Download Chapter 8 Lecture

Chapter 8 Lecture
Rotational
Motion
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Rotational Motion
• How can a star rotate 1000 times faster than a
merry-go-round?
• Why is it more difficult to balance on a stopped
bike than on a moving bike?
• How is the Moon slowing Earth's rate of
rotation?
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What's new in this chapter
• In the last chapter, we learned about the torque
that a force can exert on a rigid body.
– We analyzed only rigid bodies that were in
static equilibrium.
• In this chapter, we learn how to describe,
explain, and predict motion for objects that
rotate.
– For example, the hip joint and a car tire
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8.1 Rotational Kinematics
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Rotational Kinematics
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Rotational kinematics
• Imagine that you place small coins at different
locations on the disk:
– The direction of the velocity of each coin
changes continually.
– A coin that sits closer to the edge moves
faster and covers a longer distance than a
coin placed closer to the center.
• Different parts of the disk move in different
directions and at different speeds!
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Rotational kinematics
• There are similarities
between the motions of
different points on a
rotating rigid body.
– During a particular time
interval, all coins at the
different points on the
rotating disk turn
through the same angle.
– Perhaps we should
describe the rotational
position of a rigid
body using an angle.
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Rotational (angular) position θ
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Units of rotational position
• The unit for rotational position is the radian (rad).
It is defined in terms of:
– The arc length s
– The radius r of the circle
• The angle in units of radians is
the ratio of s and r:
• The radian unit has no dimensions; it is the ratio
of two lengths. The unit rad is just a reminder
that we are using radians for angles.
1 Radian = 57.30.
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Tip
Figure 8.4
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Tip
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Conversion Deg to Rad
• Comparing degrees and radians
 (rad )  180
• 2 (rad )  360
• Converting from degrees to radians

  rad  
  degrees 
180
• Converting from radians to degrees
 (deg rees ) 
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180

 (rad )
360
1 rad 
 57.3
2
Rotational (angular) velocity ω
• Translational velocity is the
rate of change of linear
position.
• We define the rotational
(angular) velocity v of a rigid
body as the rate of change
of each point's rotational
position.
– All points on the rigid
body rotate through the
same angle in the same
time, so each point has
the same rotational
velocity.
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Rotational (angular) velocity ω
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Tips
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Rotational (angular) acceleration α
• Translational acceleration describes an object's
change in velocity for linear motion.
– We could apply the same idea to the center of
mass of a rigid body that is moving as a
whole from one position to another.
• The rate of change of the rigid body's rotational
velocity is its rotational acceleration.
– When the rotation rate of a rigid body
increases or decreases, it has a nonzero
rotational acceleration.
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Rotational (angular) acceleration α
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Tip
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Relating translational and rotational
quantities and tip
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Angular and Linear Quantities Summary
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Quantitative Exercise 8.2
• Determine the tangential speed of a stable
gaseous cloud around a black hole.
• The cloud has a stable circular orbit at its
innermost 30-km radius. This cloud moves in a
circle about the black hole about 970 times per
second.
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Rotational motion at constant acceleration
•
•
•
•
θ0 is an object's rotational position at t0 = 0.
ω0 is an object's rotational velocity at t0 = 0.
θ and ω are the rotational position and the rotational velocity at some later time t.
α is the object's constant rotational acceleration during the time interval from time 0
to time t.
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Rotational motion at constant acceleration
• Rotational position θ is positive if the object is rotating
counterclockwise and negative if it is rotating clockwise.
• Rotational velocity ω is positive if the object is rotating
counterclockwise and negative if it is rotating clockwise.
• The sign of the rotational acceleration α depends on how
the rotational velocity is changing:
– α has the same sign as ω if the magnitude of ω is
increasing.
– α has the opposite sign of ω if the magnitude of ω is
decreasing.
