# Chapt. 10 Rotation of a Rigid Object About a Fixed Axis ```Chapt. 10 Rotation of a Rigid Object
10.1 Angular Position, Velocity, and
Acceleration
Translation: motion along a straight line (circular motion ?)
Rotation: surround itself, spins
rigid body: no elastic, no relative motion
rotation: moving surrounding the fixed axis, rotation axis, axis of rotation
s
360
, 1rev = 360 o = 2π (rad ) , 1rad =
= 57.3 o
r
2π
Angular displacement: ∆θ = θ 1 − θ 2
Angular position: θ =
An angular displacement in the counterclockwise direction is positive
θ − θ 1 ∆θ
=
Angular velocity: averaged ω = 2
t 2 − t1
∆t
instantaneous: ω =
dθ
, rpm = rev per minute
dt
Angular acceleration: averaged α =
instantaneous: α =
∆ω
∆t
dω
dt
Sample Problem:
The disk in Fig. 11-5a is rotating about its central axis like a merry-go-round. The
angular position θ(t) of a reference line on the disk is given by
θ = −1.00 − 0.600t + 0.250t 2 with t in seconds, q in radians, and the zero angular
position as indicated in the figure.
a) Graph the angular position of the disk versus time from t = -3.0 s to t = 6.0 s.
Sketch the disk and its angular position reference line at t = -2.0 s, 0 s, and 4.0 s,
and when the curve crosses the t axis.
b) At what time tmin does θ(t) reach the minimum value? What is that minimum
value?
c) Graph the angular velocity ω of the disk versus time from t = -3.0 s to t = 6.0 s.
Sketch the disk and indicate the direction of turning and the sign of ω at t = -2.0 s
and 4.0 s, and also at tmin.
1
Right-hand rule
10.2 Rotational Kinematics: The Rigid
Object Under Constant Angular
Acceleration
Sample Example: A grindstone rotates at constant angular acceleration α = 0.35 rad/s2.
At time t = 0, it has an angular velocity of ω0 = -4.6 rad/s and a reference line on it is
horizontal, at the angular position θ0 = 0.
α=
dω
= 0.35rad / s 2 , dω = 0.35dt , ω − ω 0 = 0.35t , ω = −4.6 + 0.35t
dt
dθ
= ω = −4.6 + 0.35t , θ − θ 0 = −4.6t + 0.175t 2 , θ = −4.6t + 0.175t 2
dt
a) At what time after t = 0 is the reference line at the angular position θ = 5.0 rev?
c) At what time t does the grindstone momentarily stop?
2
10.3 Angular and Translational Quantities
The position: s = rθ
The speed (velocity?): v = rω , T =
2πr
v
v2
The acceleration: a t = rα , a r =
= rω 2
r
Sample Example:
Figure shows a centrifuge used to accustom astronaut trainees to high accelerations.
The radius r of the circle traveled by an astronaut is 15 m.
(a) At what constant angular speed must the centrifuge rotate if the astronaut is to
have a linear acceleration of magnitude 11g?
11 &times; 9.8
15
(b) What is the tangential acceleration of the astronaut if the centrifuge accelerates at
a constant rate from rest to the angular speed of (a) in 120 s?
rω 2 = 15 ⋅ ω 2 = 11 &times; 9.8 , ω =
a t = rα = rα avg = 15
ω −0
120
= 15
2.68
= 0.34m / s 2
120
10.4 Rotational Kinetic Energy
K=
1
1
1
2
2
2
mi vi = ∑ miω 2 ri = ∑ mi ri ω 2
∑
2
2
2
I = ∑ mi ri (rotational inertia), K =
2
1 2
Iω
2
Example: The Oxygen Molecule
Consider an oxygen molecule (O2) rotating in the xy plane about the z axis.
The
rotation axis passes through the center of the molecule, perpendicular to its length.
The mass of each oxygen atom is 2.66 x 10-26 kg, and at room temperature the average
separation between the two atoms is d = 1.21 x 10-10 m.
particles.)
(The atoms are modeled as
(a) Calculate the moment of inertia of the molecule about the z axis.
