Rotational Motion
Definitions


Most of our discussion will concerned with
rigid bodies— objects with definite shapes
that don’t change
Purely rotational motion: all points in a
body move in circle
r
O
θ
P
ℓ
Radians

One radian is the angle created
(subtended) by an arc whose length is
equal to the radius
360o = 2 π rad
O
P
r
θ ℓ
Angular Velocity and Acceleration

Average angular velocity:
ω = Δθ / Δt


Measured in radians per second
Average angular acceleration:
α = (ω – ω0)/ Δt = Δω / Δt

Measured in radians per second squared
The Velocity of a Point

A point on a rotating
wheel has the
following linear
velocity
v = Δℓ / Δt = r (Δθ/ Δt)
or
v = rω
Different Points Can Have
Different Velocities

Despite the fact that
ω is the same for all
points, points with
different values of r
have different
velocities
Tangential Velocities Can Be
Different for Points on an Object
Tangential Velocities Can Be
Different for Points on an Object
Acceleration



Angular acceleration is related to tangential
linear acceleration by:
atan = Δv / Δt = r (Δω / Δt)
or
atan = rα
Total linear acceleration is:
a = atan + aR,
Where aR is the radial or centripetal acceleration
toward the center of the object’s path
Centripetal Acceleration

aR = v2/ r = (ωr)2 / r = ω2r
Frequency and Period



Frequency is the number of complete revolutions
(rev) per second
One revolution corresponds to an angle of 2π
radians
 Therefore, 1 rev/sec = 2π rad/ sec
f = ω/ 2π or ω = 2πf
The unit of frequency is the hertz (Hz)
1 Hz = 1 rev/s

The time required for one revolution is a period T
T = 1/f
Centripetal Force

A force is required to
keep an object
moving in a circle

If the speed is
constant the force is
directed towards the
center of the circle
∑FR = maR = mv2/r
There Is No Outward Force!
Components of Circular Motion
Force On a Revolving Ball
(Horizontal)



Estimate the force a person must exert on a string
attached to a 0.15 kg ball to make the ball revolve
in a horizontal circle of radius 0.6 m. The ball
makes 2 revolutions per second
F = mv2/r = (0.15 kg)(7.54 m/s)2/(0.6 m)
≈ 14 N
This solution ignores the fact that the ball cannot
be perfectly horizontal because it has weight due
to the force of gravity
Force On a Revolving Ball
(Vertical)

A 0.15 kg ball on the end of a 1.1 m string
is swung in a vertical circle. Determine
the minimum speed the ball must have at
the top of its arc so that it continues
moving in a circle. Calculate the tension
in the string at the bottom of the arc if the
ball is moving at twice the minimum
speed.
Solution


At the top of the arc there
are two forces on the ball:
mg and the tension FTA
∑FR = maR
FTA + mg = mvA2/r
The minimum speed
occurs when FTA = 0
mg = mvA2/r
vA= √(gr)
= 3.28 m/s
Solution (cont’d)


At the bottom of the circle the cord exerts
its tension force FTB upward, but the force
of gravity mg is downward
∑FR = maR
FTB – mg = mvB2/r
= mvB2/r + mg
For v = 6.56 m/s (2x minimum) F= 7.34 N
Ferris Wheel


A rider on a Ferris wheel moves in a
vertical circle of radius r at constant speed
v
Is the normal force the seat exerts on the
rider at the top of the circle less than,
more than, or the same as the force
exerted at the bottom of the arc?
Solution

This is exactly like the
vertical string problem
with FN replacing
tension. Therefore,
the force at the top is
less than the force at
the bottom
Centripetal Forces--Uniform
Circular Motion
Roller Coaster Physics
Demonstrations
A Right Hand Turn
Forces on Cars In Turns
Centripetal
acceleration is
horizontal, not
parallel to road
surface
Flat Road
Banked Turn
Banked vs Flat Turns
Circular Motion as the Limit of
Straight Line Motion
Ball on a String

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c8_whirligig.html
Gravitron

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c7_rotor.html
Objects on a Turntable

