# Document 13605685

```Review #3: Collisions, Rotational Motion
Momentum - Model
From Principia: Momentum is defined as “the quantity of
motion, conjointly proportional to the mass and the velocity”.
momentum is for a system of particles
system
r
r
definition ---p
=
m v
∑
i
i
i
momentum is sum of individual momenta
Dynamics (change) of momentum - from F=ma
r system forces r
dp
= ∑ Fi external
dt
i =1
Only external (through system barrier) forces !
Strategies: Momentum and External Forces 1. Identify all forces acting on the masses
2. Select your system (make troublesome forces
into internal forces; make ΣFext = 0)
3. ΣFext = 0 implies momentum is conserved
4. If there is a non-zero total external force:
r system forces r
dp
external
= ∑
Fi
dt
i
Strategy - Impulse
Impulse most useful when time is short, simplifying the
momentum change during the short time ∆t.
A. One large external force dominates ∆p:
•Ball bouncing on floor - ignore gravity
B. Finite Fext for very short ∆t --&gt; ∆psystem = 0
•Colliding cars - ignore horizontal friction
•Gun and Bullet - ignore external forces
Collisions: Momentum Conserved When the total external force on the colliding
particles is much smaller than the internal
forces, the collision duration is so short that
the impulse on the system is approximately
zero.
Then the total initial momentum of the colliding
particles equals their final momentum:
r total r total
p f
=
p0
Gives one equation each for x, y, and z
Elastic Collisions
• Kinetic Energy does not change.
K0
total
= Kf
total
1
1
1
1
2
2
2
m1v1,0 + m2 v2,0 + ... = m1v1, f + m2 v2, f 2 + ...
2
2
2
2
• Momentum conserved also
Rotation and Translation
of Rigid Body
Motion of a thrown object
Translational Motion of the Center of Mass
• Total momentum
r
r total
total
p
= m
V
cm
• External force and acceleration of center of
mass
r
r total
r
r total
dp
total dV
total
cm
F
ext
=
=m
= m A
cm
dt
dt
Rotation and Translation
of Rigid Body
• Torque produces angular acceleration about center of
mass
torques
∑τ
• I cm
cm,i
= I cmα
cm
i
is the moment of inertial about the center of mass
• α is the angular acceleration about center of mass or
any other point in a rigid body.
• This is really a vector relation; only z-component is
non-zero if problem is planar
Fixed Axis Rotation: Kinematics
Angle variable
θ
Angular velocity
dθ
ω≡
dt
Angular acceleration
d 2θ
α≡ 2
dt
These are exactly analogous to the variables x, vx, and ax for
One dimensional motion, and obey the same equations
For constant angular acceleration:
Fixed Axis Rotation: Tangential Velocity and Tangential Acceleration
Kinematics of individual mass elements:
∆mi
• Tangential velocity
• Tangential acceleration
vtan,i = r⊥ ,iω
atan,i = r⊥ ,iα
v
2
tan,i
r⊥ ,i
= r⊥ ,iω 2
A ladybug sits at the outer edge of a merry-go-round that is
turning and slowing down. The tangential component of the
QuickTime™ and a
Graphics decompressor
are needed to see this picture.
1.
in the +x direction.
2.
3.
4.
5.
in the -x direction.
in the +y direction.
in the -y direction.
in the +z direction.
6.
7.
in the -z direction.
zero.
Torque Torque about axis S: r
r
τ
S ,i = r
S ,i &times; Fi
r
• Counterclockwise +z direction
• perpendicular to the plane
τ S ,i = rS ⊥,i Ftan,i = rS ⊥,i ∆mi a y = ∆mi (rS ⊥,i ) α
2
torques
∑τ
i
S ,i
= I Sα cm = I Sα S
Energy of Rotating Mass
The mass is rotating:
Angular velocity ω
v
R
ω
Speed v
Speed v = ω R
Kinetic Energy = 1/2 m v^2 = 1/2 m R2ω2 = 1/2 Is ω2
K
rot
S
1
2
= I Sω S
2
Moment of Inertia - Idea
Mass element
∆mi
r⊥ ,i
i= N
I S = ∑ ∆mi (rS ⊥ ,i )2
S
i =1
The moment of inertia takes the place of mass in the Dynamical (F=ma) and Energy (K=mv2/2) equations for rotational motion.
