PHY2053 Lecture 15
Ch. 7.5, 7.8: Center of Mass, Collisions
Center of Mass
• Wouldn’t it be nice if complex object motion could
be simplified to that of a single point?
to Newton II, the acceleration should be:
• according
m a�
�
�
�
��
�
�
�
�
�
�i ≡
�a
mi �aCM
�
�
�F
�i =
�mi a
�i →
i i CM
Fi =� �mi a��
m�
aiCM → �aiCM = �i mi a�i
i ≡
i �
�
� =� �i mi �
mi a�i ≡ � mi aCM
→ aCM
�
� i Fi =
i
� i mi
�
�
i
ii mi a
�i �� � �
�
�
Fi(1)
∆t = ∆
mi v�i
mi a�i ≡
mi �aCM � → �aCM
=
�
�
• Recall
i
��
�
�
impulse �
�
�
m
i
momentum theorem:
��i ∆t
F
�=∆
� i ∆t = ∆
�i
F
iv
�
� m�
i
�
�
i
�
�i
�
i
�i
m
� iv
mi v�i ≡
�
mi �vCM → �vCM
(2)
�
i
�
m
v
i
i
�
�
�m v
�
�vCM
mi m
→ �vCM = �
i �i ≡
�
�
v
i i
i
to
of rCM: i mi
�vCM
v�i ≡ Which
mii �vleads
� definition (3)
i =the
CM → us
•
i
�vCM
i
mi
i
�
�i
∆�rCM
i mi r
= lim
→ �rCM = �
∆t→0 ∆t
i mi
PHY2053, Lecture 15, Center of Mass, Collisions
mi a�
= �
i mi
(1)
(1)
i
�
�
�
i
(2)
mi v�i
= �
i mi
i
(3)
(4)
2
Calc Example: Cut-out Disk
A thin disk of radius R has a hole
of radius R/2 drilled in it, as shown
in the figure. Using the coordinate
system shown in the figure, what
is the position of the center of mass
for the disk?
1)
2)
3)
4)
5)
(x=0 ,
(x=-R/2,
(x=R/2,
(x= R/6,
(x=-R/6,
y
R
x
y=0)
y=0)
y=0)
y=0)
y=0)
PHY2053, Lecture 15, Center of Mass, Collisions
3
Disk with Cutout:
PHY2053, Lecture 15, Center of Mass, Collisions
4
H-ITT: Distribution
of Masses
A set of 6 identical masses are located on the tips of
an equilateral hexagon with side length L. One of the
masses is located at (x=L, y=0). The masses are
arranged so that the CM is at (x=0, y=0). If we remove
the mass located at (x=-L, y=0), where is the CM of
the new system?
A)(x=-L/5 , y= 0 )
B) (x= 0 , y= -L/7 )
C)(x= 0 , y= 0 )
D)(x= 0 , y= L/7 )
E) (x=+L/5 , y= 0 )
PHY2053, Lecture 15, Center of Mass, Collisions
5
H-ITT: Distribution
of Masses
A set of 8 identical masses are located on the tips of
an equilateral octagon with side length L. One of the
masses is located at (x=0, y=D). The masses are
arranged so that the CM is at (x=0, y=0). If we remove
the mass located at (x=0, y=D), where is the new CM
of the system?
A)(x=-L/5 , y= 0 )
B) (x= 0 , y= -L/7 )
C)(x= 0 , y= 0 )
D)(x= 0 , y= L/7 )
E) (x=+L/5 , y= 0 )
PHY2053, Lecture 15, Center of Mass, Collisions
6
Example: M.I.R.V.
(Multiple Independently Targetable Reentry Vehicle)
A projectile splits into N pieces at the highest point of
its trajectory. The fragments land along a line so that the
distances between neighboring fragments are the same.
The first fragment was falling straight down from the
highest point of the trajectory. If the projectile had not
split up, it would have landed L away from the launch
site. Compute the
landing points for
the N fragments.
