Download October 17

Lesson 6
Physics 168
Thursday, November 5, 15
Luis Anchordoqui
1
Center of Mass
Thursday, November 5, 15
2
Center of Mass
Overview
Diver’s motion is pure translation
it is translation plus rotation
There is one point that moves in same path
a particle would take if subjected to same force as diver
This point is called center of mass (CM)
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
3
Center of Mass
Overview
For two particles center of mass reads
xCM
Where
M
mA x A + mB x B
mA xA + mB xB
=
=
mA + mB
M
is Total Mass
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
4
Center of Mass
Overview
We can generalize from two particles in one dimension
to a system of N particles in three dimensions
For a continuous distribution of mass
M xcm =
M ycm =
N
X
i
N
X
mi x i
mi y i
i
M zcm =
M =
M ~rcm =
N
X
i
N
X
i
N
X
m i zi
mi
mi ~ri
i
~rcm
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
1
=
M
Z
~rcm = xcm ı̂ + ycm |ˆ + zcm k̂
~r dm
2
5
Center of Mass of a water molecule
Overview
A water molecule consists of an oxygen atom and two hydrogen atoms
An oxygen atom has a mass of 16 u (unified mass units)
and each hydrogen atom has a mass of 1 u
Hydrogen atoms are each at an average distance of 96 pm from oxygen atom
and are separated from one another by an angle of 104.5 degrees
Find center of mass of a water molecule
xcm = 6.53 pm
ycm = 0
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
6
Center of Mass of a uniform semicircular hoop
x = r cos ✓
Use polar coordinates
Overview
y = r sin ✓
Distance of points on semicircle from origin is
~rcm
1
=
M
Z
1
(xı̂ + yˆ
|) dm =
M
mass per unit length is
~rcm
Z
r = R
R(cos ✓ı̂ + sin ✓ˆ
|)dm
= dm/ds
ds = R d✓
dm =
Z ⇡
1
=
R(cos ✓ı̂ + sin ✓ˆ
|) R d✓
M 0
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
7
Center of Mass of a uniform semicircular hoop (cont’d)
Z ⇡
⌘
2 ⇣ Z ⇡
ROverview
~rcm =
ı̂
cos ✓ d✓ + |ˆ
sin ✓ d✓
M
0
0
~rcm
Z ⇡
Z ⇡
⇣
⌘
R
=
ı̂
cos ✓ d✓ + |ˆ
sin ✓ d✓
⇡
0
0
⇡
⇡⌘
R⇣
ı̂ sin ✓
=
|ˆ cos ✓
⇡
0
0
2
= Rˆ
|
⇡
Curiously it is outside material of semicircular hoop!
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
8
Center of Mass for human body
Overview
High jumpers have developed a technique where their CM actually passes
under bar as they go over it
This allows them to clear higher bars
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
9
Center of Mass & Translational Motion
Overview
☛ Sum of all forces acting on a system is equal to total mass of system
multiplied by acceleration of center-of-mass
☛ For each internal force acting on a particle in system there is an equal
and opposite internal force acting on some other particle of system
M~aCM = F~net,ext
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
10
Changing Places in a Rowboat
Overview
Pete (mass 80 kg) and Dave (mass 120 kg) are in a rowboat (mass 60 kg) on a calm lake
Dave is near bow of boat, rowing, and Pete is at stern, 2 m from Dave
Dave gets tired and stops rowing
Pete offers to row, so after boat comes to rest they change places
How far does boat moves as Pete and Dave change places
(Neglect any horizontal force exerted by water)
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
11
M XCMi = mPete xPetei + mDave xDavei + mboat xboati
Overview
M XCMf = mPete xPetef + mDave xDavef + mboat xboatf
M XCM = mPete
M aCM
xPete + mDave xDave + mboat
X
= Fext,net =
Fexti = 0
xboat
i
0 = mPete ( xboat + L) + mDave ( xboat
xboat
L) + mboat xboat
L (mDave mPete )
=
= 0.31 m
mDave + mPete + mboat
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
12
Center of Mass Work
Overview
For a system of particles we have
F~net,ext =
M =
X
X
F~iext = M~acm
mi
i
F~net,ext · ~vcm = M~acm · ~vcm
⌘
d ⇣1
dKtrans
2
=
M vcm =
dt 2
dt
Center of mass work - translational kinetic energy relation
Z
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
1
F~net,ext · d~lcm =
Ktrans
2
13
Power
Power is rate at which work is done
Overviewwork
energy transformed
P̄ = average power =
=
time
time
In SI system, units of power are watts:
1W = 1J/s
Difference between walking and running up
these stairs is power
– change in gravitational potential energy is same-
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
14
