UNIFORM CIRCULAR MOTION

An object traveling with a constant speed in a circle is accelerating because the object’s velocity is changing in direction. The object is said to have “uniform” circular motion and it is undergoing centripetal acceleration (centrum- centre, petere- to seek (Newton) p. 123-124).

1v

2

v

1r

2

r

1v

2v

v

1r

2

r

r

These two triangles are similar and for any time interval this would be so.

r

rvv

r

r

v

v

ct

at

v

0lim

This proof is not on any test.

t

r

r

va

tr

rva

tc

tc

0

0

lim

lim

This is velocity!

For the final equation remove the absolute value notation.

r

va c

2

[towards the center]

Notice on the first diagram that half way through the interval v is pointing at the center of the circle.

This is an instantaneous acceleration so the direction of the acceleration is constantly changing.

For uniform circular motion . . .

T

rv

2 T is the period of revolution.

2

24

T

rac

224 rfac f is the frequency

rememberT

f1

An increase in the velocity of an object that is centripetally accelerating (with no change to r) is called tangential ac-celeration. A parallel push causes tangential acceleration and velocity’s magnitude changes. A perpendicular push causes centripetal acceleration and only the direction of velocity changes.

What is the centripetal acceleration of a stone being whirled in a circle, at the end of a 1.5 m string, on a smooth sheet of ice, with a frequency of 1.25 Hz?

2

22

22

53.92

)25.1)(5.1(4

4

s

ma

Hzma

rfa

c

c

c

The planet Mercury moves in an approximately circular path at an average distance of 5.8 x 1010 m, accelerating centripetally at 0.04 m/s2. What is its period of rotation about the Sun?

sT

sm

mxT

a

rT

T

ra

cc

6

2

1022

22

2

2

10566.7

04.0

)108.5(4

44

CENTRIPETAL FORCE

Centripetal acceleration (center seeking) can be extended to centripetal force. Remember centripetal acceleration is the observed motion while centripetal force is responsible for this acceleration. Finally, centripetal force is always a resultant force.

][2

centertowardr

mvFc

T

rv

2

2

24

T

mrFc

224 mrfFc

This direction is constantly changing.

textbook: p. 133 2-8, p. 138 2-7, p. 159 8, 11-13, 26-28

A 1000 kg car enters a level curve at 20 m/s. If the curve has a radius of 80 m, what centripetal force must be supplied by friction to keep the car from skidding? FN

Fg

Ff

NFm

sm

kgF

Fr

mv

FF

FFFF

f

f

f

fc

fgNnet

500080

)20(1000 2

2

Notice this would give a s of

0.5

A 1000 kg ball travels around a frictionless, banked curve having a radius of 80 m. If the bank is 20o to the horizontal, at what speed must the ball travel to maintain a constant radius?

20o

NF

NyF

NxF

gF

gNR FFF

s

mv

mkgNv

mgm

mv

Fr

mv

FF

o

o

Nyo

Nxc

90.16

)80)(/81.9(20tan

)(20tan80

)(20tan

2

2

2

A 3.5 kg steel ball is swung at a constant speed in a vertical circle of radius 1.2 m on the end of a light, rigid steel rod. If the ball has a frequency of 1.0 Hz, calculate the tension in the rod at the top and bottom of the circle.

top bottom

+ is up

gF

gF

TF

TF

gTR FFF

NF

mgmrfF

mgFmrf

FFF

bottom

T

T

T

gTc

1.200

4

422

22

NF

mgmrfF

mgFmrf

FFF

top

T

T

T

gTc

5.131

4

422

22

LAW OF UNIVERSAL GRAVITATION

221

r

mGmFG

FG -gravitational force (in 2 directions)

G -universal gravitation constant 6.67x10-11 Nm2kg-2

r -distance between the objectsm1 -mass of the larger object

near the earth’s surface . . .

