Rotational Motion

Handout

HW #1 & 2

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. 7

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

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

Handout

HW #3

Turn in Momentum Lab

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

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 radial component of the vector sum of the normal force and the weight equal the centripetal force.

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.

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

Newton’s Universal Law of Gravity

Handout

HW #4

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.

mheight 8.1

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

Kepler’s Laws of Satellite Motion

Handout

HW #5

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.

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.

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.

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 square of the period of the orbit of a planet is proportional to the cube of the semimajor axis of the orbit of the planet.

22

3

4GM

T

a

a = length of semimajor axis.

T = period of orbit.

Note: This was derived for circular orbits in example #7. The derivation for elliptical orbits is beyond the scope of the class…

Example #9: 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.

constGM

T

a

22

3

4

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 #10: 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.

22

3

4GM

T

a

2

324

GT

aM

22

211

3112

36002424.3651067384.6

1050.14

hs

dhd

kgmN

mM

kgM 301000.2

http://www.youtube.com/watch?v=6TGCPXhMLtU&edufilter=CPhKmRWIMeFn2PCPAsDuNg&safe=active

http://www.youtube.com/watch?v=KTaIPz6oC9U&edufilter=CPhKmRWIMeFn2PCPAsDuNg&safe=active

Torque and Center of Mass

Handout

HW #6

Center of Mass:

The center of mass (or mass center) is the mean location of all the mass in a system.

The motion of an object can be characterized by this point in space. All the mass of the object can be thought of being concentrated at this location. The motion of this point matches the motion of a point particle.

Finding the Center of Mass:

Uniform geometric figures have the center of mass located at the geometric center of the object.

Note that the center of mass does not have to be contained inside the volume of the object.

Collections of Point Masses:

The center of mass for a collection of point masses is the weighted average of the position of the objects in space.

Each object will have a position in space. The center of mass is found as:

321

332211

mmm

xmxmxmxcm

321

332211

mmm

ymymymycm

Example #1: A 10.0 kg mass sits at the origin, and a 30.0 kg mass rests at the 12.0 m mark on the x – axis. (a) Find the center of mass for this system.

01 x

kgm 0.101

mx 0.122

kgm 0.302

321

332211

mmm

xmxmxmxcm

kgkg

mkgkgxcm 0.300.10

0.120.3000.10

m00.9

(b) Find the center of mass for this system relative to the mass at the right.

mx 0.121

kgm 0.101

02 x

kgm 0.302

321

332211

mmm

xmxmxmxcm

kgkg

kgmkgxcm 0.300.10

00.300.120.10

m00.3

Although numerically different, it is the same point in space relative to the masses…

Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

Treat as two objects:

6 cm object:

gcmcm

gm 40.200.6

0.10

00.41

0,1 cmx

cmy cm 00.3,1

4 cm object:

gcmcm

gm 60.100.4

0.10

00.42

cmx cm 00.2,2 0,2 cmy

321

332211

mmm

xmxmxmxcm

g

cmggxcm 00.4

00.260.1040.2 cm800.0

321

332211

mmm

ymymymycm

g

gcmgycm 00.4

060.100.340.2 cm80.1

Example #3: Determine the center of mass of the following masses, as measured from the left end. Assume the blocks are of the same density.

Mm 1

21olx

Mm 82

olx 22 Mm 273 olx 5.43

321

332211

mmm

xmxmxmxcm

M

lMlMl

Mx

ooo

cm 36

5.427282

ocm lx6

23

Torque

Torque is the rotational equivalent of force. A torque is the result of a force applied to an object that tries to make the object rotate about some pivot point.

Equation of Torque:

pivot point

r distance from pivot to

applied force

F

applied force

angle between direction of force and pivot distance.

sinrFtorque

Note that torque is maximum when the angle is 90º.

The units of torque are Nm or newton · meter

The torque is also the product of the distance from the pivot times the component of the force perpendicular to the distance from the pivot.

rFrFtorque sin

The torque is also the product of the force times the lever arm distance, d.

