Rotational MotionReading: pp. 194 – 203 (sections 8.1 – 8.3)

HW #1 p. 217, question #1

p. 219, problem #1, 4, 5, 6, 15, 16, 17, 19

I. Introduction:

A rotating object is one that spins on a fixed axis. The position and direction of the rotation axis will remain constant. The position of some part of the object can be specified with standard cartesian coordinates, (x,y). All objects will be assumed to rotate in a circular path of constant radius.

y

x

(x, y)

R

Both coordinates, (x, y), change over time as the object rotates.

The object’s position can also be specified with polar coordinates, (r, ).

For this last coordinate system, only the angle changes. The radius stays constant.

II. Definitions:

Since the polar coordinates only have one changing variable, the angle, we will use this to simplify analysis of motion.

A. The ___________ is defined as where an object is in space. Here, we only need to specify the angle of the object with respect to some origin or reference line.

position

The position of the object is measured by an angle, , measured counterclockwise from the positive x – axis.

The angle will be measured in

units of ___________ rather than

___________ .

radians

degrees

Radians are defined as the ratio of the length along the arc of a circle to the position of an object divided by the radius of the circle.

Rs

R

s

s = length along the arc measured counterclockwise from the +x – axis.

R = radius of the circular path.

Since is the ratio of two lengths, the angle measurement really does not have any units. The term “radians” is just used to specify how the angle is measured.1 revolution = 360 degrees = 2 radians

B. The _____________ of the object is just the difference in its position.

We define the ____________________ of the object as the difference in

its angular position.

displacement

angular displacement

oif

A positive shows a counterclockwise {ccw} rotation, while a negative shows a clockwise {cw} rotation.

C. Motion can also be measured through a rate of rotation, a __________.velocity

The ____________________ of an object is defined as the amount of rotation of an object per time. The angular velocity is represented by the greek letter (lower case omega).

angular velocity

average angular velocity:

tto

t = elapsed time. s

radunits

D. Another measure of motion is the rate of change of velocity, called an

________________.acceleration

The ____________________ of an object is defined as the amount of change of the angular velocity of an object per time. The angular acceleration is represented by the greek letter (lower case alpha).

angular acceleration

average angular acceleration:

tto

In general, the motion may be complex, but we will again look at constant angular acceleration cases, exactly the same way as we did back in Ch. 2. The same equations of motion can be derived for circular motion.

2s

radunits

Ch. 2 Ch. 8

Linear Motion Rotational Motion

atvv o to

221 attvx o 2

21 tto

xavv o 222 222o

t

xvv o

2 t

o

2

This is only true for constant accelerations.Problem solving is the same as before.

E. The motion of an object around a circle can also be represented as actual distances along the circle and speeds tangent to the circle.

The _________________ , vt, of an object is defined as the angular velocity times the radius of the circle.

tangential speed

Rvt

The tangential speed measures the actual speed of the object as it travels around the circle.

R vt

average vt =length along arc

elapsed time

The _________________________ , at, of an object is defined as the angular acceleration times the radius of the circle.

tangential acceleration

Rat

The tangential acceleration measures how the tangential speed increases or decreases over time.

Example #1: A disk rotates from rest to an angular speed of 78.00 rpm in a time of 1.300 seconds. a. What is the angular acceleration of the disk?

o = 0, = 78.00 revolutions per minute, t = 1.300 seconds. = ?

to

sts

rado

300.1

0168.8

sradrev

rev

rad168.8

sec60

min1

1

200.78 min

2283.6s

rad

b. Through what angle does the disk turn?

221 tto

2

21 300.1283.6300.10 2 ss

srad

srad

rad309.5

c. Through what angle will the disk turn if it were to maintain the same angular acceleration up to 254.0 rad/s?

