Wednesday, September 14, 2016
Lecture 7
Lecture 7 – Wednesday, September 14, 2016
Homework: HW7-1—HW7-4 will be due at the beginning of Friday’s lecture (9/16)
OPTIONAL readings:
1) Notes from Cycles 1 and 2: 12-1 to 12-4, 12-8, 13-1 to 13-9
2) OM (Ohanian-Markert): read the chapter subsections introducing circular motion,
angular velocity, angular acceleration, torque, and statics.
3) Chaos reading Reread the Prologue, pp. 1-8, and “Life’s Ups and Downs,” pp. 57-80,
then go on to read “Universality,'' pp. 155-187.
CIRCULAR MOTION
Special 1D motion in 2D plane
We can describe the ant's position by the angle through which it has moved. Indeed,
this is one number, θ (so really 1D). Then the distance along the ring it has moved =
Rθ from the point P. Here REMEMBER that θ must be in radians,
θ (in radians) =
2π
θ (in degrees)
360
Rθ is consistent with what you remember from kindergarten:
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Wednesday, September 14, 2016
Lecture 7
Constant Angular Velocity
(Uniform Circular Motion)
If the ants' speed around the ring is uniform,
distance around ring
= constant
time interval
or
R(θ 2 − θ1 )
= constant = v
t 2 − t1
or
θ 2 − θ1
= another constant = ω , known as angular velocity or angular frequency
t 2 − t1
As a derivative,
distance = R θ ( θ =0 at t=0, say)
v=
d
d
dθ
dR
(distance) = (Rθ ) = R
= 0 therefore…
, since
dt
dt
dt
dt
v = Rω, with ω =
dθ
dt
dθ
is constant, then θ = ωt + c, where c → θ0 ≡ θ (0) for
dt
uniform motion. Note: similar to x = vt + x0
Going backwards, If ω =
Period (T), Cycle Frequency (f), and Angular Frequency (ω)
For uniform motion, if an additional time, Δt, has passed (t -> t + Δt),
then the ant will have gone through an additional angle Δθ=ωt.
One Complete revolution => Δt = T = period and Δθ=2π therefore:
2π = ωT ⇒ T =
⇒f =
ω
2π
2π
,
ω
but T =
1
f
or ω = 2π f
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Wednesday, September 14, 2016
Lecture 7
**************
HW 7-1: a) Given that the angular velocity of the earth about its axis is
7.27 x 10-5 rad/s, prove that one complete rotation of the earth takes (surprise!)
24 hours.
This angular velocity is the angular velocity of every point on and inside
the earth! All move with the same number of radians/sec (or degrees/sec, for
that matter). But the linear speed changes depending on how far you are from
the earth’s rotational axis (through the north and south poles), according to
v=ωR┴, where R┴ is the perpendicular distance (the shortest distance from
wherever you are, to the rotational axis).
b) Given that the linear speed due to the earth’s rotation of a point on the
equator is 463 m/s, find the earth’s radius.
c) What is Cleveland’s linear speed in m/s due to the earth’s rotation?
You need to look up Cleveland’s latitude and you need to calculate R┴, given
your answer in (b) and the latitude λ.
Despite the analogy in the equations to constant 1D velocity equations, we must