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QuickCheck 7.2
• Rasheed and Sofia are riding a merry-go-round
that is spinning steadily. Sofia is twice as far
from the axis as is Rasheed. Sofia’s angular
velocity is ______ that of Rasheed.
–
–
–
–
–
Half
The same as
Twice
Four times
We can’t say without
knowing their radii.
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QuickCheck 1
• Rasheed and Sofia are riding a merry-go-round
that is spinning steadily. Sofia is twice as far
from the axis as is Rasheed. Sofia’s angular
velocity is ______ that of Rasheed.
–
–
–
–
–
Half
The same as
Twice
Four times
We can’t say without
knowing their radii.
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QuickCheck 1
• Rasheed and Sofia are riding a merry-go-round
that is spinning steadily. Sofia is twice as far
from the axis as is Rasheed. Sofia’s speed is
______ that of Rasheed.
–
–
–
–
–
Half
The same as
Twice
Four times
We can’t say without
knowing their radii.
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QuickCheck 2
• Rasheed and Sofia are riding a merry-go-round
that is spinning steadily. Sofia is twice as far
from the axis as is Rasheed. Sofia’s speed is
______ that of Rasheed.
–
–
–
–
–
Half
The same as
Twice v = ωr
Four times
We can’t say without
knowing their radii.
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QuickCheck 2
• Two coins rotate on a turntable.
Coin B is twice as far from the axis
as coin A.
– The angular velocity of A is twice that of B
– The angular velocity of A equals that of B
– The angular velocity of A is half that of B
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QuickCheck 3
• Two coins rotate on a turntable.
Coin B is twice as far from the axis
as coin A.
– The angular velocity of A is twice that of B
– The angular velocity of A equals that of B
– The angular velocity of A is half that of B
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QuickCheck 4
• The fan blade is slowing down. What are the
signs of ω and ?
–
–
–
–
–
ω is positive and  is positive.
ω is positive and  is negative.
ω is negative and  is positive.
ω is negative and  is negative.
ω is positive and  is zero.
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QuickCheck 4
• The fan blade is slowing down. What are the
signs of ω and ?
–
–
–
–
–
ω is positive and  is positive.
ω is positive and  is negative.
ω is negative and  is positive.
ω is negative and  is negative.
ω is positive and  is zero.
“Slowing down” means that  and  have opposite signs, not
that  is negative.
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QuickCheck 5
• The fan blade is speeding up. What are the
signs of  and ?
A.  is positive and  is positive.
B.  is positive and  is negative.
C.  is negative and  is positive.
D.  is negative and  is negative.
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QuickCheck 5
• The fan blade is speeding up. What are the
signs of  and ?
A.  is positive and  is positive.
B.  is positive and  is negative.
C.  is negative and  is positive.
D.  is negative and  is negative.
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Three Types of Acceleration
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Three Types of Acceleration
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Three Types of Acceleration
•
We now have three accelerations associated with a rigid rotating
body: tangential, angular and centripetal, or aT, α, and aC.
•
And they are all related to each other - can you express all three of
these accelerations in terms of the angular velocity of the rotating
object?
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Total Linear Accleration
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Don’t Forget
1
𝑓=
𝑇
Therefore:
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Quick Problems
Answers 1.a 2.d and
3.c
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Quick Problems
Answers 1.d 2.b
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Quick Problems
Answers 1.d 2.c 3.c
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Quick Problems
An old phonograph record revolves at 45 rpm.
1. What is its angular
velocity in rad/sec?
ω = 45 rev/min = 45 (2π/60
radians/sec) = 4.71 rad/sec
2. Once the motor is turned off,
it takes 0.75 seconds to come
to a stop. What is its average
angular acceleration?
givens:
ωf = 0, ωo = 4.71 rad/sec, t = 0.75
seconds.
using the equation
ωf = ωo + αt
we can determine that
α = (0 - 4.71)/0.75 = -6.28 rad/sec2
Quick Problems
3. How many revolutions did it make while
coming to a stop?
using the equation
θ = ½(ωf + ωo)t
we can determine that
θ = ½ (0 + 4.71) 0.75
θ = 1.77 radians
since there are 2π radians in every revolution,
θ = 0.281 rev.