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c7_turntable.html
Vertical Circular Motion

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c8_vertical.html
Kinematic Equations
Angular
ω = ω0 + ½αt
θ = ω0t + ½αt2
ω2 = ω02 + 2αθ
ω = (ω + ω0)/2

Linear
v = v0 + at
x = v0t + ½ at2
v2 = v02 + 2ax
v = (v + v0)/2

Example


A bicycle slows down from 8.4 m/s to rest
over a distance of 115 m. Each wheel has
a diameter of 68.0 cm.
Determine the angular velocity of the
wheels at the initial moment; the total
number of revolutions each wheel makes
in coming to rest, and the time it took to
stop.
Solution



At the initial instant, points on the rim of the
wheel are moving at 8.4 m/s.
The initial angular velocity is:
ω0 = v0/r = (8.4 m/s)/ (0.34 m) = 24.7 rad/s
115m of ground passes under the bike as it
stops. Each revolution of a wheel corresponds
to a distance of 2πr so:
115 m/ 2πr = 115 m/(2πr)(0.34 m) = 53.8 rev
Solution
Angular acceleration of the wheel can be
obtained from:
ω2 = ω02 + 2αθ
α = (ω2 - ω02)/ 2θ = 0 – (24.7 rad/s)2
2(2π)(53.8 rev)
= -0.902 rad/s2
From ω = ω0 + ½αt we get that:
t = (ω - ω0)/α = (0 – 24.7 rad/s)/ -0.902 rad/s2
= 27.4 s

Rolling

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c15_rolling.html
Rotational Dynamics


Up to this point we have been studying
rotational kinematics— how things move
Now we will go on to rotational
dynamics—why things move
Ferris Wheel Kinematics

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c13_consta_ex.ht
ml
Torque

Causing an object to rotate around its axis requires a
force


The angular acceleration of an object is directly
proportional to the perpendicular distance from the axis
of rotation to the line along which the force acts


The effect of a force is greater if it is placed further from the
axis of rotation
This distance is called a lever arm
The product of the force times the lever arm is called
torque (α proportional to τ)
Only the Perpendicular Component
of Force Contributes to Torque
Torque = r┴F = rF┴
Example

The biceps muscle
exerts a vertical force
on the lower arm.
Calculate the torque
about the axis of
rotation through the
elbow joint assuming
the muscle is
attached 5.0 cm from
the elbow
Do Now (10/1/13):

The biceps muscle
exerts a vertical force
on the lower arm.
Calculate the torque
about the axis of
rotation through the
elbow joint assuming
the muscle is
attached 5.0 cm from
the elbow
Solution
F = 700 N and r┴ = 0.05m so
τ = r ┴F = (0.05 m)(700 N) = 35 m-N
Solution
r┴ = (0.05 m)(sin 60o)
Therefore
τ = (0.05 m)(sin 60o)(700N)
= (0.05 m)(0.866)(700N)
= 30 m-N
Torque on a Compound Wheel

Two thin cylindrical
wheels of radii r1 = 30
cm and r2 = 50 cm
are attached to each
other as shown.
Calculate the net
torque on this wheel
due to the two forces
shown, each of
magnitude 50 N
Solution



The two forces create torques in different directions
We can consider one to be positive an done to be
negative
Because F2 is not perpendicular to the axis of
rotation we must only use the component of the
force that is perpendicular
τ = r1F1 – r2F2sin60o
= (0.3m)(50N) – (0.5m)(50 N)(0.866)= -6.7 m-N
Balancing Torques
Force = F
B
Force = 2F
C
A
D
X
X
A force of magnitude F is applied at a distance X from the
center of a seesaw. Another force of magnitude 2F is also
applied to the seesaw at a distance X on the other side of the
fulcrum. At what location(s) and in what directions can a third
force of magnitude F be applied so that the seesaw is
balanced?
Solution
Force = F
Force = F
B
Force = 2F
C
A
D
X
X
Force = -F
The seesaw can be balanced if a force of magnitude
F is applied in the positive direction at A or in the
negative direction at D
Torques

