Download Chapter 8 Rotational Motion

Chapter 8
Rotational Motion
Rotational Work and Energy
s = rθ
W = Fs = Frθ
τ = Fr
W = τθ
Consider the work done in rotating
a wheel with a tangential force, F,
by an angle θ.
Rotational Work and Energy
DEFINITION OF ROTATIONAL WORK
The rotational work done by a constant torque in
turning an object through an angle is
WR = τθ
Requirement: The angle must
be expressed in radians.
SI Unit of Rotational Work: joule (J)
Rotational Work and Energy
According to the Work-Energy theorem:
W = KEf - KE0
So WR should be able to produce rotational kinetic energy.
Calculate the kinetic energy of a mass m undergoing rotational motion
at radius r and moving with tangential speed vT
KE = 12 mvT2 = 12 mr 2ω 2
vT = rω
For a system of rotating masses, the total kinetic energy is the sum over
the kinetic energies of the individual masses,
KE = ∑ ( 12 mr 2ω 2 ) = 12
(
)
2
2
2
1
mr
ω
=
I
ω
∑
2
Rotational Work and Energy
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy of a rigid rotating object is
KE R = 12 Iω 2
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Rotational Kinetic Energy: joule (J)
Thus, the rotational version of the Work-Energy theorem is:
WR = KERf - KER0
where
WR = τθ
{ KE
R
= 12 Iω 2
Example: A hanging mass rotating a solid disk. As seen in the figure, a
2.0 kg mass attached to a string is rotating a solid disk of mass 10.0 kg
and radius 0.20 m pivoting around its center. If the system is initially at
rest, what is the angular velocity of the disk after the mass falls 0.70 m?
M
r
∨ T
∧-T
d
{
m = 2.0 kg M = 10.0 kg r = 0.20 m d = 0.70 m
→ Find ω f
⇒ Work done on the disk:
WR = ΔKER ⇒ τθ = Trθ = 12 Iω 2f − 12 Iω 02
Since: rθ = d, I disk = 12 Mr 2 , ω 0 = 0 ⇒ Td = 14 Mr 2ω 2f
m ⇒ Work done on the hanging mass:
WNC = ΔE ⇒ −Td = ( 12 mv 2f + mgh f ) − ( 12 mv02 + mgh0 )
Since: v0 = 0 v f = rω f
h f − h0 = −d ⇒ − Td = 12 mr 2ω 2f − mgd
Add disk eq. + hanging mass eq. ⇒ 0 = 14 Mr 2ω 2f + 12 mr 2ω 2f − mgd
2
mgd
2
∴ω f = −
=−
r M + 2m
0.20
(2.0) (9.8) (0.70) = −9.9 rad/s
10.0 + 2 ( 2.0 )
Total energy of a rotating and translating rigid body in
a gravitational field

v CM
M
I
ω
X c.m.