Conservation Theorems: Momentum
Luis Anchordoqui
Thursday, November 5, 15
15
Momentum of a particle
Overview
Originally introduced by Newton as quantity of motion
momentum is defined as ☛
p
~ = m~v
Using Newton´s second law we can relate
momentum of a particle to force acting on particle
d~
p
d(m~v )
d~v
=
= m
= m~a
dt
dt
dt
~net by
Substituting force F
d~
p
~
Fnet =
dt
Net force acting on a particle equals time rate of change of particle´s momentum
In his famous treatise Principia (1687) Newton presents
second law of motion in this form
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
16
Conservation of Momentum
Overview
X
~
Total momentum of a system of particles reads
Psys =
P~sys =
mi ~vi =
i
X
X
p~i
P
i
mi ~vi = M~vcm
i
M =
X
mi
i
dP~sys
d~vcm
= M
= M acm
dt
dt
According to Newton´s second law
X
F~ext = F~net, ext
i
If
X
F~ext = 0
dP~sys
=
dt
~sys = M ~vcm = constant
then P
If sum of external forces on a system remains zero
total momentum of system is conserved
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
17
A runaway railroad car
Overview
A runaway 14,000 kg railroad car is rolling horizontally
at 4 m/s toward a switchyard
As it passes by a grain elevator 2000 kg of grain suddenly drops into the car
How long does it take car to cover 500 m distance
from elevator to switchyard?
Assume that grains falls straight down and that slowing due to
rolling friction or air drag is negligible
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
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X
F~exti
i
A runaway railroad car
~sys
d
P
Overview
= F~ggrain + F~gcar + F~n =
dt
x component of net external force is zero
Psys,xi = Psys,xf
(mcar + mgrain ) vxf = mcar vxi
Before
vxf
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
After
mcar vxi
=
)
mcar + mgrain
d
t=
= 143 s
vxf
2
19
Collisions and impulse
Overview
When two objects collide they usually exert very large forces on each other
for very brief time
Impulse of a force exerted during a time
t = tf = ti is a vector defined as
Z
I~ =
tf
F~ dt
ti
Impulse is a measure of both
strength and duration of collision force
I~ =
Z
tf
F~ dt =
ti
Z
tf
ti
d~
p
dt = p~p~rf = p~p~ii
dt
Impulse momentum theorem for a particle
I~net =
P~
Impulse momentum theorem for a system
I~net, ext =
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
Z
tf
F~net, ext dt =
P~sys
ti
2
20
Average force
Since time of collision is very short
Overview
we need not worry about exact time dependence of force
and can use average force
F̄ = hF~ i =
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
1
t
Z
tf
ti
1 ~
~
F dt =
I
t
2
21
Perfectly Elastic and Perfectly Inelastic Collisions
Conservation of momentum
m1 v1f Overview
+ m2 v2f = m1 v1i + m2 v2i
In elastic collisions kinetic energy of system is conserved
1
2
m1 v1f
+
2
2
m2 (v2f
m2 (v2f v2i )(v2f
1
1
1
2
2
2
m2 v2f = m1 v1i + m2 v2i
2
2
2
2
2
2
v2i ) = m1 (v1i v1f )
+ v2i ) = m1 (v1i v1f )(v1i + v1f )
From conservation of momentum
m2 (v2f
v2i ) = m1 (v1i
v1f )
Taking ratio of these two equations
(v2f + v2i ) = (v1f + v1i )
Rearranging we obtain relative velocities in an elastic collision
(v1i
v2i ) = (v2f
v1f )
In perfectly inelastic collisions objects have same velocity after collision
(often because they stuck together)
v1f = v2f = vcm
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
22
Conservation of Energy and Momentum in Collisions
Overview
Momentum is conserved in all collisions
Collisions in which kinetic energy is conserved as well are called elastic collisions
and those in which it is not are called inelastic
Approach
Collision
If elastic
If inelastic
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
23
Inelastic Collisions
Overview
With inelastic collisions, some of initial kinetic energy
is lost to thermal or potential energy
It may also be gained during explosions
as there is the addition of chemical or nuclear energy
A completely inelastic collision
is one where objects stick together afterwards
so there is only one final velocity
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
24
A meteor whose mass was about 108 kg struck Earth
with a speed of about 15 km/s and came to rest in Earth
ⓐ What was Earth's recoil speed?