Quick Problems
4. A fan is turning at 10 rev/min and then speeds up to 25
rev/min in 10 seconds.
a. How many revolutions does
the blade require to alter its
speed?
using the equation
θ = ½(ωf + ωo)t
we can determine that
θ = ½(1.05 + 2.63)10
θ = 18.4 radians
since there are 2π radians in every
revolution, θ = 2.92 rev
b. If the tip of one blade is 30
cm from the center, what is the
final tangential velocity of the
tip?
using the equation
v = rω
allows us to determine that
v = (0.30)(2.63)
v = 0.789 m/sec
Quick Problems
Two wheels are connected by a common cord. One
wheel has a radius of 30 cm, the other has a radius
of 10 cm
Question: When the small wheel
is revolving at 10 rev/min, how
fast is the larger wheel rotating?
Since the two wheels share the
same tangential velocity, their
angular velocities will be inversely
proportional to their radii.
vlarge=vsmall
rlargeωlarge= rsmallωsmall
ωlarge = (rsmall/rlarge) ωsmall
ωlarge = (0.10/0.30)(10) = 3.33
rev/min
8.2 Torque and Rotational
Acceleration
Rotational Dynamics or Newtons 2nd Law
for Rotation
Just like there are rotational analogs for Kinematics, there are
rotational analogs for Dynamics.
Kinematics allowed us to solve for the motion of objects, without
caring why or how they moved.
Dynamics showed how the application of forces causes motion
and this is summed up in Newton's Three Laws.
These laws also apply to rotational motion.
The rotational analogs to Newton's Laws will be presented now.
Rotational Dynamics
What causes a rigid body to have rotational acceleration?
In the previous chapters…When we discussed acceleration we
used point-like objects interactions with each other…and those
interaction cause forces which resulted in acceleration (linear
translational acclerations).
With rotational accelerations we need to look at rigid
bodies/extended body instead of point objects
Rotation is caused by torque. Last chapter we analyzed rigid
bodies in static equilibrium. In this chapter its all about net torque
not being equal to zero.
Torque and rotational acceleration
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Torque and rotational acceleration
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Torque and rotational acceleration
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Torque and rotational acceleration
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Torque and rotational acceleration
• The experiments indicate a zero torque has no
effect on rotational motion but a nonzero torque
does cause a change.
– If the torque is in the same direction as the
direction of rotation of the rigid body, the
object's rotational speed increases.
– If the torque is in the opposite direction, the
object's speed decreases.
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Torque and rotational acceleration
• Our goal is to determine which physical
quantities cause rotational acceleration of an
extended object. There are two possibilities:
1. The sum of the forces (net force) exerted on
the object or
2. The net torque caused by the forces
• Testing experiments help us determine which (if
either) of these quantities might affect rotational
acceleration.
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Torque and rotational acceleration
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Torque and rotational acceleration
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Torque and rotational acceleration
• This is similar to what we learned when studying
translational motion. A nonzero net force needs
to be exerted on an object to cause its velocity to
change. The greater the net force, the greater
the translational acceleration of the object.
• NET TORQUE IS THE CAUSE OF ANGULAR
ACCELERATION (JUST LIKE FORCE IS FOR ACCELERATION)
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Torque and rotational acceleration
• So far all experiments in this chapter have been
performed with a rigid body fixed axis at its
center of mass (Not free to move linearly)
• If a rigid body was not held fixed, then a change
in both translational and rotational motion could
occur.
• The translational acceleration of center of mass
of such an object is determined by Newtons
Second law ΣF=ma.(More on this later)
• The rotational acceleration around its center of
mass will be determined by the next few slides.