FdrFtorque sin

Example #4: Calculate the torque for the force shown below.

sinrF 0.60sin30000.2 Nm

Nm520

Example #5: Calculate the total torque about point O on the figure below. Take counterclockwise torques to be positive, and clockwise torques to be negative.

sinrF

20sin100.460sin250.2 NmNmnet

Nmnet 6.29

Example #6: The forces applied to the cylinder below are F1 = 6.0 N, F2 = 4.0 N, F3 = 2.0 N, and F4 = 5.0 N. Also, R1 = 5.0 cm and R2 = 12 cm. Determine the net torque on the cylinder.

90sin21 RFnet

90sin22 RF

90sin13 RF

0sin24 RF0

mNmNmNnet 050.00.212.00.412.00.6

Nmnet 14.0

Static Equilibrium:Torque and Center of Mass

Handout

HW #7

Extra credit due Friday

Static Equilibrium:

Static equilibrium was touched on in the unit of forces. The condition for static equilibrium is that the object is at rest. Since the object is not moving, it is not accelerating. Thus the net force is zero. Shown at right is a typical example from that unit: Find the force of tension in each rope.

A new condition can now be added into this type of problem: Since the object is at rest, it must not be rotating, as that would also require an acceleration.

If there is no rotation, there must not be a rotational acceleration. Thus the net torque must be zero. This is an application of Newton’s 2nd law to rotational motion.

0net

If the net torque is zero, then all the counterclockwise (ccw) torques must balance all the clockwise (cw) torques.

cwccw

If there is no rotation, where is the pivot point for calculating torque?

Answer: The pivot point can be put anyplace you want!

Hint: Put the pivot point at one of the unknowns. This eliminates the unknown from the torque equation.

Example #1: A meter stick has a mass of 150 grams and has its center of mass located at the 50.0 cm mark. If the meter stick is supported at each of its ends, then what forces are needed to support it?

?1 F ?2 F

gmms

Show that the two forces are equal through torque. Put the pivot point at the left end.

Force F1 does not contribute to torque. {force applied to pivot point!}

Force F2 makes a ccw torque. 90sin10022 cmF

cm100

Force mmsg makes a cw torque.

cm50

90sin0.50 cmgmmsms

Balance the net ccw and cw torque:

cmF 1002 cmgmms 0.50

gmF ms21

2 280.9150.021

smkg

NF 735.02

The other unknown must also equal half the weight, so:

NF 735.01

Example #2: Suppose the meter stick above were supported at the 0 cm mark (on the left) and at the 75 cm mark (on the right). What are the forces of support now?

?1 F ?2 F

gmms

Find the two unknown forces through torque. Put the pivot point at the left end.

Force F1 does not contribute to torque. {force applied to pivot point!}

Force F2 makes a ccw torque. 90sin0.7522 cmF

cm75

Force mmsg makes a cw torque.

cm50

90sin0.50 cmgmmsms

Balance the net ccw and cw torque:

cmF 0.752 cmgmms 0.50

gmF ms32

2 280.9150.032

smkg

NF 980.02

If force F2 holds 2/3 the weight, then F1 must hold the remaining 1/3 of the weight.

NF 490.01

Example #3: A meterstick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?

gmmsgmadded The meterstick behaves as if all of its mass was concentrated at its center of mass.cmcm 0.102.39

cm2.29 cmcm 2.397.49

cm5.10Calculate the torque about the pivot point. The support force of the fulcrum will not contribute to the torque in this case.

Force maddedg makes a ccw torque.

90sin2.29 cmgmaddedadded

Force mmsg makes a cw torque. 90sin5.10 cmgmmsms

Balance the net ccw and cw torque:

cmgmcmgm addedms 2.295.10

cm

cmgramsmms 5.10

2.290.50

gmms 139

Example #4: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and is 3.00 m long. What is the tension in each rope when the 700-N worker stands 1.00 m from one end?

