Physics 111 · Physics 111 Title page Thursday, October 28, 2004 Physics 111 Lecture 17 ... A...

Physics 111

Title page

Thursday,October 28, 2004

Physics 111 Lecture 17

• Ch 8: Work done by force at an anglePower

• Ch 10: Rotational Kinematics angular displacement angular velocity angular acceleration

• Wednesday, 8 - 9 pm in NSC 118/119• Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsThursOct

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This week’s lab will be a workshopof physics. Bring your rankingtasks book. No quiz.

labs

AnnouncementsThursOct

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Worksheet #1

CQ1: springs

Ch 8: Conservation of EnergyThursOct

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Worksheet #1

CQ1 – spring pot’l energy ?

A spring-loaded toy dart gun is used to shoot a dart straight upin the air, and the dart reaches a maximum height of 24 m.Thesame dart is shot straight up a second time from the same gun,but this time the spring is compressed only half as far beforefiring. How far up does the dart go this time, neglecting frictionand assuming an ideal spring?

PI, Mazur (1997)

≥≥

1. 96 m 2. 48 m3. 24 m 4. 28 m5. 6 m 6. 3 m7. impossible to determine

Ch 8: Conservation of EnergyThursOct

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Let’s now look at a force that isapplied at an angle to thedirection of motion.

How does our problem change?

Aristocrat at a fixed angle

Frictionless pond of ice

T

Ch 7: Work and Kinetic EnergyThursOct

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Skip to result

If the block remains on the ice, andthe man keeps the angle betweenthe rope and the ice constant, andthe man exerts a constant force,the work done on the block is….

con’t


distance = Δs

T


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We look at the component of theforce that is in the direction ofmotion of the object (i.e., alongthe direction of s)

θ

s-directions

θ

T

Ts

con’t


distance = Δs

T


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W = FsΔs = T

sΔs

= (T cosθ )Δs

con’t

θ


distance = Δs

T


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θ

T

Ts = T cosθs-directions

Skip dot product

Notice: Work is a scalar quantity, which meansthat you do NOT specify a direction associatedwith work. Work only has a magnitude.

Let’s introduce a new mathematical operationto express the type of product we need tocalculate in order to correctly compute work.

Work: scalar


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W = FsΔs

The scalar (or dot) product of any two vectors

A

Band

where

A = a

xx + a

yy + a

zz

and

B = b

xx + b

yy + b

zz

is given by the expression

Scalar product


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A•B = a

xb

x+ a

yb

y+ a

zb

z

Which mathematically says: - multiply the x-components of the two vectors; - add the result to the product of the y-components of the two vectors; and finally - add the result to the product of the z-components of the two vectors.

con’t


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skip

If we look at the distributive law ofmultiplication in evaluating our dot product, wediscover several important results.

A•B = (a

xx + a

yy + a

zz) • (b

xx + b

yy + b

zz)

= axx • (b

xx + b

yy + b

zz)

+ ayy • (b

xx + b

yy + b

zz)

+ azz • (b

xx + b

yy + b

zz)

= axb

x

+ ayb

y

+ azb

z

con’t


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This result tells us that thefollowing relationshipsmust be true:

axx • b

xx = a

xb

x

a

xx • b

yy = 0

axx • b

zy = 0

Or generally...

x • x = 1

x • y = 0

x • z = 0

math definitions


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We can analyze the other two terms (forthe y- and z- dimensions of A) and find

y • z = 0

y • y = 1

y • z = 0

z • x = 0

z • y = 0

z • z = 1

more defs


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What is also clear from our definitionof this operation is that scalar productsare commutative.

A•B =B •A

A•B = a

xb

x+ a

yb

y+ a

zb

z

B •A = b

xa

x+ b

ya

y+ b

za

z

commutative


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Finally, the scalar product also obeys thedistributive law of multiplication.