222o

2283.62

00.254

2

2222

srad

srad

o

rad5134

d. The disk has a diameter of 12.00 inches. What is the tangential speed and acceleration at the edge of the disk the moment the disk reaches 78.00 rpm?

mcm

m

in

cminR 1524.0

100

1

1

54.2

2

00.12

sm

srad

t mRv 245.1168.81524.0

22 9576.0283.61524.0s

ms

radt mRa

Centripetal Force

HW: See Schedule

III. Centripetal Acceleration and Force.

When an object moves in a circle, the direction of its velocity is always changing. This means the object is always accelerating! For an object rotating at a constant angular speed, the acceleration of the mass is always towards the center of the motion. This kind of acceleration is called a ___________ {“center seeking”} acceleration, and is represented as ac. The amount of acceleration depends on the radius of the circular path and the speed around the circle.

centripetal

rr

vac

22

= angular speed, v = tangential speed, r = radius of circular path.

Example #2: A wheel of a car has a diameter of 32.0 inches. A rock is wedged into the grooves of the tire. a. What is the centripetal acceleration on the rock if the wheel turns a rate equal to 70.0 mph?

sm

hourmile

s

hour

mile

mv 29.31

3600

1

1

344.16090.70

mcm

m

in

cminr 4064.0

100

1

1

54.2

2

0.32

22410

4064.0

29.31 22

sms

m

c mr

va

b. What is the rotation rate of the tire, in rpm?

srads

m

mr

v0.77

4064.0

29.31

rpms

rad

revs

rad 735min1

60

2

10.77

If there is a centripetal acceleration making an object move in a circle, then there must be an unbalanced force creating this acceleration.

This force is called the _____________ force, and it also points towards the center of the circular motion. This force must be made by real forces acting on an object. The centripetal force will be the sum of the radial components of the forces acting on the object. A radial component points towards the ___________ of the circle.

centripetal

rmr

mvmaF cc

22

center

Example #3: A 0.475 kg mass is tied to the end of a 0.750 meter string and the mass is spun in a horizontal circle. If the mass makes 22.0 revolutions in a time of 2.50 seconds, what is the tension in the string holding the mass to the circular motion?

Rm

T

srad

rev

rad

s

rev

t3.55

1

2

50.2

0.22

mkgrmF srad

c 750.03.55475.0 22

NFc 1090

rotation

Centripetal Force

HW #2: See Schedule

Last Chance for Make-up Tests!

Wednesday after school!

Example #4: A penny sits at the edge of a 12.00 inch diameter record. If the coefficient of static friction is 0.222 between the penny and the record, what is the maximum rotation rate of the record that will allow the penny to remain on the record?

n

mg

Fs

rotationvertical forces balance.

mgn by definition friction is:

nF ss max,

mgF ss max,

set the friction force equal to the centripetal force

mgFrmF ssc max,2

inm

inr

g sm

s

10254.0

00.6

80.9222.0 2s

rad78.3

Example #5: A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. Find the speed which m must move in order for M to stay at rest. Evaluate the speed for m = 2.00 kg,M = 15.0 kg, and r = 0.863 m.

Since M is at rest, the tension force lifting it is equal to the weight of M:

MgT

This tension is also the centripetal force on the mass m, causing it to spin in a circular path:

r

mvT

2

Set the two equations equal to one another:

Mgr

mvT

2

m

rMgv

kg

kgmv s

m

00.2

80.90.15863.0 2

smv 96.7

Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room.

Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room.

m

mg

Fs

n

set the friction force equal to the weight as the vertical forces balance.

mgFs max,

nF ss max,

Horizontal direction: set the normal force equal to the centripetal force.

rmFn c2

forces balance

definition of force of static friction

Combine all the information to solve for the rotation rate, .

ss

s mgFnrm

max,2

r

g

s 2

r

g

s

What would be the rotation rate for a room 3.00 m wide and a carpeted wall with a coefficient of friction of 0.750?

rpm

ms

radsm

2.2895.250.1750.0

80.9 2

Example #7: (Banking Angle) Determine the angle of the roadway necessary for a car to travel around the curve without relying on friction. Assume the speed of the car and the radius of the curve are given.

mg

n

cosn component of the normal force that is vertical

sinn component of the normal force that is horizontal, also becomes the centripetal force

balance the vertical forces: mgn cos

set the net horizontal force equal to the centripetal force, with towards the center of the circular path as the positive direction:

