Chapter 8
Rotational Motion
Rotational Work and Energy
θFrFsW ==
θrs =
Fr=τ
τθ=W
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
τθ=RW
Requirement: The angle must be expressed in radians. SI Unit of Rotational Work: joule (J)
Rotational Work and Energy
KE = 12mr
2ω 2( )∑ = 12 mr2∑( )ω 2 = 1
2 Iω2
22212
21 ωmrmvKE T ==
ωrvT =
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
For a system of rotating masses, the total kinetic energy is the sum over the kinetic energies of the individual masses,
Rotational Work and Energy
221 ωIKER =
DEFINITION OF ROTATIONAL KINETIC ENERGY The rotational kinetic energy of a rigid rotating object is
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 { 221 ωIKER =τθ=RW
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
m
∨ T ∧-T
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 = 1
2 Iω f2 − 1
2 Iω02
Since: rθ = d, Idisk = 12 Mr
2,ω0 = 0 ⇒ Td = 14 Mr
2ω f2
⇒ Work done on the hanging mass:
WNC = ΔE⇒−Td = 12 mvf
2 +mghf( )− 12 mv0
2 +mgh0( )
r
{d
Since: v0 = 0 vf = rω f hf − h0 = −d⇒ −Td = 12 mr
2ω f2 −mgd
Add disk eq. + hanging mass eq. ⇒ 0 = 14 Mr
2ω f2 + 1
2 mr2ω f
2 −mgd
∴ω f = −2r
mgdM + 2m
= −2
0.202.0( ) 9.8( ) 0.70( )10.0+ 2 2.0( )
= −9.9 rad/s
Total energy of a rotating and translating rigid body in a gravitational field
ωX c.m.
vCM
Mg
MI
total energy = E = Erotationabout CM
+Etranslationof CM
= 12 Iω
2 + 12 MvCM
2 +MghCM
Since a gravitational field is a conservative force ⇒ Ef = 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
mghImvE ++= 2212
21 ω
iiifff mghImvmghImv ++=++ 2212
212
212
21 ωω
iff mghImv =+ 2212
21 ω
ENERGY CONSERVATION
rv ff =ω
o o o
o
Rotational Work and Energy
iff mghrvImv =+ 22212
21
2
2rIm
mghv of +=
The cylinder with the smaller moment of inertia will have a greater final translational speed.
o
Since Isolid = ½mr2 and Ihollow = mr2 Then, Isolid < Ihollow è 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:
ωIL =Requirement: The angular speed must be expressed in rad/s. SI Unit of Angular Momentum: kg·m2/s
τ EXT = Iα = I ΔωΔt
=Δ Iω( )Δt
=ΔLΔt∑
∴ τ EXT∑( )Δt = ΔL
⇒ "angular impulse-angular momentum theorm"
If τ EXT∑( ) = 0
⇒ ΔL = 0 ⇒ Lf = L0
⇒ Conservation of angular momentum
Consider the rotational version of Newton’s 2nd Law:
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
ωIL =Since no external torques are present in this case, we have angular momentum conservation
PPAA II ωω =
rvmrI == ω2
P
PP
A
AA r
vmrrvmr 22 =
Angular Momentum
PPAA vrvr =
vA =rPvPrA
=8.37×106 m( ) 8450m s( )
25.1×106 m= 2820m s
P
PP
A
AA r
vmrrvmr 22 =
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?
r = 0.110 m
R = 0.180 m
M = 4.80 kg
m = 3.10 kg
ω0 = 2.00 rev/s = 12.6 rad/s a) Lf = L0 ⇒ I fω f = I0ω0
ω f =I0
I fω0 =
12 MR
2
12 MR
2 + 12 mr
2 ω0
=4.80( ) 0.180( )2
4.80( ) 0.180( )2+ 3.10( ) 0.110( )2 12.6( )
=10.2 rad/s =1.62 rev/s
b) KE0 −KEf
KE0
=1−KEf
KE0
=1−12 I fω f
2
12 I0ω0
2 =1−ω f
ω0
=1−10.212.6
= 0.190
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 = I ω
and write conservation of angular momentum in vector form:Lf =
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?
Lf =
L0 ⇒ −
L1 +L2 =
L1
∴L2 = 2
L1 , in upward direction
I2ω2 = 2I1ω1 ⇒ ω2 =2I1I2
ω1
ω2 = 212 5.0( ) 0.30( )2
12 90( ) 0.35( )2 20( ) =1.6 rad/s
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