Transcript 141S13-NotesCh7a-June10

Chapter 7
Circular Motion and Gravitation
Our discussion of motion up until now
has been concerned entirely with
translation, in which an object moves
along a straight or curved line. In this
chapter, we will introduce the concept of
circular motion, in which an object
follows a path with a constant radius of
curvature.
After introducing the rotational variables
(angular displacement, angular velocity,
etc.), we will develop equations to
describe the evolution of these variables
with time. Finally, we will discuss a
rotational version of Newton’s 2nd law,
and its application to gravitation.
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Slide 1
7.1 Angular Measure
In order to quantify circular motion, we need to introduce new
units of angular measure. Since a circle lies in a plane, we need
to work in two dimensions. As we
know, any point (P) in 2D can be
described by x- and y-coordinates
(these are known as Cartesian
coordinates). However, circular
motion is best described using polar
coordinates, r and 𝜃. r is the
distance of P from the origin, while
𝜃 is its angle, measured
counterclockwise from the positive
x-axis.
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Slide 2
7.1 Angular Measure
From the figure (and basic trigonometry…the same math that we
used when describing vector components in chapter 3), we see
that
𝑥 = 𝑟 cos 𝜃
𝑦 = 𝑟 sin 𝜃
𝑟=
𝜃=
𝑥2 + 𝑦2
𝑦
−1
tan
𝑥
(as usual, we need to make sure that
the correct solution of the
arctangent function is used)
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Slide 3
7.1 Angular Measure
If an object is moving in a circular path, then r is constant, and
only 𝜽 changes with time. For many problems, we are therefore
only concerned with the latter variable.
In chapter 2, we defined displacement (which, in the present
context, should really be termed linear displacement): ∆𝑥 = 𝑥 −
𝑥0 . Analagously, we can define angular displacement:
∆𝜃 = 𝜃 − 𝜃0
(often, we let the initial angle 𝜃0 be zero, in which case we
simply have ∆𝜃 = 𝜃).
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Slide 4
7.1 Angular Measure
We are all familiar with the most common unit for measuring
angles – the degree (there are 360° in one full revolution).
However, a much more useful unit is the radian (rad). This unit
arises when one considers the relation between angular
displacement and the arc length (s) over which a particle lying a
distance r from the origin travels.
By definition, the angle in radians is
the ratio of s to r. That is,
𝑠 = 𝑟𝜃
As the ratio of two distances, the
radian is actually dimensionless (the
text also calls it unitless, but this is
purely an issue of terminology. I’d
argue that “radian” is the unit).
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Slide 5
7.1 Angular Measure
Converting between radians and
degrees is quite simple. We know that
the circumference of a circle is 2𝜋𝑟.
Therefore, in one full revolution, there
𝑠
2𝜋𝑟
are 360° and 𝜃 = =
= 2𝜋
𝑟
𝑟
radians.
For angles that are simple multiples or
fractions of π, it is common to avoid
writing out the digits. For example,
you would say that 1.5 revolutions is
“3𝜋 radians” rather than “9.42
radians”, although the latter is correct.
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Slide 6
Problem #1: Car Tire Revolutions
WBL Ex 7.9
A car with a 65-cm-diameter wheel travels 3.0 km. How many revolutions does
the wheel make in this distance?
Solution: In class
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Slide 7
7.2 Angular Speed and Velocity
Angular speed and velocity are defined analogously to their
translational counterparts from chapter 2. Average angular
speed is calculated as
∆𝜃 𝜃 − 𝜃0
𝜔=
=
∆𝑡
𝑡 − 𝑡0
(as usual, the overbar indicates an average quantity. 𝜔 is the
Greek lower-case letter “omega”). The SI unit for 𝜔 (and all
other angular speeds and velocities) is the radian per second. If
we consider only an infinitesimally short time interval, we have
the instantaneous angular speed, 𝝎.
If 𝜔 is constant, then 𝜔 = 𝜔. Then, if we set 𝜃0 and 𝑡0 to be
zero, we simply have
𝜃
𝜔=
or 𝜃 = 𝜔𝑡
𝑡
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Slide 8
7.2 Angular Speed and Velocity
There is another common unit for angular speed, the revolution
per minute (rpm). To convert between the two:
1 revolution 1 min
min
60 s
2π rad
𝜋
=
rad/s ≈ 0.105 rad/s
1 revolution
30
The average and instantaneous angular velocity are vectors;
they imply both an angular speed and an angular direction.