R

mvn

2

sin substitute in:cos

mgn

R

mvmg 2

sincos

tan2 Rgv Rg

v21tan

Example #8: Conical Pendulum.A mass of m = 1.5 kg is tied to the end of a cord whose length is L = 1.7 m. The mass whirls around a horizontal circle at a constant speed v. The cord makes an angle = 36.9o. As the bob swings around in a circle, the cord sweeps out the surface of a cone. Find the speed v and the period of rotation T of the pendulum bob.

mgT cos

R

mvT

2

sin

divide…

mgR

mv

T

T 2

cos

sin

note:L

Rsin

simplify…

gR

v2

tan tanRgv

tansin gLv

os

momv 9.36tan80.99.36sin7.1 2

smv 74.2

the period, T, is the time for one revolution:

T

Rv

2

v

L

v

RT

sin22

sm

omT

74.2

9.36sin7.12

sT 34.2

Example #9: Another common amusement park ride is a rollercoaster with a loop. Determine the minimum speed at the top of the loop needed to pass through the top of the loop.

There are two forces acting on the car: the weight pulling straight downwards and the normal force pushing perpendicular to the track.

As the car goes through the loop, the normal force always points towards the center of the loop. The vector sum of the normal force and the radial component of the weight equals the centripetal force.

The net force for the mass at the top of the loop is:

m

mg

n

mgnFr

mvF netc

2

The faster the car goes, the greater the normal force to push the car into a circular path. The minimum speed for the car at the top of the loop is where the normal force goes to zero.

mgr

mv

2min

rgv min

r

Example #9: (b) Use energy conservation to find the speed of the mass at the bottom of the loop.

PEKEEtop

PEKEEbottom

rmgmvEtop 22min2

1

0221 botbottom mvE

Set the two energies equal:

rmgmvmvbot 22min2

1221 Divide by m and multiply by 2:

rgrgrggrvvbot 5442min

2 rgvbot 5

Example #9: (c) What is the necessary starting height of the mass if sliding from rest down the ramp?

PEKEE looptop PEKEE ramptop

mghE ramptop 0

rmgmvE looptop 22min2

1

Set the two energies equal:

mgrmgrrgmmgrmvmgh 25

212

min21 22

Divide both sides by m and g: drh 45

25

AP Summary Problems!

Example #10: {p. 135, problem #93} A circular curve of radius R in a new highway is designed so that a car traveling at speed vo can negotiate the turn safely on glare ice (zero friction). If the car travels too slowly, then it will slip toward the center of the circle. If it travels too fast, then it will slip away from the center of the circle. If the coefficient of static friction is increased (from zero), a car can stay on the road while traveling at any speed within a range from vmin to vmax. Derive formulas for vmin and vmax as functions of s, vo, g and R.

Start with the too fast scenario. The car will want to slide up the ramp, so friction will point to the bottom of the ramp.

Find horizontal and vertical components of the forces:

cosn

sinn

cosfF

sinfF

Vertical components of the forces balance:

cos sinfn mg F

Definition of friction: f sF n

cos n sinsn mg

cos sins

mgn

cos sins

mgn

Horizontal components of the forces make the centripetal force. Take towards the left as the positive direction:

2

sin cosf

mvn F

r

f sF n

2

sin n coss

mvn

r

2

sin coss

mvn

r

2

sin coscos sin s

s

mg mv

r

2 sin cos

cos sins

s

v rg

Next is the too slow scenario. The car will want to slide down the ramp, so friction will point to the top of the ramp.

Find horizontal and vertical components of the forces:

cosn sinn

cosfF

sinfF Vertical components of the forces balance:

cos sinfn F mg

Definition of friction: f sF n

cos n sinsn mg

cos sins

mgn

cos sins

mgn

Horizontal components of the forces make the centripetal force. Take towards the left as the positive direction:

2

sin cosf

mvn F

r

f sF n

2

sin n coss

mvn

r

2

sin coss

mvn

r

2

sin coscos sin s

s

mg mv

r

2 sin cos

cos sins

s

v rg

Example #11: A mass is tied to the end of a string and spun in a vertical circle. What is the difference in tension in the string between top and bottom of the circle?