torques
∑τ
i
S ,i
= I Sα cm
K
rot
S
1
2
= I Sω S
2
These work about ANY NON-ACCELERATING axis
Rotational
Work
r
r
Starting from
∆Wi = Fi
⋅ ∆r
i
• tangential force
r
F
tan,i = Ftan,i nθ&ouml;
• displacement vector
( )
r
∆rS ,i = rS ,⊥ ∆θ n&ouml;
i
r
r
• infinitesimal work ∆Wi = Fi
⋅ ∆r
i
= Ftan,i rS ,⊥ i ∆θ = τ i ∆θ
( )
f
• Rotational work:
W firot = ∫ τ (θ )dθ = τ avg (θ f − θ i )
i
Note: if τ is constant, it equals τavg
General Work-Kinetic Energy Rel’n
• Fixed axis passing through the c of m of the body
Wf 0
rot
1
1
2
2
= I cmω cm, f − I cmω cm,0
= K rot , f − K rot ,0 ≡ ∆K rot
2
2
• Rotation and translation combined - General Motion
⎛1 2
⎛1 2
1
1
2⎞
2⎞
K f = ⎜ mvcm, f + I cmω f ⎟ = ⎜ mvcm,0 + I cmω 0 ⎟ + W fitrans + W firot = + K 0 + W fitotal
2
⎝ 2
⎠ ⎝ 2= ∆K
W
+2 ∆K ⎠
total
trans
rot
f
W firot = ∫ τ (θ )dθ = τ avg (θ f − θ i )
i
Note: if τ is constant, it equals τavg
Strategy: Moment of Inertia
Always start from a tabulated Icm
plus the Parallel Axis Theorem
i= N
2
2
I
=
∆m
(r
)
=
dm
(r
)
Use S ∑ i S ⊥ ,i
S ⊥ ,dm
∫
i =1
S
When all else fails
r⊥
Note: I cm = ∫ dm (rcm⊥ ,dm ) = α MR
2
Where α is between 0 and 1
2
Table of Icm
Object
Hoop
Disk
Sphere
Rod
α
1.0
0.5
0.4
ML2/12
Rotational Dynamics &amp; Energy - Summary
• Dynamical Equations about axis S
Dynamics
torques
∑
τ S , i = I Sα cm = I Sα S
i
Kinetic Energy
1
K Srot =
I Sω S2
2
Requires S to be stationary
Is not true if S accelerates
Includes KE of c of m
• Most General Equations
torques
∑τ
cm, i
= I cmα cm
i
1
1
2
K tot =
I cmω cm
+ M v 2cm
2
2
KE of rotation+KE of translation
OK if c of m accelerates
PRS - kinetic energy
A disk with mass M and radius R is spinning with
angular velocity ω about an axis that passes
through the rim of the disk perpendicular to its
plane. Its total kinetic energy is:
1. 1/4 M R2 ω2
4. 1/4M R ω2
2. 1/2M R2 ω2
5. 1/2 M R ω2
3. 3/4 M R2 ω2
6. 1/2 M R ω
Dynamics: Translational &amp; Rotational
Translational Dynamics
•Total Force
r ext
∑ Fi
ri system
p
•Momentum of System
r system
r ext
dp
∑ Fi =
dt
i
•Dynamics of Translation
Rotational Dynamics of point mass about S
r
r
r
• Torque
τ S ,
i =
ri,S &times;
Fi
r
r
r
• Angular momentum
L S ,i =
ri, S &times;
p i
• Dynamics of rotation
r ext
∑τ
i
S ,i
=
r
dL S
dt
Rotational Angular Momentum &amp; Energy • Putting it Together - Spin plus Translation
Kinetic Energy
Angular Momentum
r system
r
i = N r
r
L
S =
I cmω +
∑ R
cm,S
&times;
p
i =1
L of rotation+L of translation
K
tot
1
1
2
=
I cmω
cm +
M v 2cm
2
2
KE of rotation+KE of translation
Dynamics (i.e. change)
r ext
∑τ
i
cf.
S ,i
=
r system
dL S
dt
r ext dpr
∑ Fi = dt
i
K f = K 0 + W fitrans + W firot
Physical Content of
ext
τ
∑ S ,z
i
dLsystem
z
=
dt
• In the case that IS,z does not change:
ext
τ
∑ S ,z =
i
dLsystem
z
dt
=
(
d I S , zω z
dt
)= I
dω z
S ,z
dt
= IS ,z
• But IS,z may change:
– Spinning Skater pulls in arms
– Rain falling on merry-go -round
• Conservation of Lz is richer than pz
ext
τ
∑ S ,z = 0
i
system
Lsystem
=
L
z, f
z ,i
d 2θ
= I S , zα z
2
dt
PRS - angular momentum
A disk with mass M and radius R is spinning with
angular velocity ω about an axis that passes
through the rim of the disk perpendicular to its
plane. The magnitude of its angular momentum
is:
1. 1/4 M R2 ω2
4. 1/2M R ω
2. 1/2M R2 ω2
5. 3/4 M R ω
3. 3/4 M R2 ω2
6. 3/2 M R ω
Angular Momentum is a Kepler Law
Perihelion -&gt;
&lt;- equal areas -&gt;
r
r
vtang
vorb
vtang
vorb
Angular Momentum:
Lsun = &micro;vtang r
Area: ∆Aswept = r(vtang ∆t ) / 2
∆Aswept
∆t
1
L
= rvtang =
2
2 &micro;
Circular Orbital Mechanics
-class problem
A planet of mass m1 is in a circular orbit of radius R
around a sun of mass m2.
a. Find the period, T
b. Find the ratio of kinetic to potential energy
c. As R increases, all of the physical properties of the
orbit (e.g. velocity) decrease except one; find it.
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