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Chapter Eight Lecture
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Round & Round We Go
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 Formulas:
Demonstrations:
- Conservation of Angular Momentum
- Changing Torque produces change in momentum
 Main Ideas:
- Angular Quantities
- Rotational Kinematics
- Torque and Rotational Inertial
- Rotational Kinetic Energy
- Angular Momentum
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Angular Quantities
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This chapter will again deal with rotation, like chapter 5 did. We will
look rigid bodies that have purely rotational motion. Rigid bodies are objects that have a definite shape. Purely rotational motion means that
all points in the object rotate in circles. They all rotate around a fixed axis of rotation. The concepts in this chapter are very similar to the concepts we learned in chapter 2 through 7 regarding Kinematic Motion in
one and two dimensions, Newton's Laws, Conservation of Mechanical Energy, and Conservation of Momentum. As we develop concepts in this chapter, you should try to relate the concepts to those in previous
chapters. It will help you understand this chapter.
Angular Displacement
 When an object moves along a straight path, we describe how far it has moved by its displacement. When an object rotates we describe
how far it has rotated by its angular displacement q.
The mathematics of circular motion is much simpler if we measure the angle in radians rather than degrees.
One radian as defined as an angle whose arc length is equal to its radius, or in general:
where r is the radius of the circle and l is the arc length subtended by the angleq. If the angle does a complete revolution, then l = 2pr, which is
the circumference. So 360° = 2p radians, and 1 radian is about 57.3°. Note that as the object rotates, every point on the object undergoes the exact same angular displacement. The angular displacementq, will play the same role in angular kinematics that the displacement x, played
in linear kinematics. It is customary to set counterclockwise rotations to have a positive angular displacement and clockwise rotations to have a negative angular displacement.
Angular Velocity
The angular velocity is defined in relationship to the angular
displacement in the same way that the linear velocity was defined in relationship to the linear displacement. The average angular velocity is
given by the Greek letter omega (w), and is defined as the rate of change of the angular displacement.
The instantaneous angular velocity is given by;
Angular Acceleration
Likewise, the average angular acceleration is defined as the rate of
change of the angular velocity, and is given by the Greek letter alpha (a).
and the instantaneous angular acceleration is given by;
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Problem: A compact disc player starts a disc from rest and accelerates to its
final velocity of 3.50 rev/s in 1.50s. What is the disk's average angular acceleration?
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Problem: The blades of a blender rotate at a rate of 7500rpm. When the motor is
turned off during operation, the blades slow to rest in 3.0 seconds. What is the angular acceleration?
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Relationship Between Angular & Linear Quantities
Before using these angular quantities let's look at the relationship
between them and the linear quantities we have used. We know that velocity is change in distance (l here) divided by time.
Also, the tangential component of the acceleration is;
Recall the definition of centripetal acceleration.
and since aT and ac are at right angles to each other, the magnitude of the total acceleration is given by,
Other Useful Definitions
Frequency is defined as the number of revolutions per unit time. To
change from frequency to angular frequency, I must multiply frequency by 2p since 1 revolution is 2p radians. Also (1 rev/s)*(2p radians/rev) gives
radians/sec - which is the units of angular velocity.
The period is the time required to make one complete revolution, so;
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 Problem: How fast is the outer edge of a
CD (at 6.0 cm) moving when it is rotating at its top speed of 22.0 rad/s?
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Rotational Kinematics
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The definitions of the angular quantities are analogous to the definitions of linear quantities with q playing the role of x, w playing the
role of v, and a playing the role of a. Consequently, the kinematic
equations derived in chapter 2 are valid for rotational motion with constant angular acceleration using the same derivations done in chapter 2. We get,
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 Problem: How many rotations does the
CD from the first problem make while coming up to speed from rest to wf =
22.0 rad/sec at a= 14.7 rad/s2.
Recall, we had
wf = 22.0 rad/sec, wo= 0.0 rad/sec, and a = 14.7 rad/s2
We want to find q so we use;
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Problem: A wheel with radius 0.5m makes 55 revolutions as it changes speed from
80km/h to 30 km/h. The wheel has a diameter of 1 meter. (a) What was the angular acceleration? (b) How long is required for the wheel to come to a stop if it
decelerated at that rate?
There is a bunch of work and conversions we need to do before we can answer
the questions. We need the proper units of the variables to use in the rotational kinematic equations.
Now, q = the spun angular displacement of the wheel during the revolutions;
(a) Now we can finally begin to use the rotational kinematic equations.
(b) To calculate the time this requires we can use;
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Torque & Rotational Inertia
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We have talked about the description of angular motion. Now we talk
about the causes or dynamics of angular motion. What causes something to rotate? For translational motion, we found that a force
causes an acceleration. That is a force caused an object to move. For rotational motion, it also takes a force to start something rotating.
However, there is more to it than that. The location that I apply the force is important. If I want to start a wheel rotating, I cannot apply the
force at the axel of the wheel. It will not rotate. To start something rotating, I must apply a force that is some distance from the axis of
rotation. The perpendicular distance from the axis of rotation is called the lever arm. The torque is defined as the force times the lever arm.
