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
Frictionless pond of ice
distance = Δs
T
Ch 7: Work and Kinetic EnergyThursOct
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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
Frictionless pond of ice
distance = Δs
T
Ch 7: Work and Kinetic EnergyThursOct
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W = FsΔs = T
sΔs
= (T cosθ )Δs
con’t
θ
Frictionless pond of ice
distance = Δs
T
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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.
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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)
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 7: Work and Kinetic EnergyThursOct
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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)
Ch 7: Work and Kinetic EnergyThursOct
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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...
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?
? Elevator motor power (W2)
Ch 7: Work and Kinetic EnergyThursOct
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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
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?
soln
Ch 7: Work and Kinetic EnergyThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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?
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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θi
θf
Δθ θ θ= −f i
Positive changescorrespond tocounter-clockwiserotations.
Ang. displacement
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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[ω ] = [Δθ]
[Δt]
[ω ] = rad
s= s−1
ω = Δθ
Δt
Units: angular velocity
Ch 10: Rotational KinematicsThursOct
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Worksheet #3
CQ3: Angular velocity
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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α = Δ
ωΔt
[α] = [Δ
ω ]
[Δt]
[α] = rad/s
s= s−2
Units: Angular acceleration
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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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…
Ch 10: Rotational KinematicsThursOct
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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
Ch 10: Rotational KinematicsThursOct
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