Circular Motion

Milky Way Galaxy

Orbital Speed of Solar System: 220 km/s Orbital Period: 225 Million Years

Mercury: 48 km/s

Venus: 35 km/s

Earth: 30 km/s

Mars: 24 km/s

Jupiter: 13 km/s

Neptune: 5 km/s

Equatorial Rotational Period: 27.5 days

Differential Solar Rotation Entangled Magentic Field Lines makes Sun Spots

464m/s

Precession causes the position of the North Pole to change over a period of 26,000 years.

Orbital Speed of Earth: ~ 30 km/s

Although the Moon is always lit from the Sun, we see

different amounts of the lit portion from

Earth depending on

where the Moon is

located in its month-long

orbit. Orbital Speed of Moon: ~ 1 km/s

155mph~70m/s

RADAR: RAdio Detecting And Ranging

200mph~90m/s

Electrons in Bohr Orbit: 2 million m/s

(Speed of Light: 200 million m/s)

Maximal Kerr Black Hole

Rotational Speed = Speed of Light

Translational vs Rotational

//

mx

v dx dta dv dt

∆==

//

I

d dtd dt

θω θα ω

∆==

2 /c

t

s rv ra v ra r

θω

α

∆ = ∆=

==

Connection

Translational and Rotational Kinematics

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When angular velocity ω is constant, this is uniform circular motion.

In this case, as the particle goes around a circle one time, its angular displacement is ∆θ = 2π during one period ∆t = T.

Angular Velocity in Uniform Circular Motion

Slide 4-87

: angular velocity (rad/s)ω

Angular Position: Rotation Angle

∆θ = rotation angle

∆s = arc length

r = radius

sr

θ ∆∆ =

The rotation angle is the ratio of arc length to radius of curvature. For a given angle, the greater the radius, the greater the arc length.

r

s = ? s = C = 2πr

θ = s/r θ = 2 πr /r

θ = 2 π

How many degrees in 1 radian?

How many radians in 3600 ? Ans: 3600 = 2 π radians

Consider a circle with radius r.

2 π rad = 3600

1 rad = 3600 /2 π = 57.30

d

s θ

r = 4 m

A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of 430. What is the distance, d, that the box moves?

43 43 1x=

243360

x π=

.75rad=

First: How many radians is 430 ?

d

s θ

r = 4 m

A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of .75 rad. What is the distance, d, that the box moves?

s = r θ d = r θ d = 4m (0.75 rad) d = 3 m rad d = 3 m

430 = .75rad

Radian is dimensionless and is dropped!

Angular Velocity : angular velocity (rad/s)ω

θω =ddt

( / )=

d s rdt

1=

dsr dt

t∆

vr

=

: angular velocity (rad/s)v: tangential velocity (m/s)ω

θω = =d vdt r

v rω=

Angular & Tangential Velocity

is the same for every point & v varies with r. ω

Problem Calculate the angular velocity of a 0.500 m radius car tire when the car travels at a constant speed of 25.0 m/s.

ω =vr

25 /.5m sm

=

50 /rad s=

Insert rad for angular speed. Keep it to define units.

v rω=

Tangential and Angular Acceleration changes speed

ωα =ddt

=tdvadt

( )ω=

d rdt

α=ta r

=drdtω

v rω=

Angular –Tangential Bike A bike wheel with a radius of 0.25 m undergoes a constant

angular acceleration of 2.50 rad/s2. The initial angular speed of the wheel is 5.00 rad/s. After 4.00 s

a) What angle has the wheel turned through? b) What is the final angular speed? c) What is the final tangential speed of the bike? d) How far did the bike travel?

2

012

t tθ ω α∆ = + 2 215 / 4 2.5 / (4 ) 402

rad s s rad s s rad= + =

0f tω ω α= + 25 / 2.5 / 4 15 /rad s rad s s rad s= + =

15 / .25 3.75 /v r rad s m m sω= = =

40 .25 10d s r rad m mθ= ∆ = ∆ = =

Givens:

02

5 /

2.5 /4

rad srad s

t s

ω

α

=

==

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Centripetal Acceleration

Slid 4 94

2

( / )dv vd vadt rd v r

θθ

= = =

( )ds d rvdt dt

θ= =

v rω= s rθ=

( )d rdtvθ

=

Combining:

Centripetal Acceleration Only changes Direction!!!

Centripetal and Tangential Accelerations

2

=cvar

α=ta r2( )ω

=rr

2ω=ca r

(r constant)

Total Acceleration

2 2r ta a a= +

ˆˆ θθ= − +

ra a r a

/θ α= = =ta a r dv dt2 2 /ω= = =r ca a r v r

The Earth rotates once per day around its axis as shown. Assuming the Earth is a sphere, is the rotational speed at Santa

Rosa greater or less than the speed at the equator?

366 m/s

464 m/s

Is the centripetal acceleration greater at the Equator or at Santa

Rosa?

2

cvar

= 20.034 /m s

20.027 /m s

The Earth rotates once per day around its axis. Assuming the Earth is a sphere with radius 6.38 x 106m, find the tangential speed of a person at the equator and at 38 degrees latitude (Santa Rosa!) and their centripetal accelerations.

