Download Chapter 8 Rotational Dynamics continued

Chapter 8
Rotational Dynamics
continued
Chapter 8 Rotational Kinematics/Dynamics
Angular motion variables
(with the usual motion equations)
displacement
velocity
acceleration
θ = s / r (rad.)
ω = v / r (rad./s)
Uniform circular motion
centripetal acceleration
centripetal force
α = a / r (rad./s 2 )
v2
aC = = ω 2 r
r
mv 2
FC = maC =
r
I Moment of Inertia
torque
 
(θ :  between F & r)
Newton's 2nd Law
rot. kinetic energy
angular momentum
τ = rF sin θ
K rot = Iω
L = Iω
Moments of Inertia
of Rigid Objects Mass M
Thin walled hollow cylinder
τ Net = I α
1
2
From Ch.8.4 Rotational Kinetic Energy
I = MR
2
Solid sphere through center
I = 25 MR 2
Solid sphere through surface tangent
I = 75 MR 2
Thin walled sphere through center
Solid cylinder or disk
2
I = MR
1
2
2
Thin rod length ! through center
Thin plate width !, through center
I = 121 M !2
I = 121 M !2
Thin rod length ! through end
I = 13 M !2
I = 23 MR 2
Thin plate width ! through edge
I = 13 M !2
Clicker Question 8.4
A weed wacker is a length of nylon string that rotates
rapidly around one end. The rotation angular speed is
47.0 rev/s, and the tip has tangential speed of 54.0 m/s.
What is length of the nylon string?
ω
a) 0.030 m
b) 0.120 m
c) 0.180 m
d) 0.250 m
e) 0.350 m
L
vT
Clicker Question 8.5
The sun moves in circular orbit with a radius of 2.30x104
light-yrs, (1 light-yr = 9.50x1015 m) around the center of the
galaxy at an angular speed of 1.10x10–15 rad/s. galaxy
Find the tangential speed of sun.
r
a) 2.40 × 105 m/s
b) 3.40 × 105 m/s
c) 4.40 × 105 m/s
d) 5.40 × 105 m/s
e) 6.40 × 105 m/s
ω
sun
Clicker Question 8.5
The sun moves in circular orbit with a radius of 2.30x104
light-yrs, (1 light-yr = 9.50x1015 m) around the center of the
galaxy at an angular speed of 1.10x10–15 rad/s. galaxy
Find the tangential speed of sun.
r
a) 2.40 × 105 m/s
b) 3.40 × 105 m/s
c) 4.40 × 105 m/s
d) 5.40 × 105 m/s
e) 6.40 × 105 m/s
Clicker Question 8.6 Find the centripetal force on the sun.
30
20
m
=
1.99
×
10
kg
a) 2.27 × 10 N
Sun
(
b) 3.27 × 1020 N
c) 4.27 × 1020 N
d) 5.27 × 1020 N
e) 6.27 × 1020 N
)
ω
sun
8.4 Rotational Work and Energy
Total Energy = (Translational Kinetic + Rotational Kinetic + Potential) Energy
E = 12 mv 2 + 12 Iω 2 + mgy
ENERGY CONSERVATION
1
2
E f = E0
v0 = ω 0 = 0
y0 = h0
mv 2f + 12 Iω 2f + mgy f = 12 mv02 + 12 Iω 02 + mgy0
ω =v R
1
2
mv 2f + 12 I v 2f R 2 = mgh0
v 2f (m + I R 2 ) = 2mgh0
vf =
2mgh0
=
2
m+ I R
yf = 0
2gh0
1+ I mR 2
The cylinder with the smaller moment
of inertia will have a greater final translational
speed.
I = mR
2
I = 12 mR 2
8.9 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.
Angular momentum, L
Li = I iω i ; L f = I f ω f
⇒
No external torque
⇒ Angular momentum conserved
L f = Li
I f ω f = I iω i
I = ∑ mr 2 , r f < ri
Moment of Inertia
decreases
I f < Ii
Ii
>1
If
Ii
ω f = ωi;
If
Ii
>1
If
ω f > ω i (angular speed increases)
8.9 Angular Momentum
From Angular Momentum Conservation
(
)
ω f = Ii I f ω i
⇒
because I i I f > 1
Angular velocity increases
Is Energy conserved?
K f = 12 I f ω 2f
= I f Ii I f
(
)
= Ii I f
I iω 2i
1
2
2
ω 2i
( )( )
= (I I )K ⇒
i
f
1
2
i
K i = 12 I iω 2i ;
Kinetic Energy increases
Energy is NOT conserved because pulling in the arms does (NC) work
on the mass of each arm and increases the kinetic energy of rotation.
8.9 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 8.37x106 m
from the center of the earth, and
its point of greatest distance is
25.1x106 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.
I A = mrA2 ; I P = mrP2
ω A = vA rA ; ω P = vP rP
Gravitational force along r (no torque) ⇒ Angular momentum conserved
I Aω A = I Pω P
mrA2 ( vA rA ) = mrP2 ( vP rP ) ⇒ rA vA = rP vP
vA = ( rP rA ) vP = ⎡⎣(8.37 × 106 ) / (25.1× 106 ) ⎤⎦ (8450 m/s ) = 2820 m/s
Rotational/Linear Dynamics Summary
linear
rotational
x displacement θ
velocity
v
ω
a acceleration α
m
F
inertia
force/torque
I = mr 2
τ = Fr sin θ
Potential Energies
U G = mgy
or U G = −GM E m/RE
U S = 12 kx 2
linear