bottomT

mg

topTmg

2top

top

mvmg T

r

2bottom

bottom

mvT mg

r

2 21 12 2 2bottom topmv mv mg r

energy conservation:

2bottom

bottom

mvT mg

r

2bottom

bottom

mvT mg

r

2top

top

mvmg T

r

2top

top

mvT mg

r

subtract:

22topbottom

bottom top

mvmvT T mg mg

r r

2 2

2bottom topbottom top

mv mvT T mg

r

now use energy conservation to eliminate speeds

2 2

2bottom topbottom top

mv mvT T mg

r

2 21 12 2 2bottom topmv mv mg r

2 2 2 2bottom topmv mv mg r

42bottom top

mgrT T mg

r

6bottom topT T mg

Centripetal Force

HW check tomorrow!

Example #12: Sonam is seated on the top of a frictionless hemispherical mound of ice of radius R. Why she is there, I do not know. A small breeze upsets the equilibrium and she starts sliding down the ice. At what vertical height from the ground does she leave the ice surface? Give the answer in terms of R.

h =

R c

os

n

mg

cosmg

When Sonam is at the side, as shown, the sum of the radial components gives the centripetal force:

2

cosmv

mg nR

Sonam leaves the surface when the normal force becomes zero.

0

2 cosmv mgR 2mv mgh

Now use energy conservation to find the speed of Sonam if she starts at rest at the top.

topE mgR

212side sideE mv mgh

From the previous slide:

2mv mgh

312 2mgR mgh mgh mgh

23h R

Newton’s Universal Law of Gravity

HW #3 on schedule

Turn in Rotary Motion Lab

Newton’s Law of Universal Gravity:

Any two objects are attracted to each other through the force of gravity.

The force is proportional to each mass. 1mF 2mF

The force is inversely proportional to the square of the center to center distance between the masses.

2

1

rF “inverse square law”

These two equations can be combined intoa single equation, and a proportionalityconstant can be introduced to makean equality:

221

r

mGmF

By Newton’s 3rd Law, forces are equal and opposite!

2

2111067384.6

kg

mNG

Example #1: Two students, each with mass 70 kg, sit 80 cm apart.(a) What is the force of attraction between the two students?

kgmm 7021

mcmr 80.080

221

r

mGmF

2

2

211

80.0

70701067384.6

m

kgkgkg

mN

NF 7101.5

(b) How does this compare to the weight of the students?

280.9701 smkggmweight

Nweight 686

107

107686

101.5

N

N

weight

F

The force between you and your neighbor is less than 1 part in a billion of your normal body weight.

Example #2: What is the direction of the net force on the mass at the center of the image? Why?

All forces balance out pairwise except for one:

Force on central mass points upwards.

Example #3: What is the direction of the net force on the mass at the center of the image? Why?

All forces balance out pairwise except for one:

Force on central mass points towards the left.

II. Compare the force of gravity to weight:

The weight of an object near the surface of the Earth is just the force of gravity exerted on the object by the Earth.

2E

E

R

mGMmg

Remember, r is the center to center (of the earth) distance.

kgM E241098.5 mRE

610378.6

The mass of the object cancels out, and what is left is a theoretical value for the acceleration due to gravity near the surface of the Earth.

2E

E

R

GMg

Example #4: Estimate the value of ‘g’ for the Earth.

2E

E

R

GMg

26

242

211

10378.6

1098.51067384.6

m

kgkg

mN

281.9s

mg

Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.

Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.

mwidth 4.1

Estimate volume as sphere 1.4 m in diameter.

370.03

4mvolume

344.1 m

Multiply by density to get mass:

kgmmm

kg 129344.1900 33 !2800 lbs

Estimate ‘g’ at surface:

2R

GMg

2

2

211

70.0

12931067384.6

m

kgkg

mN

27108.1

smg

III. Satellite Motion.

A satellite is an object that orbits (travels in a circular path around) some gravity source

The force of gravity provides the centripetal force to keep the satellite moving in a circular path.

The equation of motion is:

2

2

r

GMm

r

mv

m = satellite mass (divides out…)

M = mass of the gravity source

r = radius of the orbit (center to center distance!)

v = the tangential speed of the orbiting object

Example #6: The space shuttle orbits the Earth at an altitude of 400 km above the surface of the Earth. (a) Determine the speed of the space station around the Earth.

