Chapter 5 Circular Motion. MFMcGraw-PHY 1401Ch5b-Circular Motion-Revised 6/21/2010 2 Circular Motion Uniform Circular Motion Radial Acceleration Banked

Chapter 5

Circular Motion

MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010

2

Circular Motion

• Uniform Circular Motion

• Radial Acceleration

• Banked and Unbanked Curves

• Circular Orbits

• Nonuniform Circular Motion

• Tangential and Angular Acceleration

• Artificial Gravity


3

Angular Displacement

is the angular position.

Angular displacement:

if

Note: angles measured CW are negative and angles measured CCW are positive. is measured in radians.

2 radians = 360 = 1 revolution

x

y

i

f


4

x

y

i

f

r

arc length = s = r

r

s is a ratio of two lengths; it is

a dimensionless ratio!

Arc Length


5

The average and instantaneous angular speeds are:

tt t

0

av lim and

is measured in rads/sec.

Angular Speed


6

The average and instantaneous angular speeds are:

tt t

0

av lim and

Angular Velocity

The direction of is along the axis of rotation. Since we are concerned primarily with motion in a plane we will ignore the vector nature until we get to angular momentum when we will have to deal with it.

is actually a vector quantity.


7

An object moves along a circular path of radius r; what is its average linear speed?

avav timetotal

distance total r

tr

t

rv

Also, rv (instantaneous values).

x

y

i

f

r

Linear Speed

Note: Unfortunately the linear speed is also called the tangential speed. This can cause confusion.


8

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

T

rv

2

timetotal

distance totalav

Comparing to v = r: fT

22

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

Period and Frequency


9

Angular speed is independent of position.

The linear speed depends on the radius r at which the object is located.

x

y

v1

v1

v1

v1

Linear and Angular Speed

rv

r1

r2

v2

1 1

2 2

1 2

1 2

v = ωr

v = ωr

v v= = ω

r r

Everyone sees the same ω, because they all experience the same rpms.


10

Centripetal Acceleration

Here, v 0. The direction of v is changing.

If v 0, then a 0. Then there is a net force acting on the object.

Consider an object moving in a circular path of radius r at constant speed.

x

y

v

v

v

v


11

Conclusion: with no net force acting on the object it would travel in a straight line at constant speed

It is still true that F = ma.

But what acceleration do we use?



12

The velocity of a particle is tangent to its path.

For an object moving in uniform circular motion, the acceleration is radially inward.



13

The magnitude of the radial acceleration is:

vrr

var 2

2



14

The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out.

(a) What force keeps the people from falling out the bottom of the cylinder?

Draw an FBD for a person with their back to the wall:

x

y

w

N

fs

It is the force of static friction.

Rotor Ride Example


15

(b) If s = 0.40 and the cylinder has r = 2.5 m, what is the minimum angular speed of the cylinder so that the people don’t fall out?

Apply Newton’s 2nd Law: 0 2

1 2

wfF

rmmaNF

sy

rx

rad/s 13.3m 5.240.0

m/s 8.9 2

2

r

g

mgrmN

wf

s

ss

s

From (2): From (1)

Rotor Ride Example


16

A coin is placed on a record that is rotating at 33.3 rpm. If s

= 0.1, how far from the center of the record can the coin be placed without having it slip off?

We’re looking for r.

Apply Newton’s 2nd Law:

0 2

1 2

wNF

rmmafF

y

rsx

x

y

w

N

fs

Draw an FBD for the coin:

Unbanked Curve


17

rmmgNf

rmf

sss

s

2

2 :1 From

From (2)

Solving for r:

m 08.0rad/s 50.3

m/s 8.91.02

2

2

g

r s

2 g

r s What is ?

rad/s 5.3sec 60

min 1

rev 1

rad 2

min

rev3.33

Unbanked Curve


18

Banked Curves

A highway curve has a radius of 825 m. At what angle should the road be banked so that a car traveling at 26.8 m/s has no tendency to skid sideways on the road? (Hint: No tendency to skid means the frictional force is zero.)

Take the car’s motion to be into the page.

R


19

Banked Curves

The normal force is the cause of the centripetal acceleration. Nsinθ.

We need the radial position of the car itself, not the radius of the track. In either case the radius is not a uniquely determined quantity.

x

y

N

w

This is the only path by which to traverse the track as there is no friction in the problem.


20

FBD for the car:

x

y

N

w

Apply Newton’s Second Law:

0cos 2

sin 12

wNFr

vmmaNF

y

rx

Banked Curves


21

Rewrite (1) and (2): mgN

r

vmN

cos 2

sin 12

Divide (1) by (2):

22

2

26.8 m/svtanθ = = = 0.089

gr 9.8 m/s 825 m

Banked Curves

θ = 5.1°


22

Circular Orbits

r

vmam

r

MGmFF srs

esg

2

2

Consider an object of mass m in a circular orbit about the Earth.