21

221

2

r

Gmg

r

mGmgm

This equations could be applied to the surface of any planet or to the acceleration you would experience at any distance from on object.

What happens to the gravitational attraction between two particles if one mass is doubled, the other tripled and the distance between them cut in half?

221

2

21

1

2)

2(

32

rmGm

rmmG

F

F

241

2 F

F

read p. 139-142 p. 141 1-6 extra p. 143 8-13 p. 144 1-6

Planet X has a radius that is 3.5 times the radius of the earth and a mass that is 2.0 times the earth’s. Compare the acceleration due to gravity at the surface of each planet.

21

r

Gmg

ex

e

ex

e

ex

gg

r

Gmg

r

mGg

)5.3

2(

)5.3

2(

)5.3(

2

2

22

2

163.0e

x

g

g

SATELLITES

A satellite is an object or a body that revolves around another object, which is usually larger in mass.Planets, moons, space shuttles, space stations, comets, and “satellites” are satellites.Satellites remain in a constant orbit because they are acted upon by a centripetal force and display centripetal acceleration.

r

Gmv

r

vm

r

mGm

FF cG

12

22

221

remember m1 is the larger mass and the central object

Calculate the orbital speed for a satellite at Earth’s surface, and two Earth radii above Earth’s surface. Show the simpler calculation for the satellite at Earth’s surface.

What is the period of rotation of the moon about the earth?

1

324

Gm

rT

kgkg

Nm

mT

242

211

382

1098.51067.6

)1084.3(4

sT 610367.2

dT 40.27

read 145-146 p. 151 1, 3-6 extra p. 147 2-4, 6 p. 160 14-20

3221

12

12

4

2

rTGm

r

Gm

T

r

r

Gmv

What is the radius of the orbit of a geosynchronous satellite?

32

21

4TGm

r 3

2

2242

211

4

)86400(1098.51067.6

skgkgNm

r

mr 810225.4

GRAVITATIONAL FIELDS

A force field exists in the space surrounding an object in which a force is exerted on objects (e.g. gravitational, electric, magnetic). The strength of gravitational force fields is deter-mined by the Law of Universal Gravitation. If two or more gravitational fields are acting on an object then the net field is the sum of all the individual fields.

read 274-275 p.276 2-6 p.277 1-8

KEPLER’S LAWS

In 1543 Copernicus proposes the heliocentric model of the solar system in which planets revolve around the sun in circular orbits. Slight irregularities show up over long periods of study.

Tycho Brahe takes painstaking observations for 20 years with large precision instruments but dies (1600) before he can analyze them properly. A young mathematician continues Brahe’s work.

From his analysis the kinematics of the planets is fully understood.

Kepler’s First Law of Planetary Motion

Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse.

Kepler’s Second Law of Planetary Motion

The straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of time.

Planets move faster when they are closer to the Sun (centripetal force is stronger).

equal areasequal times

orbits are elliptical but are not very elongated

Kepler’s Third Law of Planetary Motion

The cube of the average radius of a planet is directly proportional to the square of the period of the planet’s orbit.

We have already proved this a few slides back. Recall.

21

2

3

4Gm

T

r

3221 4 rTGm

constant

For our solar system m1 is the mass of the sun.

Mars’ average distance from the sun is 2.28 x1011 m while its period of rotation is 5.94 x 107 s. What is Jupiter’s average distance from the sun if its period of rotation is 3.75 x 108 s ?

2

3

21

2

3

4 m

m

J

J

T

rGm

T

r

mrJ111079.7

read 278-283

p. 283 10-12 p. 284 4-7, 9

GRAVITATIONAL POTENTIAL ENERGY, AGAIN

Recall the Law of Universal Gravitation

221

r

mGmFG

for constant masses, a graph of force vs. radius would be . . .

The graph above is a F vs. d graph which means the shaded area is the work required to move an object from r1 to r2.

The shaded area is not easy to calculate but can be done with a geometric mean. In this case the work done by the lifter is equal to Ep.

