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General Physics (PHY 2130)
Lecture 21
•  Rotational dynamics
  rolling
  angular momentum
  conservation of angular momentum
  rotational kinetic energy
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Rotational dynamics:
  Equilibrium
  2nd Newton’s Law for rotation
Review Problem: Two cars, one twice as heavy as the other, are at rest on
a horizontal truck. A person pushes each car for 5 s. Ignoring friction and
assuming equal force exerted on both cars, the momentum of the light car
after the push is
1. smaller than
2. equal to
3. larger than
the momentum of the heavy car.
Newton’s Second Law for a Rotating Object
•  Recall last lecture:
•  The angular acceleration is directly proportional to the net torque
•  The angular acceleration is inversely proportional to the moment of
inertia of the object
Στ = Iα
•  There is a major difference between moment of inertia and mass: the
moment of inertia depends on the quantity of matter and its distribution
in the rigid object.
•  The moment of inertia also depends upon the location of the axis of
rotation
4
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.
5
Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp.
Both objects start from rest and from the same height. Which object reaches
the bottom of the ramp first?
Idea: the object with the largest linear velocity (v)
at the bottom of the ramp will win the race!
Apply conservation of mechanical energy:
h
Ei = E f
θ
Ui + K i = U f + K f
Since the objects are also rolling, need to remember that kinetic energy has two parts:
2
1 2 1 2 1 2 1 !v$
mgh + 0 = 0 + mv + Iω = mv + I # &
2
2
2
2 "R%
1!
I $
mgh = # m + 2 & v 2
Solving for v:
2"
R %
v=
2mgh
!
I $
#m + 2 &
"
R %
6
Now let’s figure out which object is faster! From the table 8.1:
1
mR 2
2
2
= mR 2
5
I disk =
The moments of inertia:
I sphere
For the disk:
For the sphere:
vdisk
4
=
gh
3
vsphere
Since Vsphere> Vdisk the sphere
wins the race.
10
=
gh
7
Compare these to a box sliding down the ramp:
vbox = 2 gh
7
Rolling Objects: more detail
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. This means that the object cannot rotate when
only these two forces are applied.
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Add friction:
y
∑τ = F r = Iα
∑ F = w sin θ − F = ma
∑ F = N − w cosθ = 0
s
FBD:
N
Fs
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.
It is static friction that makes an object roll.
Example:
Consider a flywheel (cylinder pulley) of mass M=5 kg and
radius R=0.2 m and weight of 9.8 N hanging from rope
wrapped around flywheel.
What are forces acting on flywheel and weight? Find
acceleration of the weight.
mg
N
Example:
I=
1
MR 2 = 0.10 kg ⋅ m 2
2
T
Mg
Given:
masses: M = 5 kg
weight: w = 9.8 N
radius: R=0.2 m
T
mg
1. Draw all applicable forces
Find:
Forces:
Forces=?
Torques:
∑ F =mg − T = ma
τ = T ⋅ R = I ⋅α
need T !
α=
T ⋅R
I
Tangential acceleration at the edge of flywheel (a=at):
TR 2
at = αR or a t =
I
or
∑ F =mg − T = ma
I
0.10 kg ⋅ m 2
T = 2 at =
a = (2.5 kg )at
R
(0.2 m)2 t
a=
mg − (2.5 kg )at = ma
mg
9.8 N
=
= 2.8 m s 2
(m + 2.5 kg ) 3.5 kg
 
Use the example:
Consider a flywheel (cylinder pulley) of mass M=5 kg and
radius R=0.2 m with weight of 9.8 N hanging from rope
wrapped around flywheel. What are forces acting on
flywheel and weight? Find acceleration of the weight.
mg
If flywheel initially at rest and then begins to rotate, a
torque must be present:
τ = I ⋅α
⎛ ω − ω 0 ⎞
= I ⎜
⎟, since ω = ω 0 + α Δt
Δ
t
⎝
⎠
Define physical quantity:
Iω = (angular momentum) ≡ L
τ=
change in ang. momentum ΔL
=
time interval
Δt
Angular Momentum
•  Similarly to the relationship between force and momentum in a
linear system, we can show the relationship between torque and
angular momentum
•  Angular momentum is defined as L = I ω
ΔL
τ=
Δt
(compare to
Δp
F=
Δt
)
•  If the net torque is zero, the angular momentum remains constant
•  Conservation of Angular Momentum states: The angular
momentum of a system is conserved when the net external
torque acting on the systems is zero.