Earth

r The only force on the satellite is the force of gravity. That is the cause of the centripetal force.

r

GMv

r

vm

r

MGm

e

ses

2

2

Solve for the speed of the satellite:

v


23

Circular OrbitsConsider an object of mass m in a circular orbit about the Earth.

Earth

r

eGMv =

r

GMe is fixed - v is proportional to

v

1

r

v and r are tied together. They are not independent for orbital motions.


24

Example: How high above the surface of the Earth does a satellite need to be so that it has an orbit period of 24 hours?

r

GMv eFrom previous slide: Also need,

T

rv

2

Combine these expressions and solve for r:3

1

224

T

GMr e

m 10225.4

s 864004

kg 1098.5/kgNm1067.6

7

31

2

2

242211

r

km 000,35 ee RrhhRr

Circular Orbits


25

31

224

T

GMr e

Kepler’s Third Law

It can be generalized to:3

1

224

T

GMr

Where M is the mass of the central body.

For example, it would be Msun if speaking of the planets in the solar system. For non-circular orbits (elliptical) the mean radius is used and so the mean velocity is obtained.

Circular Orbits


26

Nonuniform Circular Motion

There is now an acceleration tangent to the path of the particle.

The net acceleration of the body is 22

tr aaa

at

v

ar

a

Nonuniform means the speed (magnitude of velocity) is changing.

This is true but useless!


27

at

ar

aat changes the magnitude of v.

Changes energy - does work

ar changes the direction of v.

Doesn’t change energy - does NO WORK

tt

rr

maF

maFCan write:

Nonuniform Circular Motion

The accelerations are only useful when separated into perpendicular and parallel

components.


28

Example: What is the minimum speed for the car so that it maintains contact with the loop when it is in the pictured position?

FBD for the car at the top of the loop:

N w

y

x


r

vmwN

r

vmmawNF ry

2

2

r

Loop Ride


29

The apparent weight at the top of loop is:

gr

vmN

r

vmwN

2

2

N = 0 when

grv

gr

vmN

0

2

This is the minimum speed needed to make it around the loop.

Loop Ride

You lose contact when N = 0


30

Consider the car at the bottom of the loop; how does the apparent weight compare to the true weight?

N

w

y

x

FBD for the car at the bottom of the loop:


gr

vmN

r

vmwN

r

vmmawNF cy

2

2

2

Here, mgN

Loop Ride


31

Linear and Angular Acceleration

The average and instantaneous angular acceleration are:

tt t

0

av lim and

is measured in rads/sec2.


32

Recalling that the tangential velocity is vt = r means the tangential acceleration is

rrtt

tt

vat

Linear and Angular Acceleration


33

xavv

tatvxx

tavv

2

2

1

20

2

200

0

2

2

1

20

2

200

0

tt

t

Linear (Tangential) Angular

With rarv tt and

Linear and Angular Kinematics

“a” and “at” are the same thing


34

A high speed dental drill is rotating at 3.14104 rads/sec. Through how many degrees does the drill rotate in 1.00 sec?

Given: = 3.14104 rads/sec; t = 1 sec; = 0

Want .

degrees 101.80rads1014.3

sec 0.1rads/sec1014.3

2

1

64

40

00

200

t

t

tt

Dental Drill Example


35

A high speed dental drill is rotating at 3.14104 rads/sec. What is that in rpm’s?

Dental Drill Example

4 412

rad rev rev revπ × 10 × = 10 = 5000

s 2πrad s srev 60sec

5000 × = 300,000rpms min


36

Your car’s wheels are 65 cm in diameter and are rotating at = 101 rads/sec. How fast in km/hour is the car traveling, assuming no slipping?

v

X

km/hr 118cm/sec1028.3

cm 5.32rads/sec 101

2

2

timetotal

distance total

3

rT

r

NT

Nrv

Car Example


37

Artificial Gravity

A large rotating cylinder in deep space (g0).


38

x

y

N

x

y

N

rmmaNF ry2 rmmaNF ry

2

Apply Newton’s 2nd Law to each:

Artificial Gravity

Bottom position Top position

N is the only force acting. It causes the centripetal acceleration.


39

A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120m, at what frequency must it rotate so that it simulates Earth’s gravity?

Using the result from the previous slide:

rad/sec 28.0

2

r

g

mr

mg

mr

N

rmmaNF ry

The frequency is f = (/2) = 0.045 Hz (or 2.7 rpm).

Space Station Example


40

Summary

• A net force MUST act on an object that has circular motion.

• Radial Acceleration ar=v2/r

• Definition of Angular Quantities (, , and )

• The Angular Kinematic Equations

• The Relationships Between Linear and Angular Quantities

• Uniform and Nonuniform Circular Motion

rarv tt and

Documents

Chapter 5 Circular Motion. MFMcGraw-PHY 1401Ch5b-Circular Motion-Revised 6/21/2010 2 Circular Motion Uniform Circular Motion Radial Acceleration Banked