Mg
2
total energy = E = Erotation + Etranslation = 12 Iω 2 + 12 MvCM
+ MghCM
about CM
of CM
Since a gravitational field is a conservative force ⇒ E f = E0
Rotational Work and Energy
Example: Rolling Cylinders
A thin-walled hollow cylinder (mass = m, radius = r) and
a solid cylinder (also, mass = m, radius = r) start from rest at
the top of an incline.
Determine which cylinder
has the greatest translational
speed upon reaching the
bottom.
Rotational Work and Energy
E = 12 mv 2 + 12 Iω 2 + mgh
ENERGY CONSERVATION
1
2
1
2
mv 2f + 12 Iω 2f + mgh f = 12 mvoi2 + 12 Iωoi2 + mghio
mv 2f + 12 Iω 2f = mghio
ωf = vf r
Rotational Work and Energy
1
2
mv 2f + 12 I v 2f r 2 = mghoi
2mgho
vf =
m + I r2
The cylinder with the smaller moment
of inertia will have a greater final translational
speed.
Since
Isolid = ½mr2 and
Then,
Isolid < Ihollow
è
Ihollow = mr2
vf solid > vf hollow
Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a
fixed axis is the product of the body’s moment of
inertia and its angular velocity with respect to that
axis:
L = Iω
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
Consider the rotational version of Newton’s 2nd Law:
∑τ
∴
EXT
Δω Δ ( Iω ) ΔL
= Iα = I
=
=
Δt
Δt
Δt
(∑τ ) Δt = ΔL
EXT
⇒ "angular impulse-angular momentum theorm"
If
(∑ τ ) = 0
EXT
⇒ ΔL = 0 ⇒ L f = L0
⇒ Conservation of angular momentum
Angular Momentum
PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM
The angular momentum of a system remains constant (is
conserved) if the net external torque acting on the system
is zero.
Lf = L0
Angular Momentum
Conceptual Example: A Spinning Skater
An ice skater is spinning with both
arms and a leg outstretched. She
pulls her arms and leg inward and
her spinning motion changes
dramatically.
Use the principle of conservation
of angular momentum to explain
how and why her spinning motion
changes.
Angular Momentum
Example: A Satellite in an Elliptical Orbit
An artificial satellite is placed in an
elliptical orbit about the earth. Its point
of closest approach is 8.37 x 106 m
from the center of the earth, and
its point of greatest distance is
25.1 x 106 m from the center of
the earth.
The speed of the satellite at the
perigee is 8450 m/s. Find the speed
at the apogee.
Angular Momentum
L = Iω
Since no external torques are present
in this case, we have
angular momentum conservation
I Aω A = I Pω P
I = mr 2 ω = v r
vA
2 vP
mr
= mrP
rA
rP
2
A
Angular Momentum
vA
2 vP
mr
= mrP
rA
rP
2
A
rA v A = rP vP
rP vP
vA =
=
rA
6
8.37
×10
m ) (8450 m s)
(
6
25.1×10 m
= 2820 m s
Example: A potter’s wheel is rotating around a vertical axis through its center
at a frequency of 2.00 rev/s. The wheel can be considered a uniform disk of
mass 4.80 kg and diameter 0.360 m. The potter then throws a 3.10 kg chunk of
clay, approximately shaped as a flat disk of radius 11.0 cm, onto the center of
the rotating wheel. (a) What is the frequency of the wheel after the clay sticks
to it? (b) What fraction of the original mechanical energy of the wheel is lost
to friction after the collision with the clay?
ω0 = 2.00 rev/s = 12.6 rad/s
R = 0.180 m
r = 0.110 m
m = 3.10 kg
M = 4.80 kg
a) L f = L0
⇒ I f ω f = I 0ω 0
1 MR 2
I0
ω f = ω0 = 1 2 2 1 2 ω0
If
2 MR + 2 mr
=
( 4.80) (0.180)
2
2
( 4.80) (0.180) + (3.10) (0.110)
2
(12.6)
= 10.2 rad/s = 1.62 rev/s
1 I ω2
KE0 − KE f
KE f
ωf
10.2
f
2 f
b)
= 1−
= 1−
= 1−
= 1−
= 0.190
2
1
KE0
KE0
ω0
12.6
2 I 0ω 0
The Vector Nature of Angular Variables
Right-Hand Rule: Grasp the axis of rotation with
your right hand, so that your fingers circle the axis
in the same sense as the rotation.
Your extended thumb points along the axis in the
direction of the angular velocity.
∴we can express L as a vector in the direction of ω :


L = Iω
and write conservation of angular momentum in vector form:


L f = L0
Example: A person sitting on a chair that can rotate is initially at rest and
holding a bicycle wheel which is spinning with its angular momentum vector
in the vertically up direction and with magnitude 20 rad/s. The mass and
radius of the bicycle wheel are 5.0 kg and 0.30 m, respectively, approximated
as a solid disk. The mass and average radius of the person through a vertical
axis are 90 kg and 0.35 m, respectively, approximated as a solid cylinder. If
the person now flips the spinning wheel so that the angular momentum vector
is vertically down, what is the angular velocity of the person?


  
L f = L0 ⇒ − L1 + L2 = L1


∴ L2 = 2 L1 , in upward direction
2I1
I 2ω 2 = 2I1ω1 ⇒ ω 2 =
ω1
I2
ω2 = 2
1
2
1
2
(5.0) (0.30)
(90) (0.35)
2
2
(20) = 1.6 rad/s