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c14_equilibrium.ht
ml
Rotational Analogs of Mass and
Momentum

The angular acceleration of an object is
directly proportional to the perpendicular
distance from the axis of rotation to the
line along which the force acts

α is proportional to Στ
Rotational Analog of Mass





Consider a point of mass m on the end of a
string with an applied force F
From F = ma we draw the analog that
= mrα
Multiplying both sides by r we find that
Τ = rF = mr2α
The quantity mr2 is called the moment of
inertia of the mass (I)
The moment of inertia is the rotational
analog of mass
F
r
Rotational Analog of Newton’s
Second Law



From the previous page
τ = rF = mr2α
Therefore
Στ = Σ (mr2)α
But mr2 = I, therefore
Στ = Iα
Question


An ice skater in a spin turns rapidly if her
arms and legs are in line with her body,
but turns more slowly when they are
outstretched. Why?
The moment of inertia (rotational analog of
mass) is mr2. When more mass is held at
greater distance from the body, I is
increased.
Rotational Analog of Momentum



Define angular momentum as
L = Iω
The total angular momentum of a rotating
body remains constant if the net torque
acting on it is zero (∆L/∆t = 0)
Angular momentum is a vector quantity

We use the right hand rule to determine the
direction of L
Right Hand Rule



Numerous physical
phenomena have resultant
vectors that are
perpendicular to the original
vectors in the problem
In these cases a standard
convention is needed to
establish positive or negative
vector direction
The “right hand rule” is used
in these cases
Conservation of Angular
Momentum



Consider a man walking on a circular
platform that is at rest
If the person starts walking forward the
platform will turn backward
Newton’s 3rd Law provides a rationale
for this but so does conservation of
angular momentum
• Man’s angular momentum = Iω = (mr2)(v/r) where v
is his velocity relative to the earth, r is his distance
from the axis and m his mass
• The total angular momentum of the man-platform
system remains zero
Classwork


Problem 43 in textbook ch. 8
Use the rest of class to work on hw
Conservation of Momentum

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c16_collision1.html
Angular Momentum

http://webphysics.davidson.edu/physlet_re
sources/bu_semester1/c15_race.html
Do Now (10/2/13):
1.
2.
If you walked along a tall fence, why
would holding your arms out help you to
balance?
Consider two rotating bicycle wheels, one
filled with air and the other filled with
water. Which would be more difficult to
stop rotating? Explain.
Conceptual Questions




Work with your group to answer the review
questions
Submit one paper per group
Each group member must write answers!!!
Use different color inks/pencils for different
group members
*Bonus – answer the “Think and Explain”
questions
Bicycle Wheel Gyroscope




A rapidly spinning wheel has
angular momentum. Changing
the direction of the wheel will
require an applied torque or
will result in a torque to
conserve angular momentum.
Initial angular momentum of
man-gyro system is +L
After wheel is turned over the
momentum of the wheel is –L
Therefore, the man-wheel
system will have to have an
angular momentum of +2L
upward
Do Now (10/3/13):


You sit in the middle of a large, freely
rotating turntable at an amusement park. If
you crawled toward the outer rim, what
would happen to the rotation speed?
Explain.
How do clockwise and counterclockwise
torques compare when a system is
balanced?
Gravity
Universal Gravitation



Every object attracts every other object with a
force that for any two objects is directly
proportional to the mass of each object.
F = G m1 m2
d2
G = 6.67 x 10 –11 N m2/kg2
Gravity is a very weak force!!

It is the weakest of the four fundamental forces

Electromagnetic force, nuclear strong force, nuclear weak
force.
The Inverse-Square Law

The force of gravity decreases with the
square of the distance between objects


The force of gravity reduces rapidly with
distance.
However, the force of gravity never reaches
zero.
Gravitational Fields




We define the gravitational acceleration “g” as
g = F/m
For an object on Earth’s surface, it is at a radius
R from the center.
Therefore,
g = F/m = (G m M/R2)/m
= GM/R2 = 9.8m/s2
Where M is the mass of the Earth (5.98 x 1024
kg) and R = 6.37 x 106 km
Attraction Between People