A• (B +

C) =A•B +A•

C

I leave this as an exercise to you to verifyon your own. The proof is rather simpleand straightforward.

distributive


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Scalar products can also be interpreted asthe scalar product of length of one vectorand the length of the projection of the secondvector onto the first.

B

A

θ

B cosθ

A•B =

AB cosθ

AB cos θ


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Now, I’ve given you two very different lookingexpressions for the scalar product of 2 vectors.

Which one is correct?

2 forms of scalar product

A•B =

AB cosθ

A•B = a

xb

x+ a

yb

y+ a

zb

z


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(The general proof of this is a subject forlinear algebra, but here’s a specific casethat illustrates the thinking.) skip

For any two 2-D vectors, I can choose mycoordinate system such that one of thevectors lies entirely along the x-axis.

Proof

Let’s choose vector A to lie on the x-axis.


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Using the 2nddefinition of

the scalar product

A•B = A

x|B | cosθ = A

xB

x2 + B

y2( ) B

x

Bx2 + B

y2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

A•B = A

xB

x

θ B

A

B cosθ

con’t


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Using the 1stdefinition of

the scalar product

A•B = A

xB

x+ A

yB

y

con’t

A•B = A

xB

x


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θ B

A

B cosθ

These two definitions of the scalar productprovide a powerful tool with which we candetermine the angle between any two vectors!

A = 2x + 3 y

Find the angle between the followingtwo vectors:

B = −1x + 2 y

Worksheet #1a

? Find the angle… (W1a)


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? Find the angle… (W1a)

A•B = (2)(−1) + (3)(2) = 4

AB cosθ = 22 + 32( ) (−1)2 + 22( )cosθ

AB cosθ = 65( )cosθ cosθ = 4 / 65 θ = 60.3o


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A = 2x + 3 y

Find the angle between the followingtwo vectors:

B = −1x + 2 y

Now recall that our definition of work is

W F s= ( cos )θ Δwhere θ is the angle between the force (F)

and the displacement (Δs)

W F ss= Δ

Old definition of work


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This expression, however, is just that of ascalar product between the force vector andthe displacement vector!

W = • F sΔ

Valid only if the force is constant!

So, a nice, shorthand way to express work is

New definition of work


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W F s= ( cos )θ Δ

The rate at whichwork is done.

P =

ΔWΔt

Average power

Wo

rk (

J)

Time (s)

Notice the averagepower output issimply the slope

of the chord!(Look familiar?)

Power


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P(t) = lim

Δt→0

ΔWΔt

=

F • ΔsΔt

=F • v(t)

Notice that Power isalso a scalar quantity.

We can also calculate theinstantaneous power beingdelivered by a force.

Instantaneous power


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Wo

rk (

J)

Time (s)

Instantaneouspower

Average power

P =F • v

[P] = [F][v]

[P] = Nm / s

[P] = J / s = W

Units: power


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A 600 kg elevator starts from rest and is pulledupward by a motor with a constant acceleration of2 m/s2 for 3 seconds. What is the average poweroutput of the motor during this time period?

Worksheet #2

? Elevator motor power (W2)


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1) 59,920 W2) 21,240 W3) 17,640 W4) 3,600 W

W

T

FBD:

Fnet = ma = (600 kg)(2 m/s2)

Fnet = 1200 N = Fmotor - W = Fmotor - mg

1200 N = Fmotor - (600 kg)(9.8 m/s2) = Fmotor - 5880 N

Fmotor = 5880 N + 1200 N = 7080 N

Let’s first figure out the force delivered by the motor,which we’ll assume equals the tension...


? Elevator motor power (W2)


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Now we need to determine the work done by the motor...

W = F Δs But we don’t know Δs, so….

s = s0 +v0t +0.5at2 = 0 + 0 + 0.5(2 m/s2)(3 s)2 = 9 m

W = (7080 N)(9 m) = 63720 J

P = W

Δt= 63720 J

3 s= 21240 W


soln


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Rotational Kinematics

We spent the first 8 weeks of class examiningthe motion of point mass objects. We now lookmore carefully at real, extended objects thatmove along curved paths and the forcesresponsible for their motion.