 In general we write torque as:
where q is the angle between the
direction of the force and a line drawn from the axis of rotation to the force. If I were to push along the direction of F||then there would be no rotation around the hinge, since the force is directed right through the axis of rotation. If I were to push along F^ then there would be a rotation. Consequently, the only component of the force F
which causes a rotations is the component perpendicular to the lever arm (r), which is given by Fsinq. So what produces a greater torque? If I
increase the lever arm I get a greater torque. I need to do this to take off the drain plug from my car. If I increase the force, the torque will
increase. I can also maximize the torque by making the angle q=90°.
Because torque has a distance in its definition (the lever arm), the torque is only defined around a certain axis of rotation. The torque around two different axes of rotation may be very different.
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Problem: A crane picks up a heavy steel beam. The tension in the cable is 2.0´104N. What is the torque around point O?
 
in the counter-clockwise direction.
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Problem: Calculate the torque around an axis perpendicular to the paper
through (a) point O, and (b) point C.
Around O:
The lever arm is 0 from the 30 N force
Around C:
t = (.02 m)(10 N)(sin(20°)) - (.02 m)(30 N)(sin(45°)) = -.36 N×m
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Recall in our study of linear motion that we said that the property of
a body that resisted a change in velocity was called mass. Even if there were "no gravity," the mass of a body would resist a change in velocity. That is what Newton's Second Law says SF = ma. There is also a property of a body that resists a change in angular velocity. It is called
the moment of inertia. If a body has a large moment of inertia, then it is difficult to change its angular velocity. If it has a small moment of
inertia, it is easier to change its angular velocity. The moment of inertia for any object depends on a number of factors including the object's
mass, its shape, and the axis of rotation. Let's first calculate the moment of inertia for a simple object like a sphere at the end of the string. Since the sphere has much more mass than the string, we will
neglect the mass of the string. We look at the tangential forces and accelerations.
FT = maT
FT = mra
rFT = mr2a
t = mr2a
The quantity is called the moment of inertia and given the symbol I. So this becomes t = Ia. If we have more then one torque, we get
which is the rotational equivalent of Newton's Second Law. The reason we write I rather than mr2 is because, as I said, I depends on different factors and although it is mr2 for a sphere on a string, it is not mr2 for every object. In general, we can write the moment of inertia for a
collection of particles as
I = Smr2
This can be used to calculate the moment of inertia of a baton, for instance, but not for other extended objects. In that case, calculus is
needed. Page 204 lists the moment of inertia for a few shapes. In any problem you have to do, you will be given the equation for the moment of inertia.
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 Problem:
What is the moment of inertia for a baton with each end weighing 0.3kg spinning around its center. The baton is
1.0m long. Neglect the mass of the bar.
I = Smr2
I = (0.3 kg)(0.5 m)2 + (0.3 kg)(0.5 m)2
I = 0.15 kg×m2
 Note that the moment of inertia
depends on where the axis is. If I rotate around one end of the baton, then r for that end is 0, and the moment of inertia becomes:
I = Smr2 = 0 + (0.3 kg)(1.0 m)2
I = 0.3 kg×m2
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Another quantity is the radius of gyration. It is given the symbol k,
and is defined as the location on the object that would have the same moment of inertia as the original object if all the mass where located at
k. The relationship between the moment of inertia and the radius of gyration is always given by
Let's now solve some problems using Newton's second law in this
form for rotational motion. We use it the same way as we used Newton's second law. We must draw a free body diagram, determine the torques and the accelerations, and solve for the unknown quantities.
Up until now we have not been concerned about where a force acted
on an object. But because torques produce rotation around some axis a certain distance from where the force is applied, it is important to look at
where a force is applied on an object. For the force of gravity, we assume the force is applied at the center of gravity of an object. Other
forces, like tension, friction, and normal forces are applied where the objects contact.
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Problem: A cylindrical 3.00kg pulley with a radius of R=0.400m is
used to lower a 2.00kg bucket into a well. The bucket starts from rest and falls for 3.00s. (a) What is the linear acceleration of the falling
bucket? (b) How far does it drop? (c) What is the angular acceleration of the cylinder?
Draw Free Body Diagrams!
(a) Note that aT = a, and setting positive direction upward.
Forces on the bucket: SF = ma Þ T - mg = -ma
Torques on the wheel: St= IaÞTRsin(90o) = Ia = (1/2)(MR2)(a/R)
We have two equations and two unknowns (a and T), so now it is just algebra.
From torque equation, T = (1/2)Ma.
Plug this into force equation:
(b) y = vt + (1/2)at2
(c) a = (a/R) = (5.60 m/s2)/(.400 m) = 14.0 rad/s2
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So these problems are solved just like other problems using Newton's second law. Draw force diagrams. Set SF = ma and St = Ia. Break vectors into components if you need to. Solve for the unknown quantities.
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Rolling Motion
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 When an object is rolling we can note some aspects about its motion.
