Lecture 13

General Physics (PHY 2130)

•  Rotational kinematics   Non-uniform circular motion   Orbits and Kepler’s laws

http://www.physics.wayne.edu/~apetrov/PHY2130/

Lightning Review

Last lecture: 1.  Rotational kinematics

  angular displacement, angular velocity and angular acceleration   relations between angular and linear quantities

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

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

v

X

v = total distancetotal time

=2πr( )NT( )N

= 2πrT

=ωr

= 101 rads/sec( ) 32.5 cm( )= 3.28×103cm/sec =118 km/hr

Given:

d = 65 cm = 0.65 m

ω = 101 rads/sec

Find:

v = ?

Idea: since the car is moving with constant velocity, divide total path to total time!

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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

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Circular Orbits

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

The only force on the satellite is the force of gravity:

rvm

rMGm

amFrMGmFF

ses

rs

esg

2

2

2

=∴

=

==

∑

∑

the speed of the satellite: r

GMv e=

Escape Speed  The escape speed is the speed needed for an object to soar off into space and not return

 For the earth, vesc is about 11.2 km/s  Note, v is independent of the mass of the object

vesc =2GME

RE

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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?

rGMv e=Previously: Also need,

Trv π2

=

Combine these and solve for r: 31

224

⎟⎠

⎞⎜⎝

⎛= TGMr e

π

( )( )( )

m 10225.4

s 864004

kg 1098.5/kgNm1067.6

7

31

22

242211

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××=

−

πr

km 000,35=−=⇒+= ee RrhhRr

Given:

T = 24 hours

= 86400 s

Find:

h = ?

Idea: relate velocity of a satellite to both r and T!

This orbit is called geosynchronous and is used for TV satellites!

Kepler’s Laws

 All planets move in elliptical orbits with the Sun at one of the focal points.

 A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.

 The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

Kepler’s First Law

  All planets move in elliptical orbits with the Sun at one focus. • Any object bound to

another by an inverse square law will move in an elliptical path

• Second focus is empty

Kepler’s Second Law

 A line drawn from the Sun to any planet will sweep out equal areas in equal times • Area from A to B and C to

D are the same

Kepler’s Third Law  The square of the orbital period of any planet is proportional

to cube of the average distance from the Sun to the planet.

•  For orbit around the Sun, KS = 2.97x10-19 s2/m3 •  K is independent of the mass of the planet

32 KrT =31

224

⎟⎠

⎞⎜⎝

⎛= TGMr e

πor

It can be generalized to: 31

224

⎟⎠

⎞⎜⎝

⎛= TGMrπ

where M is the mass of the central body. For example, it would be Msun if speaking of the planets in the solar system.

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Example: The Hubble Space Telescope orbits Earth 613 km above Earth’s surface. What is the period of the telescope’s orbit?

Given:

h = 613 km

= 6.13 × 105 m

Find:

T = ?

F∑ = Fg =GmHsMe

r2 =mHSar =mHSv2

r

Need v! To find it, write 2nd Newton’s law:

T = 2πrv

Idea: period T is the time needed to

complete one revolution, i.e. a path of

length s = 2πr! So,

⇒ T = 2π r3

GMe

= 2π (613×103m+ 6371×106m)3

(6.674×10−11Nm2 / kg2 )(5.974×1024kg) T = 5808s =1.613h

Or with numbers (and remembering that r = RE+h):

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13

Tangential and Angular Acceleration Recall: average angular acceleration:

av ttif

Δ

Δ=

Δ

−=

ωωωα

SI unit of α is rads/sec2.

instantaneous angular acceleration:

tt Δ

Δ=

→Δ

ωα

0lim

tangential acceleration:

tangential component of velocity: ωrvt =

α

ω

ra

tt

rtv

a

t

tt

=

→ΔΔ

Δ=

Δ

Δ=

)0limit (in the

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

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Example: A child pushes a merry-go-round from rest to a final angular speed of 0.50 rev/s w i t h c o n s t a n t a n g u l a r acceleration. In doing so, the child pushes the merry-go-round 2.0 revolutions. What is the angular acceleration of the merry-go-round?

xavv

tatvx

tvvx

tavvv

xixfx

xix

ixfx

xixfxx

Δ=−

Δ+Δ=Δ

Δ+=Δ

Δ=−=Δ

221

)(21

22

2

αω rarv tt == ,

θαωω

αωθ

ωωθ

αωωω

Δ=−

Δ+Δ=Δ

Δ+=Δ

Δ=−=Δ

221

)(21

22

2

if

i

if

if

tt

t

t

Solution: given final and initial angular velocity and angle. Find acceleration.

ωf2 −ωi

2 = 2αΔθ , so α =ωf

2 −ωi2

2Δθ

=(0.50 rev s)2 −0

2(2.0 rev)2π rad

rev

#

$%

&

'(= 0.39 rad s2 .

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Example: A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2 π rad/s. After the disk rotates through 10π radians, the angular speed is 7π rad/s. (a) What is the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through 10π radians? (c) What is the tangential acceleration of a point located at a distance of 5.0 cm from the center of the disk?

xavv

tatvx

tvvx

tavvv

xixfx

xix

ixfx

xixfxx

Δ=−

Δ+Δ=Δ

Δ+=Δ

Δ=−=Δ

221

)(21

22

2

αω rarv tt == ,

θαωω

αωθ

ωωθ

αωωω

Δ=−

Δ+Δ=Δ

Δ+=Δ

Δ=−=Δ

221

)(21

22

2

if

i

if

if

tt

t

t

2 2 2 222 2 f i

f i(7 rad s) (2 rad s)2 , so 7.1 rad s .

2 2(10 rad)ω ω π π

ω ω α θ αθ π

− −− = Δ = = =

Δ(a)

(b) f if i

1 2 2(10 rad)( ) , so 2.2 s .2 7 rad s 2 rad s

t t θ πθ ω ω

ω ω π πΔ

Δ = + Δ Δ = = =+ +

(c) 2 2

2t

(7 rad s) (2 rad s)(0.050 m) 0.35 m s2(10 rad)

a r π πα

π−

= = =

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Artificial Gravity

A large rotating cylinder in deep space (g≈0).

Nautilus-X (proposed space station)

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x

y

N

x

y N

FBD for person at the bottom position

FBD for person at the top position

∑ === rmmaNF ry2ω ∑ −=−=−= rmmaNF ry

2ω

Apply Newton’s 2nd Law to each:

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Example: 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

====

===∑

rg

mrmg

mrN

rmmaNF ry

ω

ω

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

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Apparent weight

What would happen on Earth?

Our weight would change!

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

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

FBD for the car at the top of the loop:

N w

y

x

Apply Newton’s 2nd Law:

rvmwN

rvmmawNF ry

2

2

=+

−=−=−−=∑

r

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The apparent weight at the top of loop is: ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=+

grvmN

rvmwN

2

2

N = 0 when

grv

grvmN

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−= 0

2

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

Example continued:

The car will just about fall if N = 0!

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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: Apply Newton’s 2nd Law:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=−

==−=∑

grvmN

rvmwN

rvmmawNF cy

2

2

2

Here, mgN >

Example continued:

Forces in Accelerating Reference Frames

 Distinguish real forces from fictitious forces  Centrifugal force is a fictitious force  Real forces always represent interactions between objects

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Summary

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

• Radial Acceleration ar=v2/r

• Definition of Angular Quantities (θ, ω, and α)

• The Angular Kinematic Equations

• The Relationships Between Linear and Angular Quantities

• Uniform and Nonuniform Circular Motion

αω rarv tt == and