ⓑ What fraction of meteor's kinetic energy was transformed
to kinetic energy of Earth?
ⓒ By how much did Earth kinetic energy change as a result of this collision?
Overview
M = 6 ⇥ 1024 kg
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
25
mmeteor vmeteor = (mmeteor + M )vf
vf = 2.5 ⇥ 1013 m/s
Overview
ⓐ
ⓑ
Kfinal Earth
= 1.7 ⇥ 10
Kinitial meteor
ⓒ
KEarth = KEarthf
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
KEarthi
17
1
= M vf2 = 0.19 J
2
2
26
Elastic Collisions in One Dimension
Overview
Here we have two objects colliding elastically
We know masses and initial speeds
Since both momentum and kinetic energy are conserved we can write two equations
This allows us to solve for two unknown final speeds
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
27
Elastic collision of a neutron and a nucleus
A neutron of mass mOverview
n and speed vni undergoes a head-on elastic collision
with a carbon nucleus of mass initially atrest
(a) What are final velocities of both particles?
(b)What fraction
f
of its initial kinetic energy does neutron lose?
Neutron
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
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Overview
for elastic collision
v Cf
mn vni = mn vnf + mC vCf
v nf = v ni ) v C f = v ni + v nf
mn vni = mn vnf + mC (vni + vnf )
mC mn
v nf =
v ni
mn + mC
mC mn
2mn
v C f = v ni
v ni =
v ni
mn + mC
mn + mC
f=
KCf
Kn
mC
=
=
K ni
K ni
mn
✓
v Ci
v Cf
◆2
4mn mC
=
(mn + mC )2
C.
B.-Champagne
Luis
Anchordoqui
2
Thursday, November 5, 15
29
Collisions in Two or Three Dimensions
Overview
Conservation of energy and momentum can also be used to analyze collisions
in two or three dimensions but unless situation is very simple
math quickly becomes unwieldy
If moving object collides with an object initially at rest
knowing masses and initial velocities is not enough
we need to know angles as well in order to find final velocities
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
30
Overview
A novice pool player is faced with corner pocket shot shown in figure
Relative dimensions are also given
Should player be worried about this being a “scratch shot”
in which cue ball will also fall into a pocket?
p
3
Give details
1
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
2
31
~v
Overview
~vA0
~vB0
In elastic collision between two objects of equal mass with target at rest
angle between final velocities of objects is 90º
Momentum conservation
m~v = m~vA0 + m~vB0 ) ~v = ~vA0 + ~vB0
0
0
~
v
B
~v
A
~v
1
1
1
2
02
m vA + m vB02 ) v 2 = vA02 + vB02
Kinetic energy conservation ☛ mv =
2
2
2
v 2 = ~vA02 + ~vB02
2 ~vA0 ~vB0 cos ✓
2
Equating two expressions for v leads to
vA02 + vB02 = vA02 + vB02
2 vA0 vB0 cos ✓
Applying law of cosines ☛
C.
B.-Champagne
Luis
Anchordoqui
Thursday, November 5, 15
cos ✓ = 0 ) ✓ = 90
2
32
1m
✓1
p
3m
3m
✓2
Assume that target ball is hit correctly so that it goes in pocket
From geometry of right triangle
p
✓1 = arctan[1/ 3] = 30
From geometry of left triangle
p
✓2 = arctan[3/ 3] = 60
Because balls will separate at 90º ☛ if target ball goes in pocket
this does appear to be a good possibility of a scratch shot
Thursday, November 5, 15
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