realize that in uniform circular motion, there is still acceleration because v is
constantly changing direction! This is the heart of centripetal acceleration and
we will come back to that!
Analogous to how we defined translational acceleration we can now define…
AVERAGE ANGULAR ACCELERATION (α av) (over a given time interval)
net change in a given time interval
given time interval
ω − ω1
= 2
t 2 − t1
=
INSTANTANEOUS ANGULAR ACCELERATION (α)
= instantaneous rate of change of velocity with time
ω2 − ω1
Δω dω
d 2θ
= lim
≡
(t1 ) = 2 (t1 ) ≡ α (t1 )
t2 →t1 t − t
Δt →0 Δt
dt
dt
2
1
α = lim
And just like in the translational case, we look at where there is:
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Wednesday, September 14, 2016
Lecture 7
Constant Angular Acceleration
Suppose: α = constant ≡ α0 =
∴ ω = ω0 + α 0t =
dω
dt
Analogous to a0=const.
dφ
dt
Analogous to v=v0+a0t
And
1
φ = φ0 + ω0t + α 0t 2
2
Analogous to x=x0+v0t+
1
a0t2
2
Note again that these formulas, while useful, ARE NOT GENERALLY RIGHT, since
constant angular acceleration is such a special case. So please do not use these
"famous formulas" in your homework unless either the angular acceleration is really
constant OR you are estimating by sticking an average value for the acceleration in
them.
Examples:
1) α =0
0
second hand of a clock: ω =
2) α ≠ 0
2π
= 0.105 rad s
60 s
1
for f = with T = period of second hand
T
0
(see hw problem below)
*********
1
rpm. When the power to
3
the turntable is turned off, the turntable slows down at a constant rate of 0.20 rad/s2.
(a)
How many seconds elapse before the turntable stops?
(b)
How many revolutions will the turntable make before stopping?
*********
HW 7-2 An old phonograph turntable is turning initially at 33
HAVING INTRODUCED ROTATIONAL MOTION, WE NOW ASK HOW WE CAN
BRING ABOUT THE ROTATION OF A SYSTEM. WE THUS COME TO A
FUNCTION OF FORCE (IT INVOLVES NOTHING NEW – JUST A LINEAR
FUNCTION OF FORCE AND DISTANCE) THAT DIRECTLY TELLS US HOW
EFFECTIVE OUR APPLICATION OF FORCE WILL BE IN STARTING A SYSTEM
TO ROTATE. IT IS …
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Wednesday, September 14, 2016
Lecture 7
TORQUE
Q: What causes linear acceleration in a particular direction?
A: Any net force component in that direction.
e.g., Fx ⇒ max
(Fx = max )
Q: What causes angular acceleration α around some axis?
A: Again, forces, but they have to be applied away from the axis of rotation, and
there has to be a net component pushing the body around the axis
e.g., τ ⇒ α
(τ = I α for simple rotations)
EXAMPLE:
Teeter-totter
effectiveness for rotation ∝ R, F
┴
The formula (see later for more general form in terms of vectors) for calculating the
torque about a given axis is thus ± F R sinθ for a given force (magnitude F) applied at a
point a distance R from the axis, where the angle θ is between the direction of F and the
direction of R (SEE LATER). The + sign (- sign) corresponds to torque tending to
produce CCW (CW) rotations (BUT the opposite convention could also be used).
COMMENT:
Of particular relevance is that there may be important rotational motion changes even
though the total (net) force is zero. As we shall see, the question will be whether the
total (net) torque is zero, or not.
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Wednesday, September 14, 2016
Lecture 7
"STATICS"
Before going on to analyze angular accelerations due to nonzero torques, let us
consider systems that are not moving at all! The fact that nothing is moving means that
both the total (net) force, and the total (net) torque must vanish. Their vanishing tells us
wonderful things about how the whole setup must be arranged!
If everything is always at rest, then we must have Ftotal = 0
and also (about any point!) τtotal = 0
We will see that the torque vector direction is along the “rotational axis” around
which it is trying to rotate the system. But for a single, fixed axis, we only need talk
about one number (one component), which is what is used throughout this lecture (and
much of others).
THE FORMULA FOR THE COMPONENT AS NOTED ABOVE IS ± F R sinθ
TRIVIAL EXAMPLE:
balanced teeter-totter
Look at the net torque about the center where there is no torque due to F since its lever
arm is zero (the other two contributing torques each have 90o angles between the force
and the arm)
0 = τ total = −mgr + MgR ⇒ r =
M
R
m
*********
HW 7-3 A diver of weight 580 N stands at the
end of a 4.5-m diving board of negligible
weight. The board is attached by two
pedestals 1.5 m apart, as shown in the
figure. Find the values of the vertical
forces exerted by the pedestals. What
are the directions of the forces? That is,
tell whether each pedestal is under
tension or compression!
*********
Nonzero torque means nonzero rotational acceleration about the point of torque
calculation; we must have motion about that point. This is what we will discuss in the
next mechanics installment.
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Wednesday, September 14, 2016
Lecture 7
Let’s digress for the moment.
Note reference readings: OPTIONAL: 13-16
OM: Sec. 3.4
VECTOR CROSS-PRODUCT
We have an elegant way of writing the torque, angular momentum, Coriolis force, andthe
velocity of a rotating body (see later). We define a new vector given the two vectors
A, B :
  
C ≡ A × B ≡ vector cross product

 
where C ≡ | C | ≡ ABsinθ ≡ | A || B |sinθ = AB⊥ = A ⊥B if θ is the

 
 
angle between A, B . The direction of C is ⊥ to the plane of A, B using the
right-hand rule to determine to which side it points. (Recall the right-hand rule:
Lay the outside of your forefinger of your right hand along A and your middle


finger along B . Then your thumb will point along C if you let it hang out.)
 
Alternatively, you can use a right-hand screw. Place it perpendicular to the A, B


plane and turn it from A toward B The direction the screw moves (in or out) is

the direction of C
Properties (all of these follow from the above definition!)
1) Examples of
 
A×B
directions : (CW and CCW, respectively)
or
2)
 
A× A=0
3)
 
 
A×B = − B× A
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Wednesday, September 14, 2016
4)
Lecture 7

 
   
A × (B + C) = A × B + A × C
*********
HW 7-4
Prove the law of sines
Hint:
A = B
sinα sinβ
using the cross product.
C × C = 0 = C × (A + B) ⇒ C × A = -C × B
*********
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