F = ma
W = F(cosθ )x
K = 12 mv 2
W ⇒ ΔK


p = mv


FΔt = Δp
rotational


τ = Iα
Wrot = τθ
K rot = 12 Iω 2
Wrot ⇒ ΔK rot


L = Iω


τΔt = ΔL
Conservation laws
Conserved If WNC = 0, If Fext = 0, If τ = 0,
ext

quantity: E = K + U Psystem = ∑ p L = Iω
8.7 Rigid Objects in Equilibrium
EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if it has zero translational
acceleration and zero angular acceleration. In equilibrium,
the sum of the externally applied forces is zero, and the
sum of the externally applied torques is zero.
Note: constant linear speed or constant rotational speed
are allowed for an object in equilibrium.
8.7 Rigid Objects in Equilibrium
Reasoning Strategy
1.  Select the object to which the equations for equilibrium are to be applied.
2. Draw a free-body diagram that shows all of the external forces acting on the
object.
3.  Choose a convenient set of x, y axes and resolve all forces into components
that lie along these axes.
4.  Apply the equations that specify the balance of forces at equilibrium. (Set the
net force in the x and y directions equal to zero.)
5.  Select a convenient axis of rotation. Set the sum of the torques about this
axis equal to zero.
6. Solve the equations for the desired unknown quantities.
Clicker Question 8.7
A 5-kg ball and a 1-kg ball are positioned a distance L apart
on a bar of negligible mass. How far from the 5-kg mass should
the fulcum be placed to balance the bar?
L
a)
1
2
L
b)
1
3
L
c)
1
4
L
d)
1
5
L
e)
1
6
L
m1 = 1kg
m2 = 5kg
Clicker Question 8.8
A 0.280 m long wrench at an angle of 50.0° to the floor is
used to turn a nut. What horizontal force F produces a torque of
45.0 Nm on the nut?
L = 0.280m
a)
10 N
b) 110 N
c) 210 N
d) 310 N
e) 410 N
θ = 50°
F
Clicker Question 8.9
A force F1 = 38.0 N acts downward on a horizontal arm. A second
force, F2 = 55.0 N acts at the same point but upward at
the angle theta. What is the angle of F2 for a net torque = zero?
a) 23.7°
θ
b) 28.7°
c) 33.7°
d) 38.7°
e) 43.7°
F1
F2
8.7 Rigid Objects in Equilibrium
Example: A board length L lies against a wall. The
coefficient of friction with ground 0.650. What is smallest
angle the board can be placed without slipping?
1. Determine the forces acting on the board.
θ
8.7 Rigid Objects in Equilibrium
Example: A board length L lies against a wall. The
coefficient of friction with ground 0.650. What is smallest
angle the board can be placed without slipping?
θ
1. Determine the forces acting on the board.
Forces:
Forces
P
G y ground normal force
Gy = W
Gx ground static frictional force
P wall normal force
W gravitational force
L/2
Gy
θ
Gx
2. Choose pivot point at ground.
W
P = Gx = µG y = µW
8.7 Rigid Objects in Equilibrium
Example: A board length L lies against a wall. The
coefficient of friction with ground 0.650. What is smallest
angle the board can be placed without slipping?
θ
1. Determine the forces acting on the board.
Forces:
Forces
P
G y ground normal force
Gy = W
Gx ground static frictional force
P wall normal force
W gravitational force
L/2
Gy
W
P = Gx = µG y = µW
θ
P⊥
P
Gx
2. Choose pivot point at ground.
3. Find components normal to board
P⊥ and W⊥ are forces producing torque
L/2
Gy
θ
Gx
W
W⊥
8.7 Rigid Objects in Equilibrium
Example: A board length L lies against a wall. The
coefficient of friction with ground 0.650. What is smallest
angle the board can be placed without slipping?
4. Net torque must be zero for equilibrium
Torque:
τ P = + P⊥ L = ( Psin θ ) L
τ W = −W⊥ ( L / 2 ) = − (W cosθ ) ( L / 2 )
Forces:
Gy = W
P = Gx = µG y = µW
τ W + τ P = 0 ⇒ ( Psin θ ) = (W cosθ ) / 2
W = 2Psin θ / cosθ
θ
P⊥
P
L/2
θ
W⊥
8.7 Rigid Objects in Equilibrium
Example: A board length L lies against a wall. The
coefficient of friction with ground 0.650. What is smallest
angle the board can be placed without slipping?
4. Net torque must be zero for equilibrium
Torque:
τ P = + P⊥ L = ( Psin θ ) L
τ W = −W⊥ ( L / 2 ) = − (W cosθ ) ( L / 2 )
Forces:
Gy = W
P = Gx = µG y = µW
τ W + τ P = 0 ⇒ ( Psin θ ) = (W cosθ ) / 2
W = 2Psin θ / cosθ
5. Combine torque and force equations
W = 2Psin θ / cosθ = 2 µW tan θ
tan θ = 1 (2 µ ) = 1 (1.3) = 0.77
θ = 37.6°
θ
P⊥
P
L/2
θ
W⊥
Example: A flywheel has a mass of 13.0 kg and a radius of
0.300m. What angular velocity gives it an energy of 1.20 x 109 J ?
K = 12 Iω 2 ; I disk = 12 MR 2
=
1
2
(
1
2
)
MR 2 ω 2
4K
4.80 × 109 J
ω =
=
2
MR
(13.0kg)(0.300m)2
2
ω = 4.10 × 109 = 6.40 × 104 rad/s
= 6.40 × 104 rad/s ( rev/(2π )rad )
= 1.02 × 104 rev/s ( 60s/min ) = 6.12 × 105 rpm