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8.3 Rotational Inertia
(Broomstick Demo)
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Rotational inertia
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Rotational inertia
• Pull each string, and compare the rotational acceleration
for the arrangement shown on the left and the right:
– Our pattern predicts that the rotational acceleration
will be greater for the arrangement on the left
because the mass is nearer to the axis of rotation.
– When we try the experiment, we find this to be true,
consistent with our pattern.
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Rotational inertia
• Rotational inertia is the physical quantity
characterizing the location of the mass relative
to the axis of rotation of the object.
– The closer the mass of the object is to the
axis of rotation, the easier it is to change its
rotational motion and the smaller its rotational
inertia.
– The magnitude depends on both the total
mass of the object and the distribution of that
mass about its axis of rotation.
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nd
2
8.4 Newtons
Law For
Rotational Motion
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Newtons 2nd Law for Rotational Motion
• Development of relationship
between
– α (rotational accelerations)
– ∑Ƭ (net torque)
– Rotational Inertia
• Push block with finger exert a
FF/B tangent to the circular path
• Push Causes a torque that turns
the block increasing rotational
velocity.
• Torque Produces:
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Newtons 2nd Law for Rotational Motion
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Analogy between translational motion and
rotational motion
Σ𝜏 = 𝑚𝐵𝑟2𝛼
Σ𝑭 = 𝑚𝒂
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Example 8.3
• A 60-kg rollerblader holds a 4.0-m-long rope that
is loosely tied around a metal pole. You push the
rollerblader, exerting a 40-N force on her, which
causes her to move increasingly more rapidly in
a counterclockwise circle around the pole. The
surface she skates on is smooth, and the wheels
of her rollerblades are well oiled. Determine the
tangential and rotational acceleration of the
rollerblader.
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Example 8.3
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Example 8.3
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Example 8.3
Homework: Complete the try it yourself problem.
Submit tomorrow.
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Newton 2nd Law For Rotational Motion
Applied to Rigid Bodies
• An object mass is actually composed of many
small objects with different masses such that
• 𝑚𝑡 = 𝑚1 + 𝑚2 + 𝑚3 + ⋯
• Rotational Inertia is a scaler quanity so if we can
add up all the rotational inertia of all the bodies
we can determine the overall inertia of an object.
• To test this the following is presented
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Newton's second law for rotational motion
applied to rigid bodies
• The rotational inertia of a
rigid body about some
axis of rotation is the sum
of the rotational inertias of
the individual point-like
objects that make up the
rigid body.
– The rotational inertia
of this two-block rigid
body is twice the
rotational inertia of the
single block.
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We are going to use I to represent Rotational Inertia I=mr2
Therefore rotational inertia of a rigid body consists of several
point like parts located at different distances from the axis of
rotation is the sum of the mr2 terms for each part
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Quantitative Exercise 8.4
1. Each block of mass contributes differently ot
the rotational inertia of the system (further away
from axis of rotation
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Calculating rotational inertia
• The rotational inertia of the whole leg is:
• There are other ways to perform the summation
process. Often it is done using integral calculus,
and sometimes it is determined experimentally.
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Expressions for the rotational inertia of
standard-shape objects
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Expressions for the rotational inertia of
standard-shape objects
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Rotational form of Newton's second law
Σ𝜏 = 𝐼𝛼
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Tip
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Interpreting the Moment of Inertia
• The moment of inertia
is the rotational
equivalent of mass.
• An object’s moment of
inertia depends not only
on the object’s mass but
also on how the mass is
distributed around the
rotation axis.
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Interpreting the Moment of Inertia
• The moment of inertia is
the rotational equivalent
of mass.
• It is more difficult to spin
the merry-go-round when
people sit far from the
center because it has a
higher inertia than when
people sit close to the
center.
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Interpreting the Moment of Inertia
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Constraints Due to Ropes and Pulleys
• If the pulley turns without
the rope slipping on it then
the rope’s speed must
exactly match the speed of
the rim of the pulley.