Ch 10: Rotational Kinematics

Ch 10: Rotational KinematicsThursOct

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To make our life simpler, we’re going to usethe angular unit of measure known as theradian when discussing the motion of objectsin circular arcs.

Let’s spend a little time motivating ourchoice of radians over degrees as the unitof choice for measuring angles.

Radians vs degrees


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Let’s start with thefollowing question: What is π?

We all probably remember the numerical valueof π (3.14159…). But from where does thisnumerical value come?

π is exactly the ratio of thecircumference of a circleto its diameter.

True for ANY circle!

d

pi


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Cir

cum

fere

nce

Diameter

And if we plotted the circumferences of a varietyof circles against their diameters, we would see…

Slope = π

C = πd

What if we changed to a plot of circumferenceversus radius? How would the plot change?

C vs d


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Cir

cum

fere

nce

Radius

Slope = 2π C = 2πrNotice our graph is nowtwice as steep as it wasbefore.

Okay, this stuff all SEEMS pretty obvious.Just where are we headed?

How do we compute ARC-LENGTH along a circle?

C vs r


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s = θrr

θs

r

θs

How does the ratioof s/r change as weincrease the size ofour circle?

NOTE: The angle θ is still the same.

s

r= θ

No Change!

Where θ is measured in radians!

arc length


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Notice that this property – namely that thevalue of quantity s/r at a given θ does notdepend upon the size of the circle property –is shared with the trigonometric functions(sine, cosine, tangent, etc.).

r1

θy1

x1

cosθ =

x1

r1

=x

2

r2

r2

θy2

x1

trig functions


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• Here θ is measured in radians.

• Notice that the ratio s/r varieslinearly with the size of theangle θ.

•The measure of radians provides a naturalsystem with which to measure angles.

• This is in contrast with the system of degrees,which resulted from the completely arbitrarydecision to put 360 of them in one completerevolution around a circle.

radians or degrees?


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s

r= θ

Now that we’ve seen that the natural systemin which to measure in the “angular world”is the radian system, we are free to exploreangular motions.

Completely analogousto our discussion ofthe linear motionquantities of

displacementvelocityacceleration

We can now definethe angular quantities

angular displacementangular velocityangular acceleration

Angular analogs


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θi

θf

Δθ θ θ= −f i

Positive changescorrespond tocounter-clockwiserotations.

Ang. displacement


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Notice that theangulardisplacementcan be directlytranslated into alinear distancetraveled using ourarc length frombefore...

s = rΔθ = r(θ

f−θ

i)

& arc length

sr

θi

θf


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And if we knowit takes the yellowball a time Δt tomove through theangle Δθ , then wecan calculate anangular velocity...

ω θ θ θ= =

−−

ΔΔt t t

f i

f i

Angular velocity isdenoted by thesymbol ω.

Ang. velocity

sr

θi

θf


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θ

Time

slope =ωInstantaneous

angular velocity

slope =ω = Δθ

ΔtAverage

angular velocity

Just as with its linearcounterpart, we canexamine a graphicalrepresentation of thisquantity to betterunderstand its meaning.

graphical interpretation


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[ω ] = [Δθ]

[Δt]

[ω ] = rad

s= s−1

ω = Δθ

Δt

Units: angular velocity


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Worksheet #3

CQ3: Angular velocity


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Worksheet #3

CQ3 – angular velocity ?

A ladybug sits at the outer edge of a merry-go-round, and agentleman bug sits halfway between her and the axis ofrotation. The merry-go-round makes a complete revolutiononce each second.The gentleman bug’s angular speed is

PI, Mazur (1997)

≥≥

1. half the ladybug’s.2. the same as the ladybug’s.3. twice the ladybug’s.4. impossible to determine


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Just as in the linear case, where we examinedthe rate of change of the velocity, in the angularcase, we can examine the rate of change of theangular velocity, which we call theangular acceleration.

