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Chapter 8: Torque and Angular
Momentum
Definition: rigid body
rigid body, stationary axis of rotation
rotational variables (and their similarity to translations):
• angle: θ
• angular velocity: ω = ∆θ/∆t
• angular acceleration: α = ∆ω/∆t
A
B
B
A
point at distance r from axis: relationship between v and ω
rotations and kinetic energy
Distance between any two points does not change
rotational inertia
Fixed axis of rotation
Translation and Rotation
For now, we consider only the case when
the axis of rotation, wherever it is, is stationary
Rotation:
Translation:
all points rotate around
all points move
some axis by exactly
and have exactly the
the same angle
same displacements
axis of rotation
x1
θ2
θ1
x2
x-axis
Translation and Rotation
Translation
coordinate: x (meters)
Radians
Rotation
angle: θ (radians)
DO NOT USE DEGREES
(many formulas will work only in radians)
θ=
axis of rotation
r
x1
θ2
180º = π rad
s
s
r
θ
θ1
x2
s = arc length
x-axis
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Kinematics
Translation
position: x
velocity: v = ∆x/∆
∆t
acceleration: a = ∆v/∆
∆t
More equations of motion
Rotation (Ch. 5)
angle: θ
angular velocity: ω = ∆θ
θ/ ∆t
angular acc.: α = ∆ω /∆t
axis of rotation
x1
θ2
Translation
Rotation
a = const
v = vi + at
α = const
ω = ωi + αt
1
x = xi + vit + at 2
2
θ = θi + ωi t + αt 2
1
2
θ1
x2
x-axis
Acceleration and Angular Acceleration
Kinetic Energy and Moment of Inertia
Ki =
we already knew that!
v2
ar =
= rω 2
r
v
r
θ
s
at
1
1
1
2
mi vi2 = mi ( ω ri ) = ( mi ri 2 ) ω 2
2
2
2
Rotational Inertia I
- characterizes the rotational inertia of the object
- depends on the axis
ar
v
r
I
{
at = r α
K =∑
and this !
i
Kinetic Energy and Rotational Inertia
I = ∑ mi ri
2
Compare to linear motion:
M = ∑ mi
i
1
K = Iω 2
2
(pure rotation)
K=
1
Mv 2
2
1
( mi ri2 ) ω 2 = 12  ∑ mi ri2  ω 2 = 12 Iω 2
2
 i

piece i of
rigid body,
mass mi
Example 1
In a flywheel in the form of a disc, the
rotational KE is 2x108 J at peak speed of
ω=460 rad/s.
1. How much KE does the wheel have at its
minimum speed of 300 rad/s?
2. The flywheel has a mass of 20,000 kg.
What is its radius?
3. Suppose the flywheel is slowed down from
max to minimum speed in 0.2 s. How many
kW of avg power would it produce during
this interval?
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Example 1
In a flywheel in the form of a disc, the rotational
KE is 2x108 J at peak speed of ω=460 rad/s.
1. How much KE does the wheel have at its
minimum speed of 300 rad/s?
1
2
I ωmax
2
1
2
= I ωmin
2
Example 1
In a flywheel in the form of a disc, the rotational KE
is 2x108 J at peak speed of ω=460 rad/s.
2. The flywheel has a mass of 20,000 kg. What is
its radius?
K max =
K min
K max ωmax
=
K min ωmin 2
K max =
1
1 1
2
2
I ωmax = ( MR 2 )ωmax
2
2 2
⇒ R=
2
⇒
K min = K max
ωmin 2
3002
= (2 × 108 J)
= 8.5 ×107 J
460 2
ωmax 2
Example 1
2
ω
K max
2
2 ×108 J
=
= 0.435m
M
460 rad/s 2 ×104 kg
Torque (Latin for “twist”)
In a flywheel in the form of a disc, the rotational KE is
2x108 J at peak speed of ω=460 rad/s.
3. Suppose the flywheel is slowed down from max to
minimum speed in 0.2 s. How many kW of avg.
power would it produce during this interval?
To open door,
one needs to apply force
The ease of opening
depends on
• distance between the
force and the hinge
• the direction of the force
F
∆E 2 ×108 J-8.5 ×107 J
P=
=
= 5.75 ×105 kW
∆t
0.2s
Torque (Latin for “twist”):
τ = Fr sin θ
axis
r
F
θ
To twist an object,
one needs to apply force
The ease of twisting will
depend on
• force
• distance between the
force and the axis
• direction of the force
Torque: two perspectives
τ = Fr sin θ
τ = ( F sin θ )r = F⊥ r
τ = F (r sin θ ) = Fr⊥
line of action
r⊥ =r sinθ
θ
axis
axis
lever arm
r
r
F⊥ =F sinθ
θ
θ
F
θ
F
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Torque:
Torque and work
θ2
θ
axis
s
r
s
W = ( F sin φr )
r
W = τθ
W = ( F sin φ )r
r
r1
s
Units:
• N-m
φ
F sinφ
φ
θ1
F
F1
P=
torque negative
Example
Lever arm
for F2
F2=30 N
∆t
= τω
30°
F3=20 N
30°
10°
X
45°
F3=20 N
X
τ∆θ
Example continued:
Example: Calculate the torque due to the three forces shown about the left end of
the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the
bar.
F2=30 N
Note analogy with W=Fs
Proof for simple case of one particle
W = ( F sin φ ) s
axis
r2
Multiple forces:
• τnet = τ1+ τ2 + …
W = τθ
torque positive
F2
Sign convention:
• CW (clockwise): negative
• CCW(counter-clockwise):
positive
10°
45°
F1=25 N
F1=25 N
Lever arm
for F3
r1 = 0
The lever arms are:
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Mechanical Energy of Rigid Body
Example continued:
The torques are:
r2 = (2m )sin 60° = 1.73 m
r3 = (4m ) sin 10° = 0.695 m
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τ1 = 0
τ 2 = +(1.73 m )(30 N ) = +51.9 Nm
τ 3 = +(0.695 m )(20 N ) = +13.9 Nm
The mechanical energy of a rigid body with fixed
axis:
Emec = U + K rot
Gravitational Potential Energy
U = ∑ mi gyi = g ∑ mi yi = gyCM M
The net torque is +65.8 Nm and is the sum of the above results.
Conservation of Energy
Work-Energy
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Exam 2 Fall 2011: Problem 13
• A thin stick with mass M, length L, and moment
of inertia ML2/3 is hinged at its lower end and
allowed to fall freely as shown in the figure. If
its length L = 2 m and it starts from rest at an
angle θ = 20o, what is the speed (in m/s) of the
free end of the stick when it hits the table?
Answer: 7.43 m/s
% Right: 14%
Ei = MgyCM = Mg
vf =
L
cos θ
2
E f = 12 Iω 2f = 12 I
v 2f
L2
hinge
θ
h=
L
cos θ
2
Ei = E f
mgL3 cos θ
mgL3 cosθ
=
= 3 gL cos θ
2
1
I
3 mL
= 3(9.8m / s 2 )(2m) cos(20o ) ≈ 7.43m / s
University of Florida
PHY 2053
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