m2/s + Is ws = +20 kg m2/s Ls = +40 kg m2/s
c) Student supplies torque when he flips bike wheel.
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The “New” Hubble
http://hubblesite.org/
These four images are among the first
observations made by the new Wide
Field Camera 3 aboard the upgraded
NASA Hubble Space Telescope.
The image at top left shows NGC 6302,
a butterfly-shaped nebula surrounding
a dying star. At top right is a picture of
a clash among members of a galactic
grouping called Stephan's Quintet. The
image at bottom left gives viewers a
panoramic portrait of a colorful
assortment of 100,000 stars residing in
the crowded core of Omega Centauri, a
giant globular cluster. At bottom right,
an eerie pillar of star birth in the Carina
Nebula rises from a sea of greenishcolored
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Saturn
1610 - Galileo Galilei becomes the first to observe
Saturn's rings with his 20-power telescope.
On February 24, 2009, the Hubble Space Telescope took a photo of
four moons of Saturn passing in front of their parent planet. In this
view, the giant orange moon Titan casts a large shadow onto Saturn's
north polar hood. Below Titan, near the ring plane and to the left is
the moon Mimas, casting a much smaller shadow onto Saturn's
equatorial cloud tops. Farther to the left, and off Saturn's disk, are the
bright moon Dione and the fainter moon Enceladus.
These rare moon transits only happen when the tilt of Saturn's ring
plane is nearly "edge on" as seen from the Earth. Saturn's rings were
perfectly on edge to our line of sight on August 10, 2009, and
September 4, 2009. Unfortunately, Saturn was too close to the sun to
be seen by viewers on Earth at that time. This "ring plane crossing"
occurs every 14-15 years. In 1995-96 Hubble witnessed the ring plane
crossing event, as well as many moon transits, and even helped
discover several new moons of Saturn.
Scientists at NASA have discovered a nearly invisible ring
around Saturn -- one so large that it would take 1 billion
Earths to fill it.
The ring's orbit is tilted 27 degrees from the planet's main
ring plane. The bulk of it starts about 3.7 million miles (6
million km) away from the planet and extends outward
another 7.4 million miles (12 million km).
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