2
d 
d 
I = ∑ mi ri 2 what is the radius ri ? I = m  + m 
2
2
2
(b) If the angular speed of the molecule about the z axis is 4.60 X 1012 rad/s, what is
its rotational kinetic energy?
z
1
1
Ek = mv 2 -&gt; ER = Iω 2
2
2
M
Example: Four Rotating Object
a
m
b
y
m
M
x
3
Calculate the moment of inertia if the rotation is
I = ∑ mi ri 2 what is the radius ri ? I z = ma 2 + ma 2 + Mb 2 + Mb 2
I y = Mb 2 + Mb 2
z
10.5 Calculation of Moments of Inertia
Cartesian Coordinate: x, y, z
Spherical Coordinate: r ,θ , φ
z = r cos θ
x = r sin θ cos φ , y = r sin θ sin φ
x
Cylindrical Coordinate: r ,θ , z
z
z
y
y
x
x = r cosθ , y = r sin θ
(a)
M = 2πRρ
I = ∫ R 2 ρRdθ = MR 2
(b)
M = π ( R2 − R1 ) ρ
2
2
R2
2π
R2
R1
0
R1
I = ∫ r 2 ∫ rρdθ dr = ∫ 2πρr 3 dr =
M
2
2
( R1 + R2 )
2
(c)
M = πR 2 L ρ
L R
2π
0 0
0
I = ∫ ∫ r 2 ∫ rρdθ drdz
R
I = L ∫ 2πr 3 ρdr =
0
1
MR 2
2
(d)
4
M = πR 2 Lρ
I=
L
2 R 2π
∫ ∫ ∫ (z
−
2
)
+ r 2 sin 2 θ ρrdθdrdz =
L 0 0
2
(e) from (d), R &lt;&lt; L , I =
1
1
MR 2 + ML2
4
12
1
ML2
12
(f)
M =
4π 3
R ρ
3
R π 2π
I = ∫ ∫ ∫ r 2 sin 2 θρr sin θdφ rdθ dr =
0 0 0
2
MR 2
5
(g)
π
M
I = σ ∫ (R sin θ ) 2πR sin θdθ =
4πR 2
0
2
2
π
M
∫0 (R sin θ ) 2πR sin θdθ = 2
2
2
2
∫ (1 − cos θ )R d (cosθ ) = 3 MR
1
2
2
−1
(h)
M = 2πRρ
I = ∫ R 2 sin 2 θρRdθ = MR 2 =
1
MR 2
2
(i)
(
)
b/2 a/2
M
1
I=
x 2 + y 2 dxdy = M (a 2 + b 2 )
∫
∫
ab −b / 2 −a / 2
12
Parallel axis theorem:
I = I CM + Mh 2
[
]
I = ∫ r 2 dm = ∫ ( x − a ) + ( y − b ) dm
(
)
2
2
(
)
I = ∫ x 2 + y 2 dm − 2a ∫ xdm − 2b ∫ ydm + ∫ a 2 + b 2 dm
I = I cm + Mh 2
5
2
Sample Example:
(a) What is the rotational inertia Icom of this body about an
axis through its center of mass, perpendicular to the rod as
shown?
2
2
1
L
 L
I = ∑ mi ri = m  + m  = mL2
2
2
2
2
(b) What is the rotational inertia I of the body about an axis through the left end of
the rod and parallel to the first axis?
I = ∑ mi ri 2 = m ⋅ 0 2 + mL2 = mL2
2
use parallel axis theorem: I = I CM
2
1
L
L
+ M   = mL2 + 2m  = mL2
2
2
2
Example: Rotating Rod
A uniform rod of length L and mass M is free to rotate on a frictionless pin through
one end.
The rod is released from rest in the horizontal position.
(a) What is the
angular speed of the rod at its lowest position?
1.
I cm =
1
L
1
1
ML2 , I = I cm + M ( ) 2 = ML2 , Energy conservation: Iω 2 = Mgh ,
12
2
3
2
3g
11
L
1
ML2ω 2 = Mg , ω =
, vcm = Lω
23
2
L
2
2.
Mgh =
1
1
2
I cmω 2 + Mvcm
2
2
10.6 Torque
r r
Torque is a vector: N = τ = r &times; F = rF sin φ , τ = τ 1 + τ 2 + τ 3 + ...
Torque &lt;-&gt; Force, What is the difference?
Torque can increase or decrease the angular velocity.
Example: The net torque on a cylinder
τ = (R2T2 − R1T1 )kˆ
r
6
10.7 The Rigid Body Under a Net Torque
Ft = Ftan gent , Fr = Falong _ the _ radius
Ft = Ftan gent is used for angular acceleration
Ft = mat = mrα , τ = rFt = mr 2α = Iα
The torque acting on the particle is proportional to its angular acceleration.
Example: Rotating Rod
A uniform rod of length L and mass M is attached to
a frictionless pivot and is free to rotate about the
pivot in the vertical plane. The rod is released
from rest in the horizontal position.
initial angular acceleration?
