CHAPTER
CHAPTER 10
10
Rotation
Rotation Motion
Motion
Dr. Abdallah M. Azzeer
1
10-1. Angular velocity and angular acceleration
A compact disc rotating about a fixed axis
through O perpendicular to the plane of the
figure. (a) In order to define angular position for
the disc, a fixed reference line is chosen. A
particle at P is located at a distance r from the
rotation axis at O. (b) As the disc rotates,
point P moves through an arc length s on a
circular path of radius r.
Angle θ =
s
r
The SI unit of θ is radian (rad), which
is a pure number because it is a ratio.
Dr. Abdallah M. Azzeer
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2
ONE radian is the angle subtended by an arc length equal to the raduis of the arc
2π r
= 2 π rad
r
1 rad = 57.3o = 0.159 rev
1 rev = 360o =
θ (rad ) =
π
θ (deg)
180o
The angle in rad is positive if it is counterclockwise with respect to the
positive x axis.
The angular displacement is :
∆θ = θf − θi
Dr. Abdallah M. Azzeer
3
The average angular speed is :
ω=
θf − θi ∆θ
=
t f −t i
∆t
The instantaneous angular velocity is :
ω = lim
∆t →0
∆θ d θ
=
∆t d t
¾The SI unit of ω is rad/s
¾The angular velocity ω is positive if the rotation is counterclockwise (when
θ is increasing).
Dr. Abdallah M. Azzeer
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The average angular acceleration is :
α =
ωf − ωi ∆ ω
=
t f −t i
∆t
The instantaneous angular velocity is :
α = lim
∆t → 0
∆ω d ω
=
∆t
dt
The SI unit of α is rad/s2
Dr. Abdallah M. Azzeer
5
10-2. rotational kinematics
Rotational Motion with Constant acceleration
Analogy between linear and angular quantities:
r
r
x ¨ θ
r
r
v ¨ ω
r
r
a ¨ α
Dr. Abdallah M. Azzeer
Page 3
6
Example 10.1
A wheel rotates with a constant angular acceleration of 3.50 rad/s2.
(a) If the angular speed of the wheel is 2.00 rad/s at ti=0, through what angle does
the wheel rotate in 2.00s?and (b) how many rev has done during this interval?
1
θ f − θ i = ωt + αt 2
2
1
2
= 2.00 × 2.00 + 3.50 × ( 2.00 )
2
11.0
rev = 1.75 rev
=
2π
o
=630 at
(c) What is the angular speed
t=2.00 s?
ω f = ω i + αt
= 2.00 + 3.50 × 2.00 = 9.00 rad / s
READ the rest of example
Dr. Abdallah M. Azzeer
7
10-3. Relationship Between Angular and Linear
Quantities
¾ The position :
s =θ r
(radian measure )
¾ Note that for all linear-angular
relations, we must use the radian unit.
¾ The speed :
ds dθ
=
r
dt dt
v = rω
(radian measure )
¾ The period of revolution T is
T =
2π r
v
T =
2π
ω
(radian measure )
Dr. Abdallah M. Azzeer
Page 4
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¾ The acceleration
dv
dω
=r
dt
dt
at = rα
(radian measure )
¾ For a particle in a circular path, the centripetal acceleration is :
ar =
v2
= rω 2
r
(radian measure )
¾ ar points radially inward
As a rigid object rotates about a fixed axis through O, the
point P experiences a tangential component of linear
acceleration at and a radial component of linear
acceleration ar. The total linear acceleration of this point
is a = at + ar.
Dr. Abdallah M. Azzeer
9
Differences between ar and at
¾ ar is known as the radial component of linear acceleration, at is known as
the tangential component of the linear acceleration.
¾
ar = rω 2 is pointed radially inward. It is non-zero even if there is no
angular acceleration.
¾
at = rα
is tangential to the rotational path of the particle, it is zero if the
angular velocity is constant.
