General Physics (PHY 2130)
Lecture 11
•  Rotational kinematics and uniform circular motion
  Angular displacement
  Angular speed and acceleration
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Forces; laws of motion
  field forces and contact forces
  force of friction and gravitational force
  tension
Review Problem: Calculate gravitational attraction between two students (say, 70 kg
and 90 kg) that are 1 meter apart.
Example:
Question: Calculate gravitational attraction between two students (say,
70 kg and 90 kg) that are 1 meter apart
2
m1m2
−11 N m 70 kg 90 kg
−7
F = G 2 = 6.67 ×10
≈
4
.
2
×
10
N
2
2
r
kg
(1m)

Extremely small
Compare: weight of the 70 kg (154 lb) person
F = mg = 686 N
Angular Displacement
►  Recall for linear motion:
  displacement, velocity, acceleration
Δr
Δv
Δr = rf − ri , v =
, a=
Δt
Δt
►  Need
similar concepts for objects
moving in circle (CD, merry-goround, etc.)
►  As before:
  need a fixed reference system (line)
  use polar coordinate system
Angles measured CW are negative and angles measured CCW are positive.
Angular Displacement
►  Every
point on the object
undergoes circular motion
about the point O
►  Angles generally need to be
measured in radians
length of arc
s
θ=
r
►  Note:
1 rad =
θ [rad] =
radius
360°
= 57.3°
2π
π
180°
θ [degrees]
Example
A wheel has a radius of 4.1 m. How far (path length) does a point on athe
circumference travel if the wheel is rotated through angles 30°, 30 rad
and 30 rev respectively?
Angular Displacement
angular displacement is
defined as the angle the object
rotates through during some
time interval
►  The
Δθ = θ f − θ i
►  Every
point on the disc
undergoes the same angular
displacement in any given time
interval
Angular Velocity
►  The
average angular
velocity (speed), ω, of a
rotating rigid object is the
ratio of the angular
displacement to the time
interval
θ f −θi
Δθ
ω=
=
t f − ti
Δt
Angular Speed
instantaneous angular
velocity (speed) is defined as the
limit of the average speed as the
time interval approaches zero
►  The
Δθ
ω = lim
Δt →0 Δt
►  Units
of angular speed are
radians/sec (rad/s)
►  Angular speed will be
  positive if θ is increasing
(counterclockwise)
  negative if θ is decreasing
(clockwise)
Angular Acceleration
►  What
if object is initially at rest
and then begins to rotate?
►  The average angular acceleration,
α, of an object is defined as the
ratio of the change in the angular
speed to the time it takes for the
object to undergo the change:
ω f − ωi
Δω
α=
=
t f − ti
Δt
►  Units
are rad/s²
►  Similarly, instant. angular accel.:
Δω
Δt →0 Δt
α = lim
Notes about angular kinematics:
When a rigid object rotates about a fixed axis,
every portion of the object has the same angular
speed and the same angular acceleration
θ, ω, and α are not dependent upon r, distance
form hub or axis of rotation
► i.e.
Examples:
1. Bicycle wheel turns 240 revolutions/min. What is its angular
velocity in radians/second?
ω = 240
rev 1 min 2π rads
×
×
= 8π radians sec ≈ 25.1 radians sec
min 60 sec
1 rev

2. If wheel slows down uniformly to rest in 5 seconds, what is the
angular acceleration?
α=
ω f − ωi
Δt
=
0 − 25 rad sec
= −5 rad sec 2
5 sec

Examples:
Given:
1. Angular velocity:
240 rev/min
2. Time t = 5 s
3. How many revolution does it turn in those 5 sec?
Find:
1.
θ=?
1 2
x
=
v
t
+
at
Recall that for linear motion we had:
0
2
Perhaps something similar for angular quantities?
1
2
θ = ω 0t + αt 2
1
(− 5 rad sec)(5 sec)2 = 62.5 rad
2
1 rev
θ (rev) = 62.5 rad ×
= 10 revolutions