N200

m50.1

N700

m00.11F 2F

Put the pivot point on the left end. The force F1 does not contribute torque. Solve for F2.

mNmNmF 50.120000.170000.32

NF 3332

N200

m50.1

N700

m00.11F 2F

Solve F1 from Newton’s laws:

NNFF 20070021

NF 5671

Example #5: A cantilever is a beam that extends beyond its supports, as shown below. Assume the beam has a mass of 1,200 kg and that its center of mass is located at its geometric center. (a) Determine the support forces.

Put the pivot point at the left end and balance the torques.

0A

90sin0.20 mFBBccw

90sin0.25 mgmbeammgcw

Balance the net ccw and cw torque: mFB 0.20 mgmbeam 0.25

m

mkgF s

m

B 0.20

0.2580.9200,1 2

N700,14

NFA 940,2

Solve FA from Newton’s laws:

280.9200,1s

mBA kgFF

The fact FA is negative means that the force really points downwards.

When the wrong direction is chosen for a force, it just comes out negative at the end.

Static Equilibrium:Day #2

Handout

HW #7

Extra credit due Friday

Example #6: Calculate (a) the tension force FT in the wire that supports the 27.0 kg beam shown below.

Beam length is L.

LL2

1

Put the pivot point at the left end. The wall support does not contribute to torque.

sinLFTTccw

Note that and 40° are supplements, so it does not matter which is used in the sine function.

180sinsin

0.40sinLFTTccw

0.90sin2

Lgmbeambeamcw

Balance the net ccw and cw torque:

0.40sinLFT 2

Lgmbeam

0.40sin2

80.90.27 2sm

T

kgF

NFT 206

(b) Determine the x and y components to the force exerted by the wall.

yF

xF

Balance forces in each component direction.

0.40cosTx FF N158

gmFF beamTy 0.40sin

NFy 132

Example #7: A shop sign weighing 245 N is supported by a uniform 155 N beam as shown below. Find the tension in the guy wire and the horizontal and vertical forces exerted by the hinge on the beam.

Put the pivot point at the left end. The wall support does not contribute to torque.

0.35sin35.1 mFTTccw

signbeamcw

mgmm

gm signbeamcw 70.12

70.1

Balance the net ccw and cw torque:

0.35sin35.1 mFT

mNm

N 70.12452

70.1155

NFT 708

yF

xF

0.35cosTx FF N580

0.35sinTy FF

NFy 1.6

gmgm signbeam

The fact Fy is negative means that the force really points downwards.

Example #8: A person bending forward to lift a load “with his back” (see figure below) rather than “with his knees” can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in the figure below of a person bending forward to lift a 200-N object. The spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°. Find the tension in the back muscle and the compressional force in the spine.

L

2

L

3

2L

Put the pivot point at the left end. The hip support does not contribute to torque.

LNL

NL

T 2002

3500.12sin3

2

0.12sin

2002

350

2

3N

N

TNT 2705

lb610

0.12cosTRxN2646 lb600

NNTRy 2003500.12sin

NRy 5.12

Example #9: A person in a wheelchair wishes to roll up over a sidewalk curb by exerting a horizontal force to the top of each of the wheelchair’s main wheels (Fig. P8.81a). The main wheels have radius r and come in contact with a curb of height h (Fig. P8.81b). (a) Assume that each main wheel supports half of the total load, and show that the magnitude of the minimum force necessary to raise the wheelchair from the street is given by

hr

hrhmgF

22

2 2

where mg is the combined weight of the wheelchair and person. (b) Estimate the value of F, taking mg = 1 400 N, r = 30 cm, and h = 10 cm.

Example #10: A circular disk 0.500 m in diameter, pivoted about a horizontal axis through its center, has a cord wrapped around its rim. The cord passes over a frictionless pulley P and is attached to an object that weighs 240 N. A uniform rod 2.00 m long is fastened to the disk, with one end at the center of the disk. The apparatus is in equilibrium, with the rod horizontal. (a) What is the weight of the rod?