• The attached object must
have the same speed and
acceleration as the rope.
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Example 8.5
• In an Atwood machine, a block of mass m1 and
a less massive block of mass m2 are connected
by a string that passes over a pulley of mass M
and radius R.
• What are the translational accelerations a1 and
a2 of the two blocks and the rotational
acceleration α of the pulley?
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Example 8.5
1. Analyze using 3 systems: block 1, block 2 and the pulley.
2. Model blocks like point-like objects and the pulley as a rigid
body
3. Note: A string around a block or pully pulls purely
tangentially so the 𝜏 = 𝐹 ∗ 𝑟𝑝𝑢𝑙𝑙𝑒𝑦
4. Pulley non-zero mass so T differs on each side of pulley
(required of rotational acclerations)
5. Translational acceleration of both blocks are equal a1=a2
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Example 8.5
• The translational acceleration of the hanging objects is due to the
difference between the gravitational force that Earth exerts on them
and the tension force that the string exerts on them.
• The rotational acceleration of the pulley is due to a nonzero net
torque produced by the two tension forces exerted on the pulley.
• We consider the pulley to be similar to a solid cylinder. Then
according to Table 8.6, its rotational inertia around the axis that passes
through its center is I = -1MR2, where R is the radius of the pulley and
M is its mass.
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Example 8.5
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Example 8.5
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Example 8.5
Ok I want you to complete for HW tonight Exercise
8.5 Try it yourself. The answers are provide. Hand in
Tomorrow!!!
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Example 8.6 (Complete at Home its in the
book)
• A woman tosses a 0.80-kg soft-drink bottle vertically
upward to a friend on a balcony above. At the beginning
of the toss, her forearm rotates upward from the
horizontal so that her hand exerts a 20-N upward force
on the bottle. Determine the force that her biceps exerts
on her forearm during this initial instant of the throw. The
mass of her forearm is 0.65 kg and its rotational inertia
about the elbow joint is 0.044 kg•m2. The attachment
point of the biceps muscle is 5.0 cm from the elbow joint,
the hand is 35 cm away from the elbow, and the center
of mass of the forearm/hand is 16 cm from the elbow.
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8.5 Rotational Momentum
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Rotational momentum
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Rotational momentum
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Rotational momentum
• For each experiment, the rotational inertia of the
spinning person decreased (the mass moved
closer to the axis of rotation). Simultaneously,
the rotational speed of the person increased.
– We propose tentatively that when the
rotational inertia I of an extended body in an
isolated system decreases, its rotational
speed ω increases, and vice versa.
• We can test this idea with a testing experiment.
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Rotational momentum
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Rotational/Angular Momentum
We are going to substitude in the angular values for the
liner (translational ones to derive the angular momentum
formula
Linear momentum: (p)
𝒑 = 𝑚𝒗
Angluar momentum: (L)
𝐿 = 𝐼𝜔
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Rotational momentum is constant for an
isolated system
• If a system with one rotating body is isolated,
then the external torque exerted on the object is
zero.
• In such a case, the rotational momentum of the
object does not change:
0 = Lf – Li
Lf = Li
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Rotational momentum is constant for an
isolated system
• The conservation of (linear) momentum states that in the
absence of external forces, momentum is conserved. This
came from the original statement of Newton's Second Law:
Σ𝐹 =
Δ𝑝
Δ𝑡
(remember impulse!!!)
• If there are no external forces, then:
Σ𝐹 =
Δ𝑝
Δ𝑡
=0
Δ𝑝 = 0
• Thus there is no change in momentum - so it is conserved.
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Rotational momentum is constant for an
isolated system
• Newton's Second Law (rotational version):
Στ =
Δ𝐿
Δ𝑡
(remember impulse!!!)
• If there are no external torques, then:
Στ =
Δ𝐿
Δ𝑡
=0
Δ𝐿 = 0
• Thus there is no change in momentum - so it is conserved.