Total linear acceleration is
a = at2 + a r2 =
( rα )
2
+ ( rω 2 ) = r α 2 + ω 4
Dr. Abdallah M. Azzeer
Page 5
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10
Example 10.2
Audio information on compact discs are transmitted digitally through the readout
system consisting of laser and lenses. The digital information on the disc are
stored by the pits and flat areas on the track. Since the speed of readout system is
constant, it reads out the same number of pits and flats in the same time interval.
In other words, the linear speed is the same no matter which track is played.
(a) Assuming the linear speed is 1.3 m/s, find the angular speed of the disc in
revolutions per minute when the inner most (r=23 mm) and outer most tracks
(r=58mm) are read.
Using the relationship between angular and
tangential speed v = rω
Dr. Abdallah M. Azzeer
r = 23 mm
ω=
v 1.3 m / s
1.3
=
=
= 56.5 rad / s = 9.00 rev / s = 5.4 × 102 rev / min
r
23 mm
23 × 10−3
r = 58 mm
ω=
1.3 m / s
1.3
=
= 22.4 rad / s = 2.1 × 102 rev / min
58 mm
58 × 10−3
11
(b) The maximum playing time of a standard music CD is 74 minutes and 33
seconds. How many revolutions does the disk make during that time?
ω=
(ω i + ω f ) = ( 540 + 210 ) rev / min = 375 rev / min
2
2
375
θf = θi + ω t = 0 +
rev / s × 4473 s = 2.8 × 104 rev
60
(c) What is the total length of the track past through the readout mechanism?
l =vt ∆t = 1.3 m / s ×4473 s = 5.8×103 m
(d) What is the angular acceleration of the CD over the 4473 s time interval,
assuming constant α?
(ω f − ω i ) ( 22.4 − 56.5 ) rad / s
−3
2
α=
∆t
=
4473s
= 7.6 × 10 rad / s
Dr. Abdallah M. Azzeer
Page 6
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1010-4. Rotational Kinetic Energy
¾ We treat the rigid body as a collection of
particles with different speeds, mi is the mass
of the ith particle and vi is its speed.
¾ The particles move with different vi but the
same ω.
Kinetic energy of a masslet, mi, moving
at a tangential speed, vi, is
1
m i v i2
2
Ki =
Since a rigid body is a collection of masslets, the total
kinetic energy of the rigid object is
K = 12 m 1v 12 + 12 m 2v 2 2 + 12 m 3v 3 2 + ⋅ ⋅ ⋅
= ∑ 12 m i v i 2
Dr. Abdallah M. Azzeer
K
R
=
∑K
i
∑
=
i
K
R
==
1⎛
2 ⎜⎝
∑
i
1
2
m iv i 2 =
⎞
m i r i2 ⎟ ω
⎠
13
1
1⎛
⎞
m i r i 2ω 2 = ⎜ ∑ m i r i 2 ⎟ ω 2
∑
2 i
2⎝ i
⎠
2
Since moment of Inertia, I, is defined as
I ≡ ∑mi ri2
Moment of Inertia
i
The above expression is simplified as
1
K R = Iω2
2
Rotational Kinetic Energy:
Dr. Abdallah M. Azzeer
Page 7
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Example 10.3
Oxygen Molecule
d = 1.21 × 10-10 m
mi = 2.66 × 10-26 kg
d
I = ∑ i m i ri2 = m i
( 12 d )
2
+ mi
( 12 d )
2
= 1.95 × 10−46 kg ⋅ m 2
ω = 4.6 × 1012 rad/sec.
K R = 12 Iω 2 = 2.06 × 10−21 J
Cf) Average linear kinetic energy at RT: K L = 12 Mv 2 = 6.02 × 10−21 J
= 3K R
Dr. Abdallah M. Azzeer
15
Example 10.4
In a system consists of four small spheres as shown in the
figure, assuming the radii are negligible and the rods
connecting the particles are massless, compute the
moment of inertia and the rotational kinetic energy when the
system rotates about the y-axis at ω.