N240

gmrod

m00.1

mN 250.0240

mgmrod 00.1

kgmrod 12.6

(b) What is the new equilibrium direction of the rod when a second object weighing 20.0 N is suspended from the other end of the rod, as shown by the broken line in the image below? That is, what angle does the rod then make with the horizontal?

N240

gmrod

N20

mN 250.0240

cos00.1 mgmrod

cos00.220 mN

6.0cos

1.53

Inertia and Rotary MotionMoment of Inertia

Handout

HW #8

Final Schedule: Mosig’s Class

Monday 12/9 NotesTuesday 12/10 Finish notesWednesday 12/11 Study / Tower practice dayThursday 12/12 Study / Tower practice dayFriday 12/13 Final Exam {covers current unit}

Monday 12/16 Towers p. 0, 1, & 6Tuesday 12/17 Towers p. 2 & 3Wednesday 12/18 Towers p. 4 & 5Thursday 12/19 Elf Dance / Non – Academic dayFriday 12/20 No School

Definition: Inertia is the ability of an object to resist a change in its motion.

Inertia was introduced earlier in the Force unit. For straight line motion, the inertia of an object is measured through mass: The more massive an object, the more it is able to resist changes to its (straight line) motion.

Force and acceleration were related through inertia: maFnet

There is an equivalent to inertia in rotary motion. Here, inertia would try to resist a change to the angular motion. This form of inertia will depend on mass, just as before, but it will also depend on the distribution of the mass.

Demonstration:

When the mass is distributed close to the center of rotation, the object is relatively easy to turn. When the mass is held much further away, it is more difficult to rotate the object.

For example, if you had a ring (hoop) and a disk with the same radius and same mass, the ring would show more resistance to rotation than would the disk. The disk has mass uniformly distributed across its body, so some of the mass is near the center of rotation. The ring has more mass concentrated at the outside edge of the body than does the disk, so it will show more inertia (even though the mass and radius are the same).

The dependence of the inertia on mass and distribution can be built up through the kinetic energy of an object moving through a circular path.

The kinetic energy of a mass m moving at a speed v around the circle is:

221 mvKE

Let the circle have a radius r, and let the angular speed of the mass be . Write the kinetic energy in terms of the angular speed:

2221 mrKE

Define the moment of inertia I of this point mass to be:

Then: This is now the definition.

2mrI

221 IKE

Now build up to a more complicated object:

The total kinetic energy becomes:

2112

1 vmKE 2

33212

2221 vmvm

Mass m1 travels in a circle of radius r1, etc. All masses have the same angular velocity, . Write the kinetic energy in terms of this:

22332

122222

122112

1 rmrmrmKE

22122

332

222

1121 IrmrmrmKE

In general, the moment of inertia is found as: 2iirmI

The farther the mass is from the center of rotation, the higher the moment of inertia.

Example #1: (a) If m = 2.00 kg and d = 0.500 m for the image below, determine the moment of inertia of this group of objects.

233

222

211 rmrmrmI

222 32 dmdmmdI

214mdI

2500.000.214 mkg 200.7 mkg

(b) If this apparatus is rotating at 4.00 rad/s, what is its kinetic energy?

221 IKE

2221 00.400.7 s

radmkgKE

JKE 0.56

For more complicated objects, the moment of inertia is tabulated below. The calculations of these are complex and beyond the scope of the class.

Example of calculating the rotational inertia of a solid ball. You are not responsible for knowing these calculations…

Newton’s 2nd Law and Rotational Motion:

There is an equivalent to Newton’s 2nd law for rotational motion.

maFnet Inet

This can be combined with the kinematics equations from earlier in the unit to solve uniform motion problems;

to

221 tto

222o

to

2

Example #2: A force of 225.0 N is applied to the edge of a disk that can spin about its center. The disk has a mass of 240 kg and a diameter of 3.20 m. If the force is applied for 24.0 s, how fast will the disk be turning if it starts from rest?