What is the attraction between a 50 kg
person and a 75 kg person?
F = (6.67 x 10-11 Nm2/kg2)(50 kg)(75 kg)
(0.50 m)2
=1
x 10-6 N (not very much force)
Gravity on Mt. Everest

Replace r (radius of earth) by r + 8.8 km = 6389 km
g = GmE/r2 = (6.67 x 10-11 Nm2/kg2)(5.98 x 1024)
(6.389 X 108 m)2
= 9.77 m/s2 compared with 9.8 m/s2
The Falling Moon

Newton compared the moon to a cannonball
fired from the top of a high mountain.



He imagined the mountaintop to be above the
atmosphere.
He reasoned that a cannonball fired fast enough
would go into orbit around the Earth
Essentially, Newton reasoned that the moon
was falling into the Earth
Orbiting the Earth
Geosynchronous Satellites


G (msatmE)/r2 = msat (v2/r)
But v = 2πr/T where T = 1 day = 84,600 s
Therefore,
G(mE/r2) = (2πr)2/rT2
r3 = 7.54 x 1022 m3
r = 4.23 x 107 m
r = 26,283 mi from center of earth
= 22,317 mi above surface
Kepler’s Laws



50 years before Newton
Kepler had derived the
Laws of Planetary Motion
1st law: Path of each
planet about the sun is an
ellipse
2nd Law: Each planet
moves so that an
imaginary line from the
sun to the planet sweeps
out equal areas in equal
amounts of time
Kepler’s Law Animation
Kepler’s Laws



3rd Law: The ratio of the squares of the
periods (time for one revolution about the
sun) of any two planets is equal to the
ratio of their mean distances cubed from
the sun
(T1/T2)2 = (r1/r2)3
We can rewrite this as:
r13/T12 = r23/T22
Which is the same for all planets
Enter Newton
Gm1Ms/r12 = m1v12/r1
where Ms is the mass of the sun
 Since v1 = 2πr1/T1,
Gm1Ms/r12 = m1(4π2r1)/T12
 Rearranging we get:
T12/r13 = 4π2/GMs
 Since we can derive this same expression for
another planet m2, then we can set
r13/T12 = r23/T22
Using Kepler’s Laws
A Martian year is 1.88 Earth years.
Determine the mean distance of Mars
from the sun.
 Let TE = 1 year and rES = 1.5 x 1011 m
 Kepler’s 3rd Law gives us:
rMS/rES = (TM/TE)2/3 = (1.88 yr/1 yr)2/3 = 1.53
times as far from the sun as the Earth

Geosynchronous Satellites Using
Kepler’s Laws


The moon circles the Earth in roughly 27
days and rME = 380,000 km from earth
A geosynchronous satellite has a period of
1 day, therefore:
rSat = rME(Tsat/TM)2/3
= rME (1 day/27 day)2/3
= rME(1/3)2 = rME/9
Determining the Mass of the Sun


If rES = 1.5 x 1011 m determine the mass of the sun
From a previous slide:
Gm1Ms/r12 = m1(4π2r1)/T12
MS = 4π2rES3/GTE2 = 4π2(1.5 x 1011)/G(3.16 x 107)
= 2.0 x 1030 kg
 To calculate this we used:
TE = 1 yr = (365¼ d) (24 h/d) (3600 s/h)
= 3.16 x 107
Experiencing Weightlessness



The apparent weight of an object on earth
is equal to:
w = mg + ma
An object accelerating away from the
earth will appear to have increased weight
An object falling towards earth will appear
to have reduced weight or weightlessness
Riding an Elevator
Weightlessness in Orbit


Since a satellite can be considered as
“falling towards Earth”, people in a satellite
experience “free fall” which provides the
appearance of weightlessness.
This is different from the weightlessness
experienced at distances far from massive
objects
End
Do Now (10/4/13):
1.
2.
If the moon falls, why doesn’t it get closer
to the earth?
Since the planets are pulled to the sun by
gravity, why don’t they simply crash into
the sun?
Practice:

Work on homework quietly!!!