Recall from chapter 2 that in one dimension, velocities can only
be positive or negative (depending on whether the object is
moving toward increasing or decreasing values of x). Similarly,
for angular motion, there are only two possible “directions”…
toward increasing or decreasing values of 𝜃.
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Slide 9
7.2 Angular Speed and Velocity
Actually defining the direction of angular motion is a bit tricky.
Picture a few ants sitting on a turntable, which lies in the xyplane. At any moment in time, each ant is moving in a different
direction. However, all of them are rotating either clockwise or
counterclockwise (I’m pretty sure that turntables rotate
clockwise, but I haven’t owned a turntable since 1989 and I’m
not willing to hit up a club just for the sake of making these
notes).
However, we can’t really say that the motion is “clockwise”,
because this depends on which side of the turntable we’re
looking at. If we were to somehow get below the turntable and
then look back up at it, we would claim that it rotates
counterclockwise.
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Slide 10
7.2 Angular Speed and Velocity
To remove this ambiguity, we apply the
right-hand rule, which is familiar to
anyone who has learned about the crossproduct of vectors. Curl the fingers of
your right hand in the direction of the
rotation. 𝝎 points in the direction of
your right thumb. This is how we define
the “direction” of angular velocity – even
though no portion of the rotating object
actually moves in this direction at any
time!
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Slide 11
7.2 Angular Speed and Velocity
Relation between Tangential and Angular Speeds
A particle moving in a circle has an instantaneous velocity
tangential to its circular path. The direction of this velocity is
constantly changing, and is generally irrelevant. However, the
tangential speed (𝑣𝑡 ) is quite important for various reasons.
For now, we will assume that the angular velocity is constant.
Combining equations from slides 5 and 8, we find that the arc
length is
𝑠 = 𝑟𝜃 = 𝑟(𝜔𝑡)
Then, because distance is speed multiplied by time (when speed
is constant), we can also write the arc length as 𝑠 = 𝑣𝑡 𝑡.
Combining these equations produces
𝑣𝑡 = 𝑟𝜔
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Slide 12
7.2 Angular Speed and Velocity
This equation tells us that, while all particles in a rotating object
have the same angular velocity, their tangential speeds depend
on their distance from the origin.
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Slide 13
7.2 Angular Speed and Velocity
Period and Frequency
Circular motion with constant angular speed results in a periodic
repeating of the same motion – all particles in a rotating object
simply travel along the same circular path over and over again.
The time required to complete one rotation (or cycle) is called
the period (T) of the motion. For example, the period of the
Earth’s rotation about its axis is 24 hours†.
A related parameter is the frequency (f) of the circular motion.
This is the number of revolutions in a given amount of time. the
SI unit for frequency is s-1 (the “inverse second”), which is also
called the Hertz (Hz). An object that completes 10 revolutions in
2 seconds has T = 0.2 s and f = 5 Hz.
†more
or less (don’t ask)
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Slide 14
7.2 Angular Speed and Velocity
Based on their definitions, it’s clear that the two quantities
defined on the previous page are simply the inverse of each
other:
1
𝑓=
𝑇
The tangential speed is related to the period by 𝑣𝑡 = 2𝜋𝑟/𝑇.
Since 𝑣𝑡 = 𝑟𝜔, we can also write that
2𝜋
𝜔=
= 2𝜋𝑓
𝑇
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Slide 15
Problem #2: Merry-Go-Round
WBL Ex 7.27
A little boy jumps onto a small merry-go-round (radius of 2.00 m) in a park and
rotates for 2.30 s through an arc length distance of 2.55 m before coming to rest.
If he landed (and stayed) at a distance of 1.75 m from the central axis of rotation
of the merry-go-round, what was his average angular speed and average
tangential speed?
Solution: In class
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Slide 16
Problem #3: Measuring the Speed of Light
The speed of light is measured as shown in the figure.
A slotted wheel is rotated while light passes through
the slots. The light travels to a mirror a distance L
away and then returns to the wheel just in time to
pass through the next slot. For this particular wheel,
the radius is 5.0 cm and there are 500 slots around the
edge. The mirror is L = 500 m from the wheel. The
speed of light is measured at 3 x 108 m/s. What is the
constant angular speed of the wheel, and what is the
linear speed of a point on the edge of the wheel?