2

2

r

GMm

r

mv

r

GMv

mm

kgkg

mN

56

242

211

1000.410378.6

1098.51067384.6

mmr 56 1000.410378.6

smv 673,7

(b) How much time will it take for the space station to orbit the Earth?

v

r

rate

disttimeT

2

sm

mT

7673

10778.62 6

min5.925550 sT

Example #7: Write an equation that gives the period (time for one revolution) of an orbit in terms of the radius of the orbit.

2

2

r

GMm

r

mv

T

rv

2

r

GMv

T

r

2

22

22

3

4GM

T

r

Example #8: Determine the altitude (height above the surface) for a geosynchronous satellite. The period of the satellite matches the length of the Earth’s day, 24 hours.

22

23

24

T

GMTGM

r

3

2424

2

211

2

1064.81098.51067384.6

s

kgkg

mNr

sh

shT 41064.8

1

360024

3

2424

2

211

2

1064.81098.51067384.6

s

kgkg

mNr

mr 71023.4

This is the height from the center of the Earth.

Subtract the radius of the Earth to get the height from the surface.

mmh 67 10378.61023.4

mh 71059.3

Example #9: {on your own} Determine the orbital velocity of Grog’s golf ball.

66.378 10Earthr R m

Example #9: Determine the orbital velocity and period of Grog’s golf ball.

2

2

r

GMm

r

mv

GMv

r

211 24

2

6

6.67384 10 5.98 10

6.378 10

N mkg

kg

m

7,910 msv

Example #10: What is your perceived weight when you stand at the equator of the Earth?

Example #10: What is your perceived weight when you stand at the equator of the Earth?

m

mg

n

r

rota

tion2m r mg n

Your perceived weight is the normal force pushing on your feet to support you.

2n mg m r Some of the true gravity force is used to make you move in a circle. The remainder of that force is what you feel as your weight.

The percentage your weight is reduced is:

2m r

mg

2

2

626.378 10

24 3600

9.80

sh

ms

radm

h

0.003

0.3%

Example #11: {on your own!} What is maximum rotation rate for the Earth so that objects at the equator will stay on the surface and not “fly off”?

Hint: The normal force goes to zero at this maximum rotation rate. The objects are essentially in a low orbit.

2m r mg n 0

g

r 2

6

9.80

6.378 10

ms

m

31.24 10 rad

s

2T

5070 84.5mins

Potential Energy and Newton’s Law of Gravity

?

Newton’s Law of Universal Gravity:

Any two objects are attracted to each other through the force of gravity.

221

r

mGmF

2

2111067384.6

kg

mNG

This force also stores energy between the two masses. The potential energy stored in the two masses is:

1 2Gm mPE

r

Example #12: What is the minimum speed that a spaceship at the surface of the Earth must have to completely escape Earth’s gravity?

m m

Total mechanical energy is conserved.

surfaceE KE PE far awayE KE PE 0

2 21 12 2surface far awaymv PE mv

0The least speed at the surface corresponds to the least speed far away.

212 0Earth

surface

GM mmv

r

212 0Earth

surface

GM mmv

r

2 Earthsurface

GMv

r

211 24

2

6

2 6.67384 10 5.98 10

6.378 10surface

N mkg

kgv

m

41.12 10 11.2m kmsurface s sv

Kepler’s Laws of Satellite Motion

Handout

HW #4

IV. Kepler’s Laws.

Johannes Kepler was a German mathematician, astronomer and astrologer. He is best known for his eponymous laws of planetary motion, which he was able to put together by painstakingly analyzing the volumes of data collected by Tycho Brahe of the planets motions in the sky.

Kepler

Brahe

Kepler’s 1st Law:

All planets (satellites) orbit the sun (or gravity source) in elliptical orbits.

The sun sits at one focus of the ellipse. The planet will be closer to the sun for part of its “year”, and farther away for the other part of its “year”.

