Reminder: angular velocity,
acceleration
Angular Displacement
Δθ = θ 2 − θ 1
Δt = t2 − t1
Angular Velocity
ωavg ≡
Δθ
dθ
; ω ≡
Δt
dt
Angular Acceleration
LECTURE 20: ROTATIONAL
VECTORS AND ANGULAR
MOMENTUM
α avg ≡
Δω
;
Δt
Chapter 10
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3
r
α = lim
Δt →0
r
dω
dt
The angular velocity vector defines the
axis of rotation, not a direction in which
something moves.
Prof. Flera Rizatdinova
Angular Acceleration & Angular
Velocity are vectors:
α ≡
2
The Cross Product
4
r
Δω dω
=
dt
Δt
The cross product of two vectors
A and B produces a vector,
of magnitude C = ABsinθ,
that is perpendicular to the
vectors A and B.
The direction of the vector
C is defined by the right-hand
rule.
Angular
acceleration points
in the direction of
the change in the
angular velocity
Both the magnitude and direction of ω may change – this is the case
when α is neither parallel nor perpendicular to the angular velocity
Components of
vector C:
Torque and the Cross Product
r
A
r r r
C = A× B
θ
r r r
C = A × B = AB sin θ nˆ
cx = ay bz - by az
cy = az bx - bz ax
cz = ax by - bx ay
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r
B
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Clicker question 1
6
In the figure,
r rthe magnitude and direction of the vector
product B× A are closest to
Torque can be represented as a
vector using the cross product:
r
r
r
B | B |= 5 cm
r
τ = r ×F
A)26, directed out of plane
B)26, directed into the plane
C)31, directed on the plane
D)31, directed into the plane
E)31, directed out of the plane
Torque is perpendicular to both
the position and force vectors.
5
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θ = 40°
r
A | A |= 8 cm
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Angular Momentum
Angular Momentum
7
8
Start with Newton’s 2nd Law
Define angular momentum by
r
r
dp
Fnet =
dt
r r r
L=r×p
Note that
r
r
dL r dp
=r×
dt
dt
r
r
r
dL dr r r dp
=
× p+r×
⇒
dt
dt dt
Multiply by the position vector of the object
r
r r
r r dp
r × Fnet = τ = r ×
dt
r
r
=0, since v and p are parallel
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Angular Momentum & Torque
9
Conservation of Angular Momentum
10
Therefore, we can write Newton’s 2nd Law for
rotational motion
r
as
If the net external torque on a system of objects
remains zero
r
d Li
r
∑i τ i = ∑i d t = 0
r r
r dp
r × Fnet = r ×
dt
r
r dL
τ =
dt
then
r
∑L
i
= constant
i
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Angular Momentum – Example
11
Example – Clutch
(1)
12
Angular momentum of a particle on spinning disk
r r r
r r
L = r × p = mr × v
= mrv sin 900 nˆ = mrv nˆ
r
= mr 2ω nˆ = mr 2ω
r
= Iω
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Example – Clutch
(2)
Example – Clutch
13
(3)
14
Equate final and initial: Lf = Li
Since the net external torque is zero:
final angular momentum, Lf = initial angular
momentum, Li
⎛
I1 ⎞
⎟ ω1
⎝ I1 + I 2 ⎠
ωf = ⎜
Li = I1ω1
This expression is similar
to that for a perfectly
inelastic collision
L f = I1ω f + I 2ω f
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Example – Clutch
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(4)
Clicker question 2
15
16
Initial kinetic energy
A spinning ice skater is able to control the rate at which she
rotates by pulling in her arms. We can best understand
this effect by observing that in this process:
K i = 12 I1ω12
Final kinetic energy
K f = I1ω + I 2ω
1
2
2
f
1
2
A)
2
f
B)
C)
⎛ I ⎞
= ⎜ 1 ⎟ Ki
⎝ I1 + I 2 ⎠
D)
E)
Her angular momentum remains constant
Her moment of inertia remains constant
Her kinetic energy remains constant
Her total velocity remains constant
She is subject to a constant non-zero torque.
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Example 2:
Example 3: way1
17
18
In the figure, a carousel has a radius of 3 m and a moment of inertia of
8000 kg⋅m2.The carousel is rotating without friction with an angular velocity
of 1.2 rad/s. An 80-kg man runs onto carousel with a velocity of 5 m/s, on
a line tangent to the rim of the carousel, overtaking it. The man runs onto
the carousel and grabs hold of a pole on the rim. What is the change in the
angular velocity of the carousel?
A potter’s wheel, with rotational inertia 46 kg⋅m2, is spinning
freely at 40 rpm. The potter drops a lump of clay onto wheel,
where it sticks a distance 1.2 m from the rotational axis. If the
subsequent angular speed of the wheel and clay is 32 rpm,
what is the mass of the clay?
Total angular momentum is conserved in this
Solution:
system!
Lc = I1ωi ; Lm = mvr; Li = I1ωi + mvr;
Li = I1ωi ; L f = I1ω f + I 2ω f ; I 2 = mr ;
Li = Lf ; L f = I1ω f + I mω f = (I1 + mr2 )ω f
2
Li = L f ⇒ I1ωi = ( I1 + mr 2 )ω f ⇒ m =
m = 8kg
⎞
I1 ⎛⎜ ωi
− 1⎟
2 ⎜
⎟
r ⎝ωf
⎠
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I1ωi + mvr = (I1 + mr2 )ω f ⇒ ω f =
I1ωi + mvr
(I1 + mr2 )
ω f = 1.24rad/s⇒ Δω = 1.24-1.2 = 0.04rad/s
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Example 3: way 2
Gyroscopes & Precession
19
20
In the figure, a carousel has a radius of 3 m and a moment of inertia of
8000 kg⋅m2.The carousel is rotating without friction with an angular velocity
of 1.2 rad/s. An 80-kg man runs onto carousel with a velocity of 5 m/s, on
a line tangent to the rim of the carousel, overtaking it. The man runs onto
the carousel and grabs hold of a pole on the rim. What is the change in the
angular velocity of the carousel?
When the net torque is zero, the angular momentum
is constant. Consequently, the axis of rotation of will
remain fixed in space. This is the basis of a gyroscope.
1
1
1
I1ωi2 K f = K i + K m = I1ωi2 + mv 2
2
2
2
1
2
2
K f = I totω f ; I tot = I1 + I m = I1 + mr
2
Ki =
1
1
1
I totω 2f = I1ωi2 + mv 2 ⇒ ω f =
2
2
2
I1ωi2 + mv 2
I tot
ω f = 1.24rad/s ⇒ Δω = 1.24 - 1.2 = 0.04 rad/s
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Summary
21
Angular velocity, acceleration, torque and angular
momentum can all be represented as vectors.
If the net torque is zero, the angular momentum is
conserved.
Consequently, the axis of rotation remains fixed.
Otherwise the axis will precess.
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