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Rotational momentum The Skater Problem
• The standard example used to illustrate the conservation of
angular momentum is the ice skater. Start by assuming a frictionless
surface - ice is pretty close to that - although if it were totally
frictionless, the skater would not be able to stand up and skate!
• The skater starts with his arms stretched out and spins around in place.
He then pulls in his arms.
• What happens to his rotational velocity after he pulls in his
arms?
• After you discuss this in your groups, please click on the below and see
if you were correct. Why did this happen?
• https://www.youtube.com/watch?v=MjYk5TRpOlE
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Rotational momentum The Skater Problem
• By assuming a nearly frictionless surface, that
implies there is no net external torque on the skater,
so the conservation of angular
momentum can be used.
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Example 8.7
• Attach a 100-g puck to a string, and let the puck
glide in a counterclockwise circle on a
horizontal, frictionless air table. The other end of
the string passes through a hole at the center of
the table. You pull down on the string so that the
puck moves along a circular path of radius 0.40
m. It completes one revolution in 4.0 s. If you
pull harder on the string so that the radius of the
circle slowly decreases to 0.20 m, what is the
new period of revolution?
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Example 8.7
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Example 8.7
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Example 8.7
OK. Homework tonight. Complete the try it yourself question.
Submit tomorrow.
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Rotational momentum of an isolated system
is constant
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Example 8.8
• Imagine that our Sun ran out of nuclear fuel and
collapsed. The Sun's current period of rotation is
25 days.
• What would the Sun's radius have to be for its
period of rotation to be the same as during the
pulsar described earlier?
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Right-hand rule for determining the
direction of rotational velocity and rotational
momentum
• Curl the four fingers of your
right hand in the direction of
rotation of the turning object.
Your thumb points in the
direction of both the object's
rotational velocity and its
rotational momentum.
• Curl the fingers of your right
hand in the direction of the
object rotation caused by
that torque. Your thumb
shows the direction of this
torque.
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Stability of rotating objects
• If the rider's balance shifts a bit,
the bike + rider system will tilt and
the gravitational force exerted on it
will produce a torque.
– The rotational momentum of
the system is large, so torque
does not change its direction
by much.
– The faster the person is riding
the bike, the greater the
rotational momentum of the
system and the more easily the
person can keep the system
balanced.
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8.6 Rotational Energy
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Rotational kinetic energy
• We are familiar with the kinetic energy of a
single particle moving along a straight line or in
a circle.
– It would be useful to calculate the kinetic
energy of a rotating body. Doing so would
allow us to use the work-energy approach to
solve problems involving rotation.
• We will test an analogous expression for
rotational kinetic energy:
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Rotational kinetic energy
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Rotational kinetic energy
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Rotational kinetic energy
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Physics 1D03
Rolling Motion
• Combined translational and rotational motion
• “Rolling without slipping”
• Dynamics of rolling motion
Rolling Motion
• Rolling is a combination motion in which an
object rotates about an axis that is moving along
a straight-line trajectory.
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General Motion of a Rigid Body
 F  ma
CM
 CM  ICM 
Kinetic energy:
Gives linear acceleration of the position of the
center of mass.
Gives angular acceleration of the body about
the center of mass.
2
K  K trans  K rot  12 mvCM
 12 I CM  2
These are special properties of the centre of mass only
(although   I is also true for any stationary axis of rotation).
Physics 1D03
Rolling Motion
Different points on a rolling object
have different velocities (blue
vectors). The velocity of each
particle (blue) is the velocity of the
centre (red) plus the velocity r
relative to the centre (green):

C
v = vCM + rω
The point in contact with the ground is momentarily
stationary. The point on the top of the wheel moves
forward at twice the speed of the centre.
Physics 1D03
vC
Rolling Motion
• The figure above shows exactly one revolution for a
wheel or sphere that rolls forward without slipping.
• The overall position is measured at the object’s
center.