221 mrIdisk

Inet

90sinrF

2

2 mrIrF

2

2mrrF

mr

F2

mkg

N

60.1240

2252 217.1

srad

to ss

rad 0.2417.10 2

srad1.28

Example #3: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm. What is the acceleration of the hanging mass downwards?

T

T

212

1 rmIdisk

2m

gm2

The tangential acceleration of the disk is equal to the linear acceleration of the falling mass.

ra The net torque on the disk gives one equation:

r

armITr 2

121

amT 121

Next use Newton’s 2nd law on the falling mass:

Tgmam 22

Combine the equations: Tam 121

TTgmamam 2121

2

121

2

2

mm

gma

Inertia and Rotary MotionDay #2

Handout

HW #8

Rolling and Energy Conservation.

When an object rolls along the ground, the tangential speed of the outside edge of the object is the same as the speed of the center of mass of the object relative to the ground.

The rolling object has two parts to its motion. First is the motion of the center of mass, and second is the rotation around the center of mass. The total kinetic energy is the sum of the kinetic energy associated with each part.

The kinetic energy associated with the center of mass moving in a straight line is given by the term:

2

21

cmcm vmKE

The portion associated with rotating around the center of mass is:

221 cmIKE Icm is the moment of inertia of the object about its

center of mass. Refer to the given table for values.

Example #4: A solid ball of radius 10.0 cm and mass 10.0 kg rolls at a given speed vo of 5.00 m/s. (a) What is the total kinetic energy of this rolling ball?

2

21

cmtot vmKE 221 cmI

For a rolling ball: &252 mrIcm

r

vcm

2

252

212

21

r

vmrvmKE cm

cmtot

2

1072

51

21

cmcmtot vmvmKE

2

107 00.50.10 s

mtot kgKE

JKEtot 175

(b) What percentage of the total kinetic energy is rolling?

2107

221

cmtot

roll

vm

I

KE

KE 2

107

22

52

21

cm

cm

vm

rv

mr

7

2

Example #5: Four different objects are placed at the top on an incline, as shown below. A point particle can slide down without friction. The other three objects will roll down the incline. In what order will the objects reach the bottom, from fastest to slowest? (a) What is the speed of the sliding

point particle when it reaches the bottom?

Energy conservation!

bottomtop EE

bottombottomtoptop PEKEPEKE

0 0

221 mvmgh ghvpoint 2

(b) Solve for the speed of the sphere (solid ball) at the bottom.

Energy conservation!

bottomtop EE

2212

21 Imvmgh

Note that there is a fixed starting energy, and this is split between linear motion and rotation. The rolling objects are slower!

252 mrIcm

r

vcm

22

52

212

21

r

vmrmvmgh 2

107 mv

ghv 710

(c) Solve for the speed of the hoop at the bottom.

Energy conservation!

bottomtop EE

2212

21 Imvmgh

The hoop has the slowest speed, and thus takes the longest to reach the bottom. The disk will be between the ball and the hoop.

2mrIcm

r

vcm

22

212

21

r

vmrmvmgh 2mv ghv

hoopdiskballpoint tttt

Example #6: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm.(a) How fast will the mass be traveling after it falls a distance of h = 4.00 m downwards? Energy conservation!

bottomtop EE

2212

21 Imvmgh

221 mrIcm

Speed of falling mass equals tangential speed of disk:

r

v

r

v objectt

2212

21 Imvmgh

2

221

212

21

r

vrmmvmgh disk

221

21 vmmmgh disk

diskmm

mghv

21

2

kgkg

mkgs

m

0.2000.6

00.480.900.62

21

2

smv 42.5

(b) Solve for the acceleration of the hanging mass:

221

21 vmmmgh disk

hmm

mgv

disk21

2 2

Shortcut: Remember the kinematics equation… xavv o 222

diskmm

mga

21

Same result as before!