At the equator, r = 6.38 x 106m: 62 2 (6.38 10 ) 464 /

86,400r x mv m st sπ π

= = =∆

( )222

6

464 /.034 /

6.38 10= = =c

m sva m sr x m

At Santa Rosa, r = 6.38 x 106m cos38: 62 2 (6.38 10 )cos38 366 /

86,400r x mv m st sπ π

= = =∆

( )222

6

366 /.027 /

cos38 6.38 10 cos38= = =c

m sva m sr x m

ca

ca

What is the total acceleration acting on a person in Santa Rosa?

cag

The vector sum.

Is your apparent weight as measured on a spring scale more at the Equator or at Santa Rosa?

cag

Since you are standing on the Earth (and not in the can) the centrifugal force tends

to throw you off the Earth. You weigh less where the centripetal force is greatest because

that is also where the centrifugal force is greatest – the force that tends to throw you

out of a rotating reference frame.

Centrifugal Force & Acceleration “Center FLEEing”

Factual = Centripetal Force Ffictitious = Centrifugal Force

Centrifugal Force & Acceleration

Suppose there is a lady bug in the can. There is a centripetal force acting on the bug, transmitted to her feet by the can. Her feet push back on the can producing a centrifugal “fictitious” force, that acts

like artificial gravity.

“Center FLEEing”

Centripetal & Centrifugal Force Depends on Your Reference Frame

Inside Observer (rotating reference frame) feels Centrifugal Force pushing them against the can.

Outside Observer (non-rotating frame) sees Centripetal Force pulling can in a circle.

Center-seeking

Center-fleeing

Centrifugal Force is Fictitious? The centrifugal force is a real effect. Objects in a rotating frame feel a centrifugal force acting on them, trying to push them out. This is due to your inertia – the fact that your mass does not want to go in a circle. The centrifugal force is called ‘fictitious’ because it isn’t due to any real force – it is only due to the fact that you are rotating. The centripetal force is ‘real’ because it is due to something acting on you like a string or a car.

In Physics, we use ONLY

CentriPETAL acceleration NOT

CentriFUGAL acceleration!

Artificial Gravity How fast would the space station segments A and B have to rotate in

order to produce an artificial gravity of 1 g?

56 / ~ 115Av m s mph=

104 / ~ 210Bv m s mph=

Can the two segments be connected?

Space Station Rotation

A space station of diameter 80 m is turning about its axis at a constant rate. If the acceleration of the outer rim of the station is 2.5 m/s2, what is the period of revolution of the space station? a. 22 s b. 19 s c. 25 s d. 28 s e. 40 s

td

adt

=v

2

r Cva ar

= − = −

2 2r ta a a= +

Nonuniform Circular Motion

Circular Motion Total Acceleration • The tangential

acceleration causes the change in the speed of the particle

• The radial acceleration

comes from a change in the direction of the velocity vector

td

adt

=v

2

r Cva ar

= − = −2 2r ta a a= +

Total Acceleration:

Total Acceleration The total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain of time is 15.0m/s2. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration.

( ) ( )o 2 2cos30.0 15.0 m s cos30 13.0 m sca a= = ° =

2

cvar

= ( )2 2 2 22.50 m 13.0 m s 32.5 m scv ra= = =

32.5 m s 5.70 m sv = =2 2 2t ra a a= +

( ) ( )2 22 2 2 2 215.0 m s 13.0m s 7.50 m s= − = − =t ra a a

“Coriolis Force”

• This is an apparent force caused by changing the radial position of an object in a rotating coordinate system

• The result of the rotation is the curved path of the ball

“Coriolis Force”

• This is an apparent force caused by changing the radial position of an object in a rotating coordinate system

• The result of the rotation is the curved path of the ball

Coriolis Effect

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A ball rolls around a circular track with an angular velocity of 4π rad/s. What is the period of the motion?

A. s

B. 1 s

C. 2 s

D. s

E. s

QuickCheck 4.11

Slide 4-88

12

1

2π

1

4π

© 2013 Pearson Education, Inc.

A ball rolls around a circular track with an angular velocity of 4π rad/s. What is the period of the motion?

A. s

B. 1 s

C. 2 s

D. s

E. s

QuickCheck 4.11

Slide 4-89

12

1

2π

1

4π

T =

2πω

© 2013 Pearson Education, Inc.

The fan blade is slowing down. What are the signs of ω and α? A. ω is positive and α is positive.

B. ω is positive and α is negative.

C. ω is negative and α is positive.

D. ω is negative and α is negative.

E. ω is positive and α is zero.

QuickCheck 4.15

Slide 4-105

© 2013 Pearson Education, Inc.

The fan blade is slowing down. What are the signs of ω and α? A. ω is positive and α is positive.

B. ω is positive and α is negative.

C. ω is negative and α is positive.

D. ω is negative and α is negative.

E. ω is positive and α is zero.

QuickCheck 4.15

Slide 4-106

“Slowing down” means that ω and α have opposite signs, not that α is negative

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s angular velocity is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.12

Slide 4-95

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s angular velocity is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.12

Slide 4-96

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s speed is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.13

Slide 4-97

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s speed is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.13

Slide 4-98

v = ωr

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s acceleration is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.14

Slide 4-99

© 2013 Pearson Education, Inc.

Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s acceleration is ______ that of Rasheed.

A. half

B. the same as

C. twice

D. four times

E. We can’t say without knowing their radii.

QuickCheck 4.14

Slide 4-100

Centripetal acceleration a = v2

r= ω 2r