PHY2053, Lecture 15, Center of Mass, Collisions
7
Example: MIRV
PHY2053, Lecture 15, Center of Mass, Collisions
8
�
The CM Reference Frame
mi �aCM
•
= v1,i − vCM =
(v1 + v2 )
m1 v1,i − m2vv1,i
2,i
m1 +(5)
m2
m1
vCM =
�
v2,i = v2,i − vCM = −
(v
m1 + m2
�
m1 + m2
m
1
�
�
m
a
i i
i
= v2,i − vCM = −
(v1 + v2 )
m2v2,i
→
a
=
(1)
�
� �
CM
v1,i = v1,i − vCM
=
(v1 + v2 )
m1 +(6)m2 m m − m m
i mi m + m
1 2
2
�
ptotal = m1 v1,i + m2 v2,i =
recall
definition
of
CM
velocity,
compute
velocities
as
m
m
−
m
m
m
+
m2
1
2
2
1
1
�
m
1
�
v =
(v1 + v2 ) = 0
1 v(v
1,i1 +
v2,i
+ vm
(7)
�i= v2,i − vCM =p−
=∆
m
(2)
total = m
2 )2 2,i
iv
�
m
+
m
m
+
m
1
2
observed
in
CM
reference
frame
(S’):
1
2
i
vi −
j�=i mj (�
�
�
�
�
�
�
v
=
v
v
=
�
i−�
CM
i
m
m
−
m
m
1
2
2
1
�
m
(�
v
−
v
)
�
�=i
m
j
i
j
j�=i (8)
m
iv
i
�
j
p
=
m
v
+
m
v
(v
+
v
)
=
0
j
1
1,i
2
2,i
1
2
total
�
�
�
v
=
v
−
v
=
�vCM → �vCM = �
(3)
�
i
CM
i
m
+
m
1
2
i mi
j mj
�
�
� � mi mj
�
�
�
�
vi�
− �vj ) =
j�=i mj (�
�pi =
p
m
vi =
�
�
total
�v = �v − �v
=
(9)i�
�
•
�
�
1
�
2
i
i frame:
j�=i
total momentum observed
in thei CM reference
j mj
i
�p�total =
�
i
i
�p�i =
CM
�
i
mi�vi� =
��
i
j�=i
mi mj (�vi − �vj )
=0
�
j mi
�
�
pi
(10) �
=0
i
• this will be a useful property in collision analysis and
related calculations
PHY2053, Lecture 15, Center of Mass, Collisions
9
Collisions
PHY2053, Lecture 15, Center of Mass, Collisions
10
Collision Concepts
• Collision event lasts a very short amount of time
external forces acting upon m
thea�
• EvenF�if =theremare
a� ≡
m �a
→ �a
=
�
�
�
�
�
�
i i
i
�
system of icolliding objects
(friction
etc),
during
this
m
i
i
i
extremely short time, their effects are negligible
��
i
�
� i ∆t = ∆
F
i i
�
�
i
i
CM
CM
�
mi v�i � 0 → �ptotal ≡
�
�
mi v�i = const
i
m v�
• Total m v� ≡ m �v →objects
�
v
=
• Momentum is always conserved in collisions m
(but it gets redistributed among the colliding objects)
∆�r
m r�
�
�
v
=
lim
→
r
=
Kinetic
energy
is
not
necessarily
conserved:
divide
•
∆t
m
collisions into elastic, partially elastic, inelastic, explosive
• each category redistributes momenta in a different way
�
�
�
�
momentum
of�all colliding
�
i i
i
i
CM
�
is conserved
i
CM
i
i
CM
CM
∆t→0
PHY2053, Lecture 15, Center of Mass, Collisions
�
CM
�
i
�
i
i i
i
i i
i
11
�=
v
=
v
−
v
(v
+
v
)
1,i
CM
1
2
1,i
m1 v1,i −∆�
m
v
�1i + m2
rCM
2 2,i
ir
i mm
m
1
vCM
= = lim
(5)
�= �
�vCM
→ �rCM
(4)
v2,i = vi2,i
m1 + m
∆t→0
2
∆t
mi− vCM = − m + m (v
�
�
m
1
2
1
�
�
m
a
i
i
i
=
v−
− vvCM = −
(v1 + v2 )
m2vm
2,i m
2,i
mi �aCM →
a
=
(1)
�
� �
CM
v
1(v1,i
2 2,i
v1,i = v1,i − vCM
=
+
v
)
m1 +(6)m2 m m (5)
1
2
m
i vm
i
=
1 2 − m2 m
CM
�
1 + m2
m
ptotal
=2 m1 v1,i + m2 v2,i =
1+m
�recall �
definition �of CM
velocity,
compute
velocities
as
m
m
−
m
m
m
+
m2
� �
1
2
2
1
1
m
1
(v1 + v2 ) = 0
1 v(v
1,i1 +
v2,i
+mvm
(7)
�i= v2,i − v�CM =p−
=∆
m
(2)
total = m
22 )2 v2,i =
iv
� (6)
m
+
m
v
=
v
−
v
=
(v
+
v
)
m
+
m
1
2
observed
in
CM
reference
frame
(S’):
1,i
CM
1
2
1
2
1,i
i
vi −
m1 + m2 �
j�=i mj (�
�
�
�
�
�
v
=
v
v
=
�
i−�
CM
i
m
m
−
m
m
1
2
2
1
�
m
(�
v
−
v
)
�
�=i
m
j
i
j
j�=i (8)
m
iv
i
�
m
j
p
=
m
v
+
m
v
(v
+
v
)
=
0
j
1
1
1,i
2
2,i
1
2
total
�
�
�
�
v
=
v
−
v
=
�vCM → �vCM = �
(3)
�
CM
mv1 CM
+ m=2i − i
v2,im= v2,i −
(v1 + v2 )
(7)
m
i
j
i
j
m
+
m
1
2
�
�
� � mi mj
�
�
�
�
vi�
− �vj ) =
j�=i mj (�
p
pi =
m
vi =
�
�
m − m m�
�v = �v − �v
=
(9)i�
� m total
•
•
The CM Reference Frame
ptotal = m1 v1,i + m
(v1 +reference
(8)
i CM
iv2 ) = 0
i frame:
j�=
i
total momentum
observed
in