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Rolling Motion
• In one revolution, the center moves forward by
exactly one circumference (Δx = 2πR).
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Rolling Motion
• Since 2π/T is the angular velocity, we find
• This is the rolling constraint, the basic link
between translation and rotation for objects that roll
without slipping.
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Rolling Motion
• The point at the bottom of the wheel has a
translational velocity and a rotational velocity in
opposite directions, which cancel each other.
• The point on the bottom of a rolling object is
instantaneously at rest.
• This is the idea behind “rolling without slipping.”
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Rolling Without Slipping Radian Calculations
If the body rolls without slipping, then its angular velocity and
angular acceleration are related to the linear velocity and tangential
acceleration of the centre:
v=r
and
a=r
Proof: When the object turns through an angle , it moves forward
a distance s = R (if it doesn’t slip).
s = r ;
r
and

s = r
Physics 1D03
vcentre = r
acentre = r
Quiz
There is friction between a wheel and the road,
otherwise a car could not accelerate. This friction is
what causes a rolling wheel to stop.
a) True
Physics 1D03
b) False
Friction and Rolling Resistance
Friction is necessary to create a torque
which can cause rotation and rolling.
This is static friction if the wheel
rolls without slipping. At the point
of contact between the road and
wheel, there is no relative motion,
and friction does no work.
pull
R
f
  Rf
“Rolling resistance” which slows the wheel results from the tire
(and the road surface) flexing inelastically at the contact point.
This can be a small effect, so wheels roll a long way.
Physics 1D03
Symmetrical rolling objects (CM in centre):
1. Dynamics: Use aCM  R
CM motion:
F  maCM
  I
Rotation about CM:
2. Energy:
these are related by
the condition for rolling
Use vCM  R
Translational energy from CM:
Ktrans  mvCM
1
2
Rotational energy about CM axis: K rot  12 I
2
2
K  mvCM  12 I 2
1
2
2
These are related by
the condition for rolling
Physics 1D03
Example: A cylinder (mass m, radius R) rolls down a ramp
without slipping. What is its speed after it has descended a
vertical height h?
n
f
h
mg
Which forces do work?
Physics 1D03
How would you calculate the force of static friction?
forces (x components): mg sin  – f = maCM
torques (clockwise):
aCM= R
rolling:
solve (exercise for student!) ...
y
N
f
a
mg
Physics 1D03
Rf = ICM 
x
f 
mg sin 
1  mR 2 I CM 
Example: A solid cylinder and a
hollow pipe roll down a ramp
without slipping.
a) Which gets to the bottom first?
b) If they have equal masses,
which has the greatest kinetic
energy at the bottom?
c) If the surface were slippery, would the time increase or
decrease?
d) For a moderate coefficient of friction, which can roll down the
steepest slope without slipping?
Physics 1D03
d)
mg sin 
f 
1  mR 2 I CM 
Icyl = ½ mR2
Ipipe = mR2
So f is larger if I is larger.
So the pipe needs more
friction than the solid
cylinder.
y
N
f
a
mg
Physics 1D03
x
Summary
For arbitrary motion of a rigid body, divide the motion into a linear
motion of the centre of mass, plus rotation about the centre of
mass. Then:
 F  ma
  I
CM
CM
CM

2
K  K trans  K rot  12 mvCM
 12 I CM  2
Rolling without slipping:
Physics 1D03
v=R
at = R 
Flywheels for storing and providing energy
• In a car with a flywheel, instead of rubbing a
brake pad against the wheel and slowing it
down, the braking system converts the car's
translational kinetic energy into the rotational
kinetic energy of the flywheel.
• As the car's translational speed decreases, the
flywheel's rotational speed increases. This
rotational kinetic energy could then be used later
to help the car start moving again.
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Example 8.9
• A 1600-kg car traveling at a speed of 20 m/s
approaches a stop sign. If it could transfer all of
its translational kinetic energy to a 0.20-m-radius,
20-kg flywheel while stopping, what rotational
speed would the flywheel acquire?
