# Angular Acceleration

```Angular Acceleration
Angular acceleration α
measures how rapidly the
angular velocity is changing:
Slide 7-10
Linear and Circular Motion Compared
Slide 7-11
Linear and Circular Kinematics Compared
Slide 7-12
Tangential Acceleration
at = α r
Slide 7-14
Torque
Which force would be most effective in opening the door?
Slide 7-15
Interpreting Torque
Torque is due to the component of the force
τ = rF⊥ = rF sin φ
Slide 7-16
A Second Interpretation of Torque
τ = r⊥ F = rF sin φ
Slide 7-17
Example (text problem 8.12): The pull cord of a lawnmower
engine is wound around a drum of radius 6
6.00
00 cm
cm, while the
cord is pulled with a force of 75.0 N to start the engine. What
magnitude torque does the cord apply to the drum?
F 75 N
F=75
R=6.00 cm
τ = r⊥ F
= rF
= (0.06 m )(75.0 N ) = 4.5 Nm
8
Example
Revolutionaries attempt to pull down a statue of the Great
Leader by pulling on a rope tied to the top of his head. The
statue is 17 m tall, and they pull with a force of 4200 N at an
angle
l off 65&deg; to
t the
th h
horizontal.
i
t l Wh
Whatt iis th
the ttorque they
th exertt on
the statue? If they are standing to the right of the statue, is the
torque positive or negative?
Slide 7-18
Newton’s Second Law for Rotation
α =τ /I
I = moment of inertia. Objects
j
with larger
g moments of
inertia are harder to get rotating.
I = ∑ mi ri
2
Slide 7-24
Moments of Inertia of Common Shapes
Slide 7-25
Example (text problem 8.2): What is the rotational inertia of a
solid iron disk of mass 49
49.0
0 kg with a thickness of 5
5.00
00 cm
and a radius of 20.0 cm, about an axis through its center and
perpendicular to it?
From table 8.1:
1
1
2
2
I = MR = (49.0 kg )(0.2 m ) = 0.98 kg m 2
2
2
12
Example: (a) Find the moment of inertia of the system below.
Th masses are m1 and
The
d m2 and
d th
they are separated
t db
by a
distance r. Assume the rod connecting the masses is
massless.
ω
m1
r1
r2
m2
r1 and r2 are the distances
between mass 1 and the
rotation axis and mass 2
and the rotation axis (the
dashed, vertical line)
p
y
respectively.
13
Example continued:
Take m1 = 2.00 kg, m2 = 1.00 kg,
r1= 0.33 m , and r2 = 0.67 m.
2
I = ∑ mi ri 2 = m1r12 + m2 r22
i =1
= (2.00 kg
k )(0.33 m ) + (1.00 kg
k )(0.67 m )
2
2
= 0.67 kg m 2
(b) What is the moment of inertia if the axis is moved so
that is passes through m1?
2
I = ∑ mi ri 2 = m1r12 + m2 r22
i =1
= (2.00 kg )(0.00 m ) + (1.00 kg )(1.00 m )
2
= 1.00 kg m 2
2
14
Example (text problem 8.2): What is the rotational inertia of a
solid iron disk of mass 49
49.0
0 kg with a thickness of 5
5.00
00 cm
and a radius of 20.0 cm, about an axis through its center and
perpendicular to it?
1
1
2
2
I = MR = (49.0 kg )(0.2 m ) = 0.98 kg m 2
2
2
15
Rotational and Linear Dynamics Compared
Slide 7-26
Example
The motor in a CD player exerts a torque of
7.0 x 10-4 N &middot; m. What is the disk’s angular
acceleration? (A CD has a diameter of 12.0 cm
and
d a mass off 16 g.))
Slide 7-28
Example
A baseball bat has a mass of 0
0.82
82 kg and is 0
0.86
86 m long
long. It’s
It s
held vertically and then allowed to fall. What is the bat’s
angular acceleration when it has reached 20&deg; from the
vertical?
ti l? (Model
(M d l th
the b
batt as a uniform
if
cylinder).
li d )
Slide 7-29
Constraints Due to Ropes and Pulleys
Slide 7-30
Example
How long does it take the small mass to fall 1
1.0
0 m when
released from rest?