Another method involves calculus and integration over a range from r1 to r2.

2

21

1

21

1221

21

1222

212

1

21

1221

)(

)(

)(

r

mGm

r

mGmE

rrrr

mGmE

rrr

mGm

r

mGmE

rrFFW

p

p

p

geometric mean of force

r

mGmE

r

mGm

r

mGmE

p

p

21

1

21

2

21

Know these two equations, you are not required to know the previous development.

Which preceding equation can be simplified to mgh, the potential energy change near the earth’s surface?

Potential energy is a negative function!

It increases until it is zero.

PE stops here because the objects come into contact and cannot get closer.

Recall

)(22

22

12

21

22

2

22

1

21

mghmghmvmv

or

mghmv

mghmv

)(22

22

1

21

2

2121

22

2

2122

1

2121

r

mGm

r

mGmmvmv

or

r

mGmmv

r

mGmmv

so . . .

read p. 285-287 p. 287 1-5

Escape from a Gravitational Field

To escape a gravitational field an object must have at least a total mechanical energy of zero!!

PKM EEE for escape 0MEEscape energy - the minimum EK needed to project a mass (m2)

from the surface of another mass (m1) to escape the gravitational force of m1

Escape speed - the minimum speed needed to project a mass (m2) from the surface of another mass (m1) to escape the gravitational force of m1

Binding energy - the additional EK needed by a mass (m2) to escape the gravitational force of m1 (similar to escape energy but applies to objects that possess Ek i.e. satellites).

To calculate the escape energy or the escape velocity of a mass (m2): 0 PK EE

0 BPK EEETo calculate the binding energy of a mass (m2):

binding energy

Calculate the escape velocity of any object on the Earth’s surface.

0 PK EE

e

e

r

Gmv

r

mGmvm

12

212

2

2

02

m

kgkgNmv

6

242211

1038.6

)1098.5)(1067.6(2

s

mv 11182

The escape velocity is the same for all objects on the Earth’s surface while the escape energy is different for different massed object.

What is Ek and EM of an orbiting body (satellite)?

22

21

22

221

vmr

mGmr

vm

r

mGm

FF cG

this is always true of satellites

2

22

2221

pK

EE

vm

r

mGm

2

2

pM

pp

M

PKM

EE

EE

E

EEE

for an orbiting satellite !!

Note that the total energy is negative since the satellite is “bound” to the central body.

read p.288-293 p. 293 6-11 #12 is interesting!

extra p. 294 1-8 p. 300 1-17 25,26 look fun

a) What is the speed of Earth in orbit about the Sun?b) What is the total energy of Earth?c) What is the binding energy of Earth?d) If Earth was launched from the surface of the Sun to its present orbit then what velocity must it be launched with (Ignore the radius of Earth.)?e) If Earth came to rest and fell to the Sun then what velocity would it have when it hit the Sun (Ignore radius of Earth.)?

me= 5.98x1024 kg

ms= 1.99x1030 kg

re= 1.49x1011 m (of orbit)

rs= 6.96x108 m (of the body)

G= 6.67x10-11 Nm2kg-2

sT

T

rv

710156.3

2

s

mv 410966.2

a)

or 22

21 vmr

mGm

s

mv 410985.2

e

esM

pM

r

mGmE

EE

2

2

b) JEM

3310664.2

c) The binding energy is 2.664 x 1033 J

1

21

2

2121

2

21

1

2121

2

2122

1

2121

22

22

22

r

mGm

r

mGmmv

r

mGm

r

mGmmv

r

mGmmv

r

mGmmv

d)

s

mv

Jmv

51

3621

10169.6

10138.12

s

mv

Jmv

mv

r

mGm

r

mGm

r

mGmmv

r

mGmmv

52

3622

22

2

21

1

21

2

2122

1

2121

10161.6

10135.12

2

22

e)