Solution: In class
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Slide 17
7.3 Uniform Circular Motion and Centripetal Acceleration
Uniform circular motion occurs when an object moves at a
constant speed in a circular path. Note that the object does not
need to complete any full revolutions.
Although the object’s speed is constant, its velocity is not, since it
is constantly
changing direction.
Therefore, there must
be an acceleration
associated with this
motion.
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Slide 18
7.3 Uniform Circular Motion and Centripetal Acceleration
What is the direction of this acceleration? It’s definitely not in
the direction of the instantaneous velocity (tangent to the
circular path), since this would imply that the object is speeding
up – the circular motion wouldn’t be uniform. In fact, as the
figure indicates, ∆v, and therefore the acceleration, are directed
radially inward, toward
the center of the circle.
This is termed centripetal
acceleration, a𝒄 (the
term “centripetal” means
“center-seeking”).
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Slide 19
7.3 Uniform Circular Motion and Centripetal Acceleration
The magnitude of centripetal acceleration, ac, is derived on p.
231 of the text. It can be expressed as a function of tangential
speed and the circle’s radius:
𝑣2
𝑎𝑐 =
𝑟
or as a function of angular speed and radius:
𝑎𝑐 = 𝑟𝜔2
(the text drops the subscript on tangential speed at this point).
These equations beg the question: is centripetal acceleration
directly proportional, or inversely proportional to the radius?
We’ll discuss it in class.
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Slide 20
7.3 Uniform Circular Motion and Centripetal Acceleration
Centripetal Force
Newton’s 2nd law tells us that where there’s an acceleration,
there must be a force. Since F = ma, we can define the
centripetal force:
𝑚𝑣 2
𝐹𝑐 = 𝑚𝑎𝑐 =
= 𝑚𝑟𝜔2
𝑟
(the direction of Fc is the direction of ac radially inward). Since
the centripetal force is perpendicular to the instantaneous
velocity (which is in the tangential direction), it can do no work
on the object. Therefore, the object’s KE does not change, and
the object’s speed remains constant. Which is a requirement for
uniform circular motion. Everything checks out!
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Slide 21
7.3 Uniform Circular Motion and Centripetal Acceleration
Centripetal Force cont’
It is important to note that the centripetal force is not a result of
the uniform circular motion, it is the cause. If an object is
moving along a circular path, there must be a reason. For
example, a satellite orbiting the Earth maintains a circular path
because of the gravitational force (which points inward, toward
the center of the Earth). A rock that is swung at the end of a
rope is subjected to a (radially inward) tension force along the
rope. A car that travels around a curve experiences (radially
inward) static frictional forces between the tires and the road.
Should fs,max be reached, the car will skid in a straight, tangential
line. If the rope breaks, the rock is flung along a straight,
tangential line. If the Earth disappears, we’re all screwed the
satellite will travel in a straight line.
PC141 Intersession 2013
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Slide 22
Problem #4: Uniform Circular Motion
WBL LP 7.9
In uniform circular motion, there is a…
PC141 Intersession 2013
A
…constant velocity
B
…constant angular velocity
C
…zero acceleration
D
…nonzero tangential acceleration
Day 14 – June 10 – WBL 7.1-7.3
Slide 23
Problem #5: Stunt Flying
WBL Ex 7.33
An airplane pilot is going to demonstrate flying in a tight vertical circle. To ensure
that she doesn’t black out at the bottom of the circle, the acceleration must not
exceed 4.0g. If the speed of the plane is 50 m/s at the bottom of the circle, what
is the minimum circle radius so that the 4.0g limit is not exceeded?
Solution: In class
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Slide 24
Problem #6: Loop-the-Loop
WBL Ex 7.41
A block of mass m slides down an inclined plane
into a loop-the-loop of radius r.
a) Neglecting friction, what is the minimum
speed the block must have at the highest
point of the loop in order to stay in the loop?
b) At what vertical height on the inclined plane
(in terms of r) must the block be released if it
is to have the required minimum speed at the
top of the loop?
Solution: In class
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Slide 25
Problem #7: Swinging a Rock
A string 1.00 m in length has a breaking
strength of 50.0 N. A rock with mass 0.10 kg is
tied to one end. The other end is held tightly,
and the rock is swung in circular motion along
a horizontal plane. What is the maximum
angular velocity, in rpm, that the rock can be
swung without breaking the string?
Solution: In class
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Slide 26