Slide 7-31
Work done from Torque
The work done by a torque τ is W = τΔθ .
where Δθ is the angle (in radians) the object turns through.
21
Example (text problem 8.25): A flywheel of mass 182 kg has a
0 62 m (assume the flywheel is a hoop)
hoop).
(a) What is the torque required to bring the flywheel from
rest to a speed of 120 rpm in an interval of 30 sec?
ω f = 120
d ⎞⎛ 1 min
i ⎞
⎜
⎟⎜
min ⎝ 1 rev ⎠⎝ 60 sec ⎠
⎛ Δω ⎞
τ = rF = r (ma ) = rm(rα ) = mr ⎜
⎟
⎝ Δt ⎠
ω f − ωi ⎞
ωf ⎞
2⎛
2⎛
⎟⎟ = mr ⎜⎜
⎟⎟ = 29.4 Nm
= mr ⎜⎜
⎝ Δt ⎠
⎝ Δt ⎠
2
22
Example continued:
(b) How much work is done in this 30 sec period?
W = τΔθ = τ (ωav Δt )
⎛ ωi + ω f
= τ ⎜⎜
⎝ 2
⎞
⎛ωf
⎟⎟Δt = τ ⎜⎜
⎝ 2
⎠
⎞
⎟⎟Δt = 5600 J
⎠
23
Rotational KE and Inertia
For a rotating solid body:
K rot
n
1
1
1
1
2
2
2
= m1v1 + m2 v2 + … + mn vn = ∑ mi vi2
2
2
2
i =1 2
For a rotating body vi = ωri where ri is the distance from the
rotation axis to the mass mi.
1
1⎛ n
1 2
2
2⎞ 2
= ∑ mi (ωri ) = ⎜ ∑ mi ri ⎟ω = Iω
2 ⎝ i =1
2
i =1 2
⎠
n
K rot
24
Equilibrium
The conditions for equilibrium are:
∑F = 0
∑τ = 0
25
Example (text problem 8.35): A sign is supported by a uniform
horizontal boom of length 3
3.00
00 m and weight 80
80.0
0N
N. A cable
cable,
inclined at a 35&deg; angle with the boom, is attached at a
distance of 2.38 m from the hinge at the wall. The weight of
th sign
the
i iis 120
120.0
0 N.
N What
Wh t iis th
the ttension
i iin th
the cable
bl and
d what
h t
are the horizontal and vertical forces exerted on the boom by
the hinge?
g
26
Example continued:
y
FBD for the bar:
Fy
T
X
θ
Fx
x
wbar
Apply the conditions for
equilibrium to the bar:
Fsb
(1) ∑ Fx = Fx − T cos θ = 0
(2) ∑ Fy = Fy − wbar − Fsb + T sin θ = 0
⎛L⎞
(3) ∑τ = − wbar ⎜ ⎟ − Fsb (L ) + (T sin θ )x = 0
⎝2⎠
27
Example continued:
Equation (3) can be solved for T:
⎛L⎞
wbar ⎜ ⎟ + Fsb (L )
⎝2⎠
T=
x sin
i θ
= 352 N
Equation (1) can be solved for Fx:
Fx = T cos θ = 288 N
Equation (2) can be solved for Fy:
Fy = wbar + Fsb − T sin θ
= −2.00 N
28
Equilibrium in the Human Body
Example (text problem 8.43): Find the force exerted by the
biceps
p muscle in holding
g a one liter milk carton with the
forearm parallel to the floor. Assume that the hand is 35.0
cm from the elbow and that the upper arm is 30.0 cm long.
The elbow is bent at a right angle and one tendon of the
biceps is attached at a position 5.00 cm from the elbow and
the other is attached 30.0 cm from the elbow. The weight of
the
h fforearm and
d empty h
hand
d iis 18
18.0
0 N and
d the
h center off
gravity is at a distance of 16.5 cm from the elbow.
29
Example continued:
y
Fb
“hinge””
“hi
(elbow
joint)
x
w
∑τ = F x
b 1
Fca
− wx2 − Fca x3 = 0
wx2 + Fca x3
= 130 N
Fb =
x1
30
Rolling Objects
An object that is rolling combines translational motion (its
center of mass moves) and rotational motion (points in the
body rotate around the center of mass).