Suppose the object is moving to the right with a speed of v. Look at the linear speed of various points on the object. Where the object
touches the ground its linear speed is zero. At the center of the wheel the linear speed is v and at the top of the wheel the velocity is 2v. The angular velocity, then, is given by
where v is the linear speed of the wheel. What causes the wheel to
rotate? It is the force of static friction on the bottom. It is static friction because the place where the wheel touches does not move sideways with
respect to the ground. The static friction produces a torque around the middle point of the wheel. That is the axis of rotation.
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Rotational Kinetic Energy
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We have learned that when something is moving it has a kinetic energy which is equal to (1/2)mv2. When something is rotating it has a kinetic energy which is equal to (1/2)Iw2. If it has both translational and
rotational motion, then it has both forms of kinetic energy, as well.
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 Problem: Two bicycles
roll down a hill that is 20m high. Both bicycles have a total mass of 12kg and 700 mm diameter wheels (r=.350 m). The first bicycle has
wheels that weigh 0.6kg each, and the second bicycle has wheels that weigh 0.3kg each. Neglecting air resistance, which bicycle has the faster speed at
the bottom of the hill? (Consider the wheels to be thin hoops).
 The only friction is static friction, so there
are no non-conservative forces. (Static friction involves no motion and since work is defined as W = Fd, when there is no
distance involved, there is no work/energy used).
Mechanical Energy is Conserved, so:
SEi = SEf
KEi +PEi = KEf +PEf
(1/2)mvi2 + (1/2)Iwi2 + mghi = (1/2)mvf2 + 2*(1/2)Iwf2 + mghf
mghi = (1/2)mvf2 + 2(1/2)Iwf2 (The two for two wheels).
Now w is given by w=v/r where v is the velocity of the rim of the
wheel and is the same as the velocity of the bike. I is given by Mr2 where M is the mass of the tire.
mgh = (1/2)mvf2 + Mr2vf2/r2= (1/2)mvf2 + Mvf2
So for the first bike vf = 19.3 m/s
For the second bike, the wheels have a mass of 0.6 kg, and we get
vf = 18.9 m/s
Why is this true? More of the potential energy goes into rotational
kinetic energy and less into translational kinetic energy when the wheel has more mass.
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Angular Momentum
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In the same way that linear momentum is defined as p = mv, so the angular momentum of a rotating object is defined as L = Iw. Similarly, Newton's law for rotational motion can be written as
and just as linear momentum is conserved if there is no net external force, so angular momentum is conserved (it does not change) if there is
no net external torque. The law of the conservation of angular moment says:
The total angular momentum of a rotating body remains constant if the net external torque acting on it is zero.
Some examples of this include an ice skater who is rotating and pulls her
arms in or a diver who goes into a tucked position. Why do they rotate faster?
SLi = SLf
SIwi = SIwf
Smri2wi = Smrf2wf
So as the distance from the axis of rotation decreases, the angular velocity (w) must increase for L to be the same.
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Problem: A student is sitting on a swivel seat and is holding a 2.0kg
weight in each hand. If she is rotating at 1 rev/s (6.28 rad/s) when the weights are held in outstretched arms 0.75m from the axis of rotation, how fast is
she rotating when she pulls the weights in to the axis of rotation? (The rest of her body can be approximated as a cylinder with mass of 72kg and radius of
.25m).
SLi =SLf
IBwi + Iwwi = I Bwf + Iwwf
(I B+Iw)wi = (IB+I w)wf
((1/2)MR2 + 2mri2)wi = ((1/2)MR2 + 2mrf2)wf
wf = {((1/2)MR2 + 2mri2)wi} / {((1/2)MR2 + 2mrf2)}
wf = {(1/2)(72kg)(.25m)2+2(2.0kg)(.75)2)(6.28rad/s)}¸{(1/2)(72kg)(.25m)2+0}
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Vector Nature of Angular Quantities
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The angular equivalent of all of the linear quantities - which are
vectors - are also vectors, (which is all of the quantities we have talked about except moment of inertia). The direction is chosen arbitrarily by
the right hand rule. Your fingers rotate in the direction of the rotational motion and your thumb points in the direction of the vector. Notice that
the relationship between angular and linear quantities (like v =rw, aT =ra, and aC =w 2r) are not vector equations. The linear quantities point
in the direction of motion which changes as the object rotates, but the angular quantities point in the direction given by the right hand rule.
Demonstration: Rotating wheel using L = Iw.
Demonstration: Changing momentum using St = DL/Dt or DL = tDt
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Inertial & Non-Inertial Frames: The Coriolis "Force"
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Any frame of reference where Newton's laws appear to work is called
an inertial frame of reference. Sometimes it may appear that his laws do not work. For instance, in a rotor carnival ride, we may think that we are
being thrown outward. However, there is no force that is throwing us outward, instead we are being accelerated inward by the centripetal
acceleration provided by the wall of the ride. This reference frame, where we think there is a force, but there really is not one is called a
non-inertial reference frame, and the force we think we feel is called a pseudo-force. The Coriolis force is one such force. Suppose I have a
plate rotating around its center. Points on the inside of a plate have a lower tangential velocity than those on the outside of the plate. Consequently, something thrown outward appears to curve in the
opposite direction of rotation. This is an important pseudo-force in understanding weather and in understanding the path of ballistic shells.
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