•  That is, when
Στ = 0, Li = L f or I iω i = I f ω f
Return to our example once again:
Consider a flywheel (cylinder pulley) of mass M=5 kg and
radius R=0.2 m with weight of 9.8 N hanging from rope
wrapped around flywheel. What are forces acting on
flywheel and weight? Find acceleration of the weight.
mg
Each small part of the flywheel is moving with some
velocity. Therefore, each part and the flywheel as a
whole have kinetic energy!
∑ KE
pulley i
=∑
mass pulley i
2
v
2
pulley i
KE pulley =
1 2
Iω
2
1
1
Thus, total KE of the system: KEtot = I pulleyω 2 + mv 2
2
2
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Angular Momentum
Linear
Rotational
Δp
Fnet = lim
Δt → 0 Δt
p = mv
ΔL
τ net = lim
Δt →0 Δt
L = Iω
Units of p are kg m/s
Units of L are kg m2/s
When no net external forces
act, the momentum of a system
remains constant (pi = pf)
Conservation of linear
momentum
When no net external torques
act, the angular momentum of
a system remains constant (Li
= Lf).
Conservation of angular
momentum
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Example: A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a
frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform
disk.
Given:
m = 5.00 kg
R = 0.100 m
f = 0.500 rev/s
Shape: disk (I = MR2/2)
Find:
Angular momentum
L =?
Idea: angular momentum is obtained when we know
moment of intertia and the angular velocity.
Determine angular velocity:
rev ⎛ 2π rad ⎞
ω = 0.500
⎜
⎟ = 3.14 rad/sec
sec ⎝ 1 rev ⎠
Knowing the shape of the turntable (disk), determine
angular momentum L:
⎛ 1
⎞
L = Iω = ⎜ MR 2 ⎟ω = 0.079 kg m 2 /s
⎝ 2
⎠
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The Vector Nature of Angular Momentum
Angular momentum is a vector. Its direction is defined with a right-hand rule.
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.
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Conservation of Angular Momentum
Consider a person holding a spinning wheel. When viewed from the
front, the wheel spins CCW.
Holding the wheel horizontal, he steps on to a platform that is free to rotate about a
vertical axis.
nothing happens.
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Conservation of Angular Momentum
He then moves the wheel so that it is over his head. As a result, the
platform turns CW (when viewed from above).
Initially there is no angular
momentum about the vertical
axis.
When the wheel is moved so that
it has angular momentum about
this axis, the platform must spin
in the opposite direction so that
the net angular momentum stays
zero.
Lf = Li
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What will happen when he moves hands closer?
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Example: A skater is initially spinning at a rate of 10.0 rad/sec with I=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?
Idea: Use angular momentum conservation! The skater is on ice, so we
can ignore external torques.
Li = L f
I i ωi = I f ω f
⎛ I i
ω f = ⎜⎜
⎝ I f
⎞
⎛ 2.50 kg m 2 ⎞
⎟ωi = ⎜
⎟⎟(10.0 rad/sec) = 15.6 rad/sec
2
⎜
⎟
⎝ 1.60 kg m ⎠
⎠
Note on problem solving:
•  The same basic techniques that were used in linear
motion can be applied to rotational motion.
•  Analogies: F becomes τ , m becomes I and a becomes α , v
becomes ω and x becomes θ
•  Techniques for conservation of energy are the same as
for linear systems, as long as you include the rotational
kinetic energy
•  Problems involving angular momentum are essentially
the same technique as those with linear momentum
•  The moment of inertia may change, leading to a change in
angular momentum
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