For a rolling object:
K tot = K T + K rot
=
1 2 1 2
mvcm + Iω
2
2
If the object rolls without slipping then vcm = Rω.
31
32
Example: Two objects (a solid disk and a solid sphere) are
rolling down a ramp
ramp. Both objects start from rest and from
the same height. Which object reaches the bottom of the
ramp first?
h
θ
The object with the largest linear velocity (v) at the bottom
of the ramp will win the race.
33
Example continued:
Apply conservation of mechanical energy:
Ei = E f
U i + Ki = U f + K f
1 2 1 2 1 2 1 ⎛v⎞
mgh + 0 = 0 + mv + Iω = mv + I ⎜ ⎟
2
2
2
2 ⎝R⎠
2
1⎛
I ⎞
mgh = ⎜ m + 2 ⎟v 2
2⎝
R ⎠
Solving for v:
2mgh
v=
I ⎞
⎛
m
+
⎜
2 ⎟
R ⎠
⎝
34
Example continued:
I disk
The moments of inertia are:
I sphere
For the disk: vdisk
4
=
gh
3
For the sphere: vsphere
10
=
gh
7
1
= mR 2
2
2
= mR
R2
5
Since Vsphere&gt; Vdisk the
sphere wins the race
race.
Compare these to a box sliding down the ramp. vbox = 2 gh
35
How do objects in the previous example roll?
y
N
FBD:
w
x
Both the normal force and the weight act through the center
of mass so Στ = 0.
0 This means that the object cannot rotate
when only these two forces are applied.
36
y
FBD:
s
N
Fs
∑τ = F r = Iα
∑ F = w sin θ − F = ma
∑ F = N − w cosθ = 0
x
s
cm
y
θ
w
Also need acm = αR and
x
v 2 = v02 + 2aΔx
The above system of equations can be solved for v at the
bottom of the ramp. The result is the same as when using
energy methods. (See text example 8.13.)
It is static friction that makes an object roll.
37
Angular Momentum
Δp
Fnet = lim
Δt →0 Δt
p = mv
Units of p are kg m/s
When no net external
forces act, the momentum
of a system remains
constant (pi = pf)
ΔL
τ net = lim
Δt →0 Δt
L = Iω
Units of L are kg m2/s
When no net external
torques act, the angular
momentum of a system
remains constant (Li = Lf).
38
Example (text problem 8.69): A turntable of mass 5.00 kg has
0.100
100 m and spins with a frequency of 0
0.500
500
rev/sec. What is the angular momentum? Assume a uniform
disk.
ω = 0.500
⎜
sec ⎝ 1 rev ⎠
⎛1
2⎞
L = Iω = ⎜ MR ⎟ω = 0.079 kg
g m 2 /s
⎝2
⎠
39
Example (text problem 8.75): A skater is initially spinning at a
rate of 10.0
2 50 kg m2 when her arms are
extended. What is her angular velocity after she pulls her
arms in and reduces I to 1.60 kg m2?
The skater is on ice, so we can ignore external torques.
Li = L f
I iωi = I f ω f
⎛ Ii
ω f = ⎜⎜
⎝If
⎞
⎛ 2.50 kg m 2 ⎞
⎟ωi = ⎜
⎟
⎝
⎠
⎠
40
The Vector Nature of Angular
Momentum
Angular momentum is a vector. Its direction is defined with
g
rule.
a right-hand
41
Curl the fingers of your right hand so that they curl in the
direction a point on the object moves, and your thumb will
point in the direction of the angular momentum.
42
Consider
C
id a person
holding a spinning
wheel. When viewed
from the front, the
wheel spins CCW.
Holding
g the wheel horizontal, they
y step on to a platform
that is free to rotate about a vertical axis.
43
Initially, nothing happens. They then move the wheel so
that it is over their head
result, the platform turns
CW (when viewed from above).
This is a result of conserving angular momentum.
44
Initially there is no angular momentum about the vertical axis.
When the wheel is moved so that it has angular
g
momentum
direction so that the net angular momentum stays zero.
Is angular momentum conserved about the
direction of the wheel’s initial, horizontal axis?
45
It is
i not.
t The
Th floor
fl
exerts
t a torque
t
on the
th system
t
(platform
( l tf
+
person), thus angular momentum is not conserved here.
46
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