the
2 v2,i =
j mj
m1 + m2
i
i�
�p�total =
�
�p�i =
i
1
CM
�
i
2
��
2
1
mi m�
vi − �vj )
j (�
�
m�i�vi =
=−0�
� j�=i mj (�
v
vj )
i
m
�vi = �vi −i �vj�=CM
= j i�
i
mj
�
�
pi
(10) �
i
=0
(9)
in CM : the redistributed
momenta
• useful �pproperty
m m (�v − �v )
�p =
=
m �v =
=0
(10)
�
total
�
�
i
�
�
i i
j
��
i
j
i
j
� zero:
after the collision
have
to
add
up
to
j mi
i
i
i j�=i
�
k
�p�k,initial
=
�
�p�k,f inal = 0;
k
PHY2053, Lecture 15, Center of Mass, Collisions
�
→ �vk,f inal = �vk,f
vCM
inal + �
(11)
12
1D Collisions
v1,i
v2,i
v1,f
v2,f
y
x
• Simplify problem to 1D to focus on underlying
principles, later move to 2D (more realistic)
• Further simplification: 2 body collisions
• Center of mass velocity:
• Initial velocities of colliding objects in CM frame:
PHY2053, Lecture 15, Center of Mass, Collisions
13
(Fully) Elastic Collisions
• Code language for: Total Kinetic Energy is conserved
• Utilize momentum conservation + K.E. conservation
(slow, potentially cumbersome calculation)
• Move to CM reference frame - there are useful
properties to take advantage of
• “approach speed equals separation speed”
PHY2053, Lecture 15, Center of Mass, Collisions
14
Discussion
PHY2053, Lecture 15, Center of Mass, Collisions
15
Discussion
PHY2053, Lecture 15, Center of Mass, Collisions
16
Demo: Basketball
Styrofoam Ball Cannon
Example: Basketball Ball Cannon
PHY2053, Lecture 15, Center of Mass, Collisions
18
Example: Basketball Ball Cannon 2
PHY2053, Lecture 15, Center of Mass, Collisions
19
(Fully) Inelastic Collisions
• Code language for: objects are stuck together after
the collision
• Utilize momentum conservation + property:
v1,f = v2,f = vCM
• problems usually ask for system energy or velocity
after collision
PHY2053, Lecture 15, Center of Mass, Collisions
20
Demo: Ballistic
Pendulum
Ballistic Pendulum
PHY2053, Lecture 15, Center of Mass, Collisions
22
Partially Elastic Collisions
• Code language for: dynamics are not simply elastic or
inelastic, a fraction of kinetic energy is lost
• Only tool you can rely on: momentum conservation
• illustration: generic 2-body collision problem
• 4 “degrees of freedom” - 2 initial, 2 final velocities
• typical problem will define initial velocities and one
final velocity, ask for other velocity and energy loss
• or vice versa - define final, ask for initial velocity
PHY2053, Lecture 15, Center of Mass, Collisions
23
Explosive Collisions
• Code language for: total kinetic energy of the system
is larger after the collision than it was before
• similar treatment to partially elastic collisions: can
only rely on momentum conservation to solve
PHY2053, Lecture 15, Center of Mass, Collisions
24
Demo: Soda Cans and
Firecrackers
Demo: Milk Jug Rocket
2D Collisions
• momentum is now conserved as a 2D vector
quantity, need to take this into account
• special case of snooker / pool shot ignoring spin:
after collision the velocities of the two balls are
perpendicular to each other
PHY2053, Lecture 15, Center of Mass, Collisions
27
Demos:
2D Collision Table
2D Collision Video
Next Lecture
Ch. 8.1-8.3:
Rotational Kinetic Energy,
Rotational Inertia,
Torque
Demos
• map of state of Florida - actually that is torque
• spunge with center-of-mass point
• basketball styrofoam ball cannon
• elastic collisions and inelastic collisions
• soda cans and firecrackers
• milk jug rocket
• ballistic pendulum
• demonstration of impulse (is this mom. conserv?)
PHY2053, Lecture 15, Center of Mass, Collisions
30