Since the rotation is about y axis, the moment of inertia
about y axis, Iy, is
I y = ∑mi ri2 = Ma 2 + Ma 2 + m ⋅ 02 + m ⋅ 02 = 2Ma 2
i
K
R
=
1
1
Iω 2 = ( 2 M a 2 ) ω 2 = M a 2ω 2
2
2
Dr. Abdallah M. Azzeer
Page 8
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Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.
I z = ∑ m i r i 2 = Ma 2 + Ma 2 + mb 2 + mb 2 = 2 ( Ma 2 + mb 2 )
i
1
1
K R = Iω 2 = ( 2Ma 2 + 2mb 2 ) ω 2 = ( Ma 2 + mb 2 ) ω 2
2
2
READ THE REST OF THE EXAMPLE
Dr. Abdallah M. Azzeer
17
1010-5. Calculation of Moment of Inertia
We defined the moment of inertia as
I ≡ ∑ m i ri2
UNIT; kg.m2
i
Moments of inertia for large objects can be computed, if we assume the
object consists of small volume elements with mass, ∆mi.
The moment of inertia for the large rigid object is
I = lim
∆m i →0
∑r
i
2
i
∆m i = ∫ r 2dm
Using the volume density, ρ, replace dm in the above
equation with dV.
ρ=
dm
dV
d m = ρd V
Dr. Abdallah M. Azzeer
Page 9
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The moments of inertia becomes
I = ∫ ρr 2dV
Example 10.5
Find the moment of inertia of a uniform hoop of mass M and radius R about an axis
perpendicular to the plane of the hoop and passing through its center.
The moment of inertia is
I = ∫ r 2dm = R 2 ∫ dm = MR 2
What do you notice from this result?
The moment of inertia for this object is the same
as that of a point of mass M at the distance R.
Dr. Abdallah M. Azzeer
19
Example 10.6
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about
an axis perpendicular to the rod and passing through its center of mass.
The line density of the rod is
so the masslet is
λ=
M
L
dm = λdx =
M
dx
L
The moment of inertia is
I y = ∫ r 2dm =
L /2
x 2M
M ⎡1 3⎤
∫− L / 2 L dx = L ⎢⎣ 3 x ⎥⎦ − L / 2
L /2
3
3
M ⎡⎛ L ⎞ ⎛ L ⎞ ⎤
⎢⎜ ⎟ − ⎜ − ⎟ ⎥
3L ⎣⎢ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦
M ⎛ L3 ⎞ M L2
=
⎜
⎟=
3L ⎝ 4 ⎠
12
=
Dr. Abdallah M. Azzeer
Page 10
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What is the moment of inertia when the rotational axis is at one end of the rod.
L
x 2M
M ⎡1 3⎤
∫0 L dx = L ⎢⎣ 3 x ⎥⎦ 0
M ⎡
M
3
=
( L ) − 0 ⎤⎦ = ( L 3 )
3L ⎣
3L
M L2
=
3
I y ' = ∫ r 2dm =
L
Will this be the same as the above. Why or why not?
Since the moment of inertia is resistance
to motion, it makes perfect sense for it to
be harder to move when it is rotating
about the axis at one end.
Dr. Abdallah M. Azzeer
21
Example 10.7
Inertia of Moment of a uniform solid cylinder
R
Density ρ
r
l
M
ρ ⋅ πR ⋅ l = M ⇒ ρ =
πR 2 l
2
dI = dm ⋅ r 2 = ρ ⋅ 2πr ⋅ dr ⋅ l ⋅ r 2 = 2πρlr 3dr
R
I = 2πρl ∫ r 3dr = 12 ρπlR 4 = 12 MR 2
0
I = 12 MR 2
Dr. Abdallah M. Azzeer
Page 11
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Moments of
Inertia of
Homogeneous
Rigid Objects
with Different
Geometries
Dr. Abdallah M. Azzeer
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