Rp = perihelion = closest distance of planet to sun

Ra = aphelion = farthest distance of planet to sun

Kepler's first law. An ellipse is a closed curve such that the sum of the distances from any point P on the curve to two fixed points (called the foci, F1 and F2) remains constant. That is, the sum of the distances, F1P + F2P, is the same for all points on the curve. A circle is a special case of an ellipse in which the two foci coincide, at the center of the circle.

Kepler’s 2nd Law:

As planets orbit around the sun, an imaginary line from the planet to the sun will sweep out equal areas in equal times.

Kepler's second law. The two shaded regions have equal areas. The planet moves from point 1 to point 2 in the same time as it takes to move from point 3 to point 4. Planets move fastest in that part of their orbit where they are closest to the Sun. Exaggerated scale.

Potential energy from gravity still has the same behavior on this scale:

The farther the planet moves from the sun, the higher the potential energy of the planet. But total energy is still constant:

constPEKEE

As the planet moves farther from the sun, it slows down. The closer the planet to the sun, the faster it moves in its orbit.

Measurable Application: Earth’s weather is due to the tilt of the Earth’s axis relative to its orbital plane.

Earth at perihelion Earth at aphelion

Since summer (for the northern hemisphere) occurs when Earth is farthest from the sun, the Earth spends slightly more time in this part of the orbit as compared to winter (again, for northern hemisphere). The effect is that summer is about 3 days longer than winter. This is sometimes called the “summer analemma”.

The analemma is the location of the sun in the sky at the same time each day of the year plotted on a graph or photograph. The uneven shape is due to the elliptical orbit of the Earth around the sun.

Kepler’s 3rd Law:

The ratio of the cube of the length of the semi major axis, to the square of the period of revolution, is constant for all planetary satellites.

3

2

rK

T

r = length of semi major axis.

T = period of orbit.

3

2 24

r GM

T Newton later figured out how this

constant fit into his gravitational scheme.

Example #13: Earth’s orbital information is sometimes used as a standard: The orbital period is one year and its semimajor axis is one AU (astronomical unit). If the orbital period for Jupiter is 11.9 years, determine the semimajor axis distance for Jupiter.

3

2 24

r GMconst

T

2

3

2

3

E

E

J

J

T

a

T

a

32

32

2

E

JE

E

JEJ T

Ta

T

Taa

3

2

00.1

9.1100.1

y

yAUaJ

AUaJ 21.5

Example #14: Earth’s orbit has a semimajor axis length of 1.50×108 km and a period of 365.24 days. Determine the mass of the sun.

3

2 24

r GM

T

2 3

2

4 rM

GT

22

211

3112

36002424.3651067384.6

1050.14

hs

dhd

kgmN

mM

kgM 301000.2

Example #15: A satellite moves in a circular orbit around Earth at a speed of 5,000 m/s. Determine (a) the satellite’s altitude above the surface of Earth and (b) the period of the satellite’s orbit.

2

2

r

GMm

r

mv

2

GMr

v

211 24

2

2

6.67384 10 5.98 10

5000 ms

N mkg

kgr

71.60 10r m subtract the radius of Earth to get the altitude:

69.59 10h m

How much time will it take for the space station to orbit the Earth?

v

r

rate

disttimeT

2

72 1.596 10

5000 ms

mT

42.01 10 334min 5.57T s hours

Example #16: Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 × 105 km. From this data, determine the mass of Jupiter.

3

2 24

r GM

T

2 3

2

4 rM

GT

32 8

2211

2

4 4.22 10

6.67384 10 1.77 24 3600h sd h

mM

N md

kg

271.90 10M kg

Example #17: Assume the Earth has a uniform density of . Determine the strength of Earth’s gravitational field as a function of its radius for the interior of the Earth.

density is mass per unit volume! 3 34

3

3

4

M M M

V R R

When inside a spherically symmetric object, the pull of gravity at that point depends only on the amount of mass contained in a ball whose radius matches that location. The shell of mass on the outside of that position does not contribute to the pull of gravity.

2insideGM

g rr

Determine the mass located inside:

33

3 4

4 3inside

MM V r

R

3

3inside

rM M

R

Plug into the formula for the acceleration due to gravity:

2insideGM

g rr

3

2 3

G rM

r R

3

GMr

R