© 2014 Pearson Education, Inc.
Example 8.9
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Example 8.9
© 2014 Pearson Education, Inc.
Example 8.9
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Rotational motion: Putting it all
together—Tides and Earth's day
• Point A is closer to the Moon than point B, so the
gravitational force exerted by the Moon on point A is
greater than that exerted on point B.
– Due to the difference in forces, Earth elongates along
the line connecting its center to the Moon's center.
– This makes water rise to a high tide at point A and
(surprisingly) at point B. The water "sags" a little at
points C and D, forming low tides at those locations.
© 2014 Pearson Education, Inc.
Summary
© 2014 Pearson Education, Inc.
Summary
© 2014 Pearson Education, Inc.
Summary
© 2014 Pearson Education, Inc.
OK Lets Put It All Togheter
•
•
•
•
•
Torque
Torque and Forces and Static Equilibrium
Axis of Rotation and Agular Properties
Rotational Kinematics (Fab four for rotation)
Rotational Dynamics (translational Force and
rotational motion torque). This includes rotational
inertia.
• Rotational Kinetic Energy
• Angular/Rotational Momentum
© 2014 Pearson Education, Inc.
Questions
© 2014 Pearson Education, Inc.
Example 7.5 Finding the speed at two points
on a CD
The diameter of an audio compact disk is
12.0 cm. When the disk is spinning at its
maximum rate of 540 rpm, what is the
speed of a point (a) at a distance 3.0 cm
from the center and (b) at the outside
edge of the disk, 6.0 cm from the center?
PREPARE Consider two points A and B
on the rotating compact disk in
FIGURE 7.7. During one period T, the disk rotates
once, and both points rotate through the same angle,
2π rad. Thus the angular speed, ω = 2π/T, is the same
for these two points; in fact, it is the same for all points
on the disk.
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Example 7.5 Finding the speed at two points
on a CD (cont.)
But as they go around one time,
the two points move different
distances. The outer point B goes
around a larger circle. The two
points thus have different speeds.
We can solve this problem by first
finding the angular speed of the
disk and then computing the speeds at the two
points.
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Example 7.5 Finding the speed at two points
on a CD (cont.)
We first convert the
frequency of the disk to rev/s:
SOLVE
We then compute the angular speed using
Equation 7.6:
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Example 7.5 Finding the speed at two points
on a CD (cont.)
We can now use Equation 7.7
to compute the speeds of points
on the disk. At point A,
r = 3.0 cm = 0.030 m, so the
speed is
At point B, r = 6.0 cm = 0.060 m, so the speed at
the outside edge is
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Example 7.5 Finding the speed at two points
on a CD (cont.)
The speeds are a few
meters per second, which seems
reasonable. The point farther
from the center is moving at a
higher speed, as we expected.
ASSESS
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QuickCheck 7.7
• This is the angular velocity graph of a wheel.
How many
revolutions does the wheel make in the first 4 s?
–
–
–
–
–
1
2
4
6
8
© 2015 Pearson Education, Inc.
QuickCheck 7.7
• This is the angular velocity graph of a wheel.
How many
revolutions does the wheel make in the first 4 s?
–
–
–
–
–
1
2
4
6
8
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 = area under the angular velocity curve
QuickCheck 7.9
• Starting from rest, a wheel with constant angular
acceleration turns through an angle of 25 rad in
a time t. Through what angle will it have turned
after time 2t?
–
–
–
–
–
25 rad
50 rad
75 rad
100 rad
200 rad
© 2015 Pearson Education, Inc.
QuickCheck 7.9
• Starting from rest, a wheel with constant angular
acceleration turns through an angle of 25 rad in
a time t. Through what angle will it have turned
after time 2t?
–
–
–
–
–
25 rad
50 rad
75 rad
100 rad
200 rad
© 2015 Pearson Education, Inc.