2π
= 25 rad sec (5 sec) +
Analogies Between Linear and
Rotational Motion
Rotational Motion About a
Fixed Axis with Constant
Acceleration
Linear Motion with
Constant Acceleration
ω = ωi + αt
v = vi + at
1 2
Δθ = ω i t + αt
2
1 2
Δx = vi t + at
2
2
2
i
ω = ω + 2αΔθ
2
2
i
v = v + 2aΔx
Relationship Between Angular
and Linear Quantities
►  Displacements
►  Speeds
►  Accelerations
Δs
Δθ =
r
Δθ 1 Δs
=
Δt r Δt
or
1
ω= v
r
a = αr
Relationship Between Angular
and Linear Quantities
►  Displacements
►  Speeds
s = θr
v = ωr
►  Accelerations
a = αr
►  Every
point on the
rotating object has the
same angular motion
►  Every point on the
rotating object does
not have the same
linear motion
ConcepTest
A ladybug sits at the outer edge of a merry-go-round, and
a gentleman bug sits halfway between her and the axis of
rotation. The merry-go-round makes a complete revolution
once each second.The gentleman bug’s angular speed is
1.
2.
3.
4.
half the ladybug’s.
the same as the ladybug’s.
twice the ladybug’s.
impossible to determine
ConcepTest
A ladybug sits at the outer edge of a merry-go-round, and
a gentleman bug sits halfway between her and the axis of
rotation. The merry-go-round makes a complete revolution
once each second.The gentleman bug’s angular speed is
1.
2.
3.
4.
half the ladybug’s.
the same as the ladybug’s.
twice the ladybug’s.
impossible to determine
Note: both insects have an angular speed of 1 rev/s
Period and frequency
The time it takes to go one time around a closed path is called the period (T).
total distance 2πr
vav =
=
total time
T
Comparing to v = rω:
2π
ω=
= 2πf
T
f is called the frequency, the number of revolutions (or cycles) per second.
19
Centripetal Acceleration
►  An
object traveling in a circle,
even though it moves with a
constant speed, will have an
acceleration (since velocity
changes direction)
►  This acceleration is called
centripetal (“center-seeking”).
►  The acceleration is directed
toward the center of the circle
of motion
Centripetal Acceleration
and Angular Velocity
►  The
angular velocity and the
linear velocity are related
(v = ωr)
►  The centripetal acceleration
can also be related to the
angular velocity
Δv Δs
v
=
⇒ Δv = Δs, but
v
r
r
a=
Δv
v Δs
⇒ a=
Δt
r Δt
Thus:
Similar
triangles!
v2
aC =
r
or aC = ω 2 r
Total Acceleration
►  What
happens if linear velocity
also changes?
►  Two-component acceleration:
  the centripetal component of the
acceleration is due to changing
direction
  the tangential component of the
acceleration is due to changing
speed
►  Total
acceleration can be found
from these components:
slowing-down car
2
t
2
C
a = a +a
Vector Nature of Angular
Quantities
►  As
in the linear case,
displacement, velocity
and acceleration are
vectors:
►  Assign a positive or
negative direction
►  A more complete way is
by using the right hand
rule
  Grasp the axis of rotation
with your right hand
  Wrap your fingers in the
direction of rotation
  Your thumb points in the
direction of ω
Forces Causing Centripetal
Acceleration
►  Newton’s
Second Law says that the centripetal
acceleration is accompanied by a force
v2
∑ F = maC = m r
  F stands for any force that keeps an object following a
circular path
► Force
of friction (level and banked curves)
► Tension in a string
► Gravity
Example1: level curves
Consider a car driving at 20 m/s (~45
mph) on a level circular turn of
radius 40.0 m. Assume the car’s
mass is 1000 kg.
1.
2.
What is the magnitude of
frictional force experienced by
car’s tires?
What is the minimum coefficient
of friction in order for the car to
safely negotiate the turn?
Example1:
Given:
masses: m=1000 kg
velocity: v=20 m/s
radius: r = 40.0m
1. Draw a free body diagram, introduce
coordinate frame and consider vertical
and horizontal projections
∑F
y
= 0 = N − mg
N = mg
Find:
1.
2.
f=?
µ=?
∑F
x
= ma = − f
2
(20 m s ) = −1.0 ×10 4 N
v2
f = −ma = −m = −1000 kg
r
40 m

2. Use definition of friction force:
v2
f = µmg = m = 10 4 N , thus
r
1.0 × 10 4 N
µ=
≈ 1.02
2
1000 kg 9.8 m s

Lesson: µ for rubber on dry concrete is 1.00!
rubber on wet concrete is 0.2!
driving too fast…
ConcepQuestion
Is it static or kinetic friction that is responsible for the fact
that the car does not slide or skid?
1. Static
2. Kinetic
Example2: banked curves
Consider a car driving at 20 m/s (~45
mph) on a 30° banked circular
curve of radius 40.0 m. Assume
the car’s mass is 1000 kg.
1.
2.
What is the magnitude of
frictional force experienced by
car’s tires?
What is the minimum coefficient
of friction in order for the car to
safely negotiate the turn?
A component of the normal force adds to the
frictional force to allow higher speeds
v2
tan θ =
rg
Example2:
Given:
masses: m=1000 kg
velocity: v=20 m/s
radius: r = 40.0m
angle: α = 30°
Find:
1.
2.
f=?
µ=?
1. Draw a free body diagram,
introduce coordinate frame and
consider vertical and
horizontal projections
v2
∑ Fx = −m r cos 30 = − f − mg sin 30
v2
f = m cos 30  − mg sin 30  = 3760 N
r
2
v
∑ Fy = m r sin 30 = N − mg cos 30
v2
N = m sin 30 + mg cos 30  = 1.3 ×10 4 N
r
2. Use definition of friction force:
f = µ s N , thus minimal µs is
µs =
fs
3760 N
=
≈ 0.28
4
N 1.3 ×10 N


Lesson: by increasing angle of banking,
one decreases minimal µ or friction with
which one can take curve!