Lecture 11

General Physics (PHY 2130)

•  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

( )N

mkgkg

kgmN

rmmGF 7

22

211

221 102.4

190701067.6 −− ×≈×==  

Extremely small

Compare: weight of the 70 kg (154 lb) person

NmgF 686==

Angular Displacement

►  Recall for linear motion:   displacement, velocity, acceleration

►  Need similar concepts for objects moving in circle (CD, merry-go-round, etc.)

►  As before:   need a fixed reference system (line)   use polar coordinate system

tva

trvrrr if Δ

Δ=

Δ

Δ=−=Δ ,,

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

►  Note:

rs

=θ

°=°

= 3.5723601π

rad

]degrees[180

]rad[ θπ

θ°

=

length of arc

radius

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

►  The angular displacement is defined as the angle the object rotates through during some time interval

►  Every point on the disc undergoes the same angular displacement in any given time interval

if θθθ −=Δ

Angular Velocity

►  The average angular velocity (speed), ω, of a rotating rigid object is the ratio of the angular displacement to the time interval

ttt if

if

Δ

Δ=

−

−=

θθθω

Angular Speed

►  The instantaneous angular velocity (speed) is defined as the limit of the average speed as the time interval approaches zero

►  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)

tt Δ

Δ=

→Δ

θω

0lim

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:

►  Units are rad/s² ►  Similarly, instant. angular accel.:

ttt if

if

Δ

Δ=

−

−=

ωωωα

tt Δ

Δ=

→Δ

ωα

0lim

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

► i.e. θ, ω, and α are not dependent upon r, distance form hub or axis of rotation

Examples:

1. Bicycle wheel turns 240 revolutions/min. What is its angular velocity in radians/second?

secradians1.25secradians8rev1rads2

sec60min1

minrev240 ≈=××= π

πω  

2. If wheel slows down uniformly to rest in 5 seconds, what is the angular acceleration?

2secrad5sec5

secrad250−=

−=

Δ

−=

tif ωω

α  

Examples:

Given: 1. Angular velocity: 240 rev/min 2. Time t = 5 s Find: 1.  θ = ?

3. How many revolution does it turn in those 5 sec?

( ) ( )( )

srevolution102rev1rad5.62)(

rad5.62sec5secrad521sec5secrad25

21

2

20

=×=

=−+=

+=

πθ

αωθ

rev

tt

 

Recall that for linear motion we had: Perhaps something similar for angular quantities?

20 2

1 attvx +=

Analogies Between Linear and Rotational Motion

Rotational Motion About a Fixed Axis with Constant Acceleration

Linear Motion with Constant Acceleration

ti αωω +=

2

21 tti αωθ +=Δ

θαωω Δ+= 222i xavv i Δ+= 222

2

21 attvx i +=Δ

atvv i +=

Relationship Between Angular and Linear Quantities

► Displacements

► Speeds

► Accelerations

rsΔ

=Δθ

vr

ts

rt

1or

1

=

Δ

Δ=

Δ

Δ

ω

θ

ra α=

Relationship Between Angular and Linear Quantities

► Displacements

► Speeds

► Accelerations

► Every point on the rotating object has the same angular motion

► Every point on the rotating object does not have the same linear motion

rs θ=

rv ω=

ra α=

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. half the ladybug’s. 2. the same as the ladybug’s. 3. twice the ladybug’s. 4. 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. half the ladybug’s. 2. the same as the ladybug’s. 3. twice the ladybug’s. 4. impossible to determine

Note: both insects have an angular speed of 1 rev/s

19

The time it takes to go one time around a closed path is called the period (T).

Trv π2

timetotaldistance total

av ==

Comparing to v = rω: fT

ππ

ω 22==

f is called the frequency, the number of revolutions (or cycles) per second.

Period and frequency

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

ts

rva

tv

srvv

rs

vv

Δ

Δ=⇒

Δ

Δ=

Δ=Δ⇒Δ

=Δ

a

but,

rarva CC

22

or ω==Thus:

Similar triangles!

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:

22Ct aaa +=

slowing-down car

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

  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

rvmmaF C

2

==∑

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.  What is the magnitude of

frictional force experienced by car’s tires?

2.  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 Find: 1.  f=? 2.  µ=?

1. Draw a free body diagram, introduce coordinate frame and consider vertical and horizontal projections

mgNmgNFy

=

−==∑ 0

( ) Nmsmkg

rvmmaf

fmaFx

422

100.140201000 ×−=−=−=−=

−==∑

  2. Use definition of friction force:

02.18.91000

101.0

thus,10

2

4

42

≈×

=

===

smkgN

Nrvmmgf

µ

µ

  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.  What is the magnitude of

frictional force experienced by car’s tires?

2.  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 rg

v2tan =θ

Example2:

Given: masses: m=1000 kg velocity: v=20 m/s radius: r = 40.0m angle: α = 30° Find: 1.  f=? 2.  µ=?

1. Draw a free body diagram, introduce coordinate frame and consider vertical and horizontal projections

Nmgrvmf

mgfrvmFx

376030sin30cos

30sin30cos

2

2

=−=

−−=−=∑

NmgrvmN

mgNrvmFy

42

2

103.130cos30sin

30cos30sin

×=+=

−==∑

 

2. Use definition of friction force:

28.0101.3

3760is minimalthus,

4

s

≈×

==

=

NN

NfNf

ss

s

µ

µµ

  Lesson: by increasing angle of banking,

one decreases minimal µ or friction with which one can take curve!