α ω ω ω= =

−−

ΔΔt t t

f i

f i

Angular accelerationis denoted by thesymbol α.

Ang. acceleration


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As we did for the linearquantity and for angularvelocity, let’s look at thegraph.

ω

Time

slope =α

Instantaneousangular

acceleration

slope =

α = Δ

ωΔt

Averageangular

acceleration

graphical interpretation


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α = Δ

ωΔt

[α] = [Δ

ω ]

[Δt]

[α] = rad/s

s= s−2

Units: Angular acceleration


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Equations for Systems InvolvingRotational Motion with

Constant Angular Acceleration

Again, these are completely analogous to whatwe derived for the kinetic equations of a linearsystem with constant linear acceleration!

θ θ ω α= + +0 012

2t t

ω ω α= +0 t

Rot. Kinematic equations


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A wheel rotates with constant angular accel-eration of α0 = 2 rad/s2. If the wheel starts fromrest, how many revolutions does it make in 10 s?

θ = θ

0+ω

0t + 1

2αt 2

Δθ =ω

0t + 1

2αt2 = 0 + 1

2(2 rad

s2)(10 s)2

Δθ = 100 rad 1 rev = 2π rad

Δθ = (100 rad) 1 rev

2π rad= 50

π rev = 15.9 rev

? Use kinematic eqn (W4)

Worksheet #4


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s r r f i= = −Δθ θ θ( )

We’ve seen how arc lengthrelates to an angle swept out:

Let’s look at how our angular velocityand acceleration relate to linear quantities.

Relating to linear quantities


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For an object movingin a circle with a constantlinear speed (a constantangular velocity), theinstantaneous velocityvectors are always tangentto the circle of motion.

The magnitude of the tangential velocity can befound from our relationship of arc length to angle...

s r= Δθ

sΔt

= rΔθΔt

v rtan = ω

tangential velocity vs omega


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Two wheels, A and B, are rotated with constantangular acceleration of α = 2 rad/s2. Bothwheels start from rest. If the radius of wheel Ais twice the radius of wheel B, how does theangular velocity of wheel A compare to wheelB at time t = 10 s?

1) 1/4 as great2) 1/2 as great3) Same4) 2 times as great5) 4 times as great

? Rotating wheels (W5)

Worksheet #5


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Two wheels, A and B, are rotated with constantangular acceleration of α = 2 rad/s2. Bothwheels start from rest. If the radius of wheel Ais twice the radius of wheel B, how does thetangential velocity of wheel A compare towheel B at time t = 10 s?

1) 1/4 as great2) 1/2 as great3) Same4) 2 times as great5) 4 times as great

? Rotating wheels (W6)

Worksheet #6


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For an object movingaround a circle with achanging angular velocity,and hence a changingtangential velocity, theinstantaneous tangentialacceleration is NON-ZERO!

Let’s look at how the tangential velocity changeswith time in such a case:

v rtan = ω

Δvtan

Δt=

vtan f

− vtan i

tf− t

i

=rω

f− rω

i

tf− t

i

changing omega


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Let’s look at how the tangential velocity changeswith time in such a case:

changing omega con’t

ar

trtan = =Δ

Δω α


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For an object movingaround a circle with achanging angular velocity,and hence a changingtangential velocity, theinstantaneous tangentialacceleration is NON-ZERO!

The tangential acceleration,however, is not the onlyacceleration we need toconsider in problems ofcircular motion...

The purple arrows represent the direction ofthe CENTRIPETAL ACCELERATION,which always points towards the centerof the circle.

Don’t forget centripetal…


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Recall our definition ofcentripetal acceleration:

av

r

r

rrc

t= = =2 2

2( )ω ωNotice:

Our new form usesangular velocity!

Cent. Acc. and omega


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