What is the
2
I CM
1
ML2
L
= ML2 , I pivot = I CM + M   =
12
3
2
τ=
L
3g
3
Mg = Iα , α =
, at = Lα = g at right end
2
2L
2
Example: Angular Acceleration of a Wheel
A wheel of radius R, mass M, and moment of inertia I is
mounted on a frictionless horizontal axle.
A light cord
wrapped around the wheel supports an object of mass m.
Calculate the angular acceleration of the wheel, the linear
acceleration of the object, and the tension in the cord.
I=
1
MR 2 , mg − T = ma , τ = RT = Iα , a = Rα
2
mg −
I a
= ma , a =
RR
mg
m+
I
R2
, T=
Iα Ia
mgI
= 2 =
, α=
R R
mR 2 + I
7
Example: Atwood Machine Revisited
Two blocks having masses m1 and m2 are
connected to each other by a light cord
that passes over two identical frictionless
pulleys, each having a moment of inertia I
and radius R. Find the acceleration of
each block and the tension T1, T2, and T3
in the cord. (No slipping)
m1 g − T1 = m1a , R(T1 − T2 ) = Iα , R(T2 − T3 ) = Iα
T3 − m2 g = m2 a , Rα = a
a=
(m1 − m2 )g
m1 + m2 +
2I
R2
10.8 Energy Considerations in Rotational
Motion
r r
dW = F ⋅ ds = (F sin φ )rdθ
dW = τ ⋅ dθ
P=
dW
dθ
=τ
= τω
dt
dt
∑τ = Iα = I
dω
dω dθ
dω
=I
=I
ω
dt
dθ dt
dθ
work-kinetic energy theorem: ∑τ ⋅ dθ = Iωdω ,
ωf
ωi
Rotational motion
θf
W = ∫τ ⋅ dθ
θi
KR =
1
∑W = ∫ Iωdω = 2 Iω
2
f
1
− Iωi2
2
Linear motion
xf
W = ∫ Fx dx
xi
1 2
Iω
2
P = τω
p = Iω
1
K = mv 2
2
P = Fv
p = mv
dL
∑τ = dt
dp
∑ F = dt
8
Example: Rotating Rod Revisited
A uniform rod of length L and mass M is free to rotate on a frictionless pin passing
through one end. The rod is released from rest in the horizontal position.
Determine the tangential speed of the CM and the tangential speed of the lowest point
on the rod when it is in the vertical position.
Mgh = Mg
Mg
L 1 2
1
L2
= Iω , I CM = ML2 , I pivot = I CM + M
2 2
12
4
3g
L 1 ML2 2
=
ω , ω=
2 2 3
L
vCM =
3 gL
L
ω=
2
2
vend = Lω = 3gL
10.9 Rolling Motion of a Rigid Object
Rolling as rotation and translation combined
θ
vcm = Rω , vedge = Rω sin θ ⋅ iˆ − Rω cos θ ⋅ ˆj
vedge = Rω (1 + sin θ )iˆ − Rω cos θ ⋅ ˆj
Rolling as pure rotation
9
1.
K=
1
1
Mv 2 + Iω 2
2
2
2.
K=
1
1
1
I pω 2 , I P = I CM + MR 2 , K = I CM ω 2 + MR 2ω 2
2
2
2
Sample Problem:
A uniform solid cylindrical disk, of mass M = 1.4 kg and radius R = 8.5 cm, rolls
smoothly across a horizontal table at a speed of 15 cm/s. What is its kinetic energy K?
1
1
2
K = I CM ω 2 + MvCM
2
2
2
1 MR 2  vCM  1
3
2
2
=

 + MvCM = MvCM = 24
2 2  R  2
4
mg sin θ − f s = ma , a = Rα
Iα = R ⋅ f s , f s = I
a=
a
R2
mg sin θ
I
m+ 2
R
Sample Example:
A uniform ball, of mass M = 6.00 kg and radius R, rolls smoothly from rest down a
ramp at angle θ = 30.0&deg;.
(a) The ball descends a vertical height h = 1.20 m to reach the bottom of the ramp.
What is its speed at the bottom?
1
1
2
2
MvCM + I CM ω 2 = Mgh , I = MR 2 , Rω = vCM
2
2
5
2
v
10 gh
1
12
2
MvCM +
MR 2 CM2 = Mgh , vCM =
= 4.1m / s
2
25
7
R
(b) What are the magnitude and direction of the friction force on the ball as it rolls
down the ramp?
10
a=
g sin θ
4.9
I
2
=
= 3.5m / s 2 , f s = 2 a = Ma = 8.4 N
I
2
5
R
1+
1+
2
5
MR
11
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