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Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html
Lecture 16
Chapter 9
Linear Momentum.Center of Mass.
11.06.2013Physics I
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Chapter 9
2-D collisions Systems of particles (extended objects) Center of mass
Outline
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Collisions in 2D
(1D) You know how to figure out the results of a collision between objects in 1D: -use conservation of momentum- and, -if collision is ellastic, conservation of mech. energy.
AAvm BBvm
conserved isdirection -in x momentum then ,0extxFIf
finalx
initialx pp
finaly
initialy pp
finalinitialmvmv )
2()
2(
22
conserved isdirection -yin momentum then ,0extyFIf
dimension)each in (not collisions elasticin conserved isEnergy Mech.
You can continue to use the same rules in 2D collisions as follows:
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Collisions in 2D: Momentum Conservation (I)A projectile (mA) moves along the x-axis and hits a target (mB) at rest.
Since net external forces in x and y directions are zero, px and py are conserved
After the collision, the two objects go off at different angles.
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Collisions in 2D: Momentum Conservation (II)
BBxBAAAAA vmvmvm cos'cos'
BByBAAyA vmvm 'sin''sin'0
12
mAvA2 1
2mAv 'A
2 12
mBv 'B2
Two equations, can be solved for two unknowns
conservation of x-momentum:
conservation of y-momentum:
If collision is elastic, we get a third equation (conservation of mechanical energy)
Three equations, can be solved for three unknowns
afterx
beforex pp
aftery
beforey pp
)( BABA vvvv in 2-D collisions
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Example: 2D collisionBall A moving at 4 m/s strikes ball B (of equal mass) at rest. After the collision, ball A travels forward at an angle of +45º, and ball B travels forward at -45º. What are the final speeds of the two balls?
vA vB 2.83ms
)45cos('45cos')4( BAsm mvmvm
)45sin('45sin'0 BA mvmv
conservation of x-momentum
conservation of y-momentum
v 'A v 'B 2 24 ( 12
)v 'A ( 12
)v 'B
4 2 v 'A v 'B
v 'A v 'B
)45sin('45sin'0 BA vv
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Center of Mass (CM)
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Center of Mass (CM)
The general motion of an object can be considered as the sum of the translational motion of a certain point, plus rotational motion about that point.
How to describe motions like these?
That point is called the center of mass point
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Center of Mass (CM)
Pure translational motion
Translationalplus rotational
motion
This allows us to find a translational motion; however, it doesn’t tell us anything about rotation about the CM.
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
How to find the center of mass?
A CM point depends only on the mass distribution of an object.
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Center of Mass: Definition
321 mmmM
n
iiiCM rm
Mr
1
1 Position vector of the CM:
),,( CMCMCMCM zyxr
1r
2r
3r
n
iiiCM xm
Mx
1
1
n
iiiCM ym
My
1
1
n
iiiCM zm
Mz
1
1
total mass of the systemComponent form:2m
1m
3m
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Center of Mass (2 particles, 1D)
x-axisx=0 x1
m1 m2
x2
MxmxmxCM
2211
M m1 m2
where
MvmvmvCM
2211
MamamaCM
2211
Position of the CM:
Velocity of the CM:
Acceleration of the CM:
n
iiiCM xm
Mx
1
1
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Example: Center of Mass (2 particles)What is the center of mass of 2 point masses (mA=1 kg and mB=3 kg), at two different points: A=(0,0) and B=(2,4)?
rCM 1.5i 3 j
)( BA
BBAACM mm
xmxmx
xCM (10) (32)13
1.5
yCM (1 0) (3 4)13
3
B=(2,4)
CM
A=(0,0)
Or in a vector form:
By definition:
)( BA
BBAACM mm
ymymy
mA=1 kg and mB=3 kg
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
CM motion
m mv vX
X
For unequal masses
X
CM
m mv v/2CM
m 2mv v/3CM
M m1 m2MvmvmvCM
2211 Consider CM velocity :
02
)(
mvmmvvCM
2/2
)0( vmmmvvCM
3/2
)0(2 vmm
mmvvCM
Two equal-mass particles (A and B) are located at some distance from each other. Particle A is held stationary while B is moved away at speed v. What happens to the center of mass of the two-particle system?
A) it does not move
B) it moves away from A with speed v
C) it moves toward A with speed v
D) it moves away from A with speed v
E) it moves toward A with speed v
Let’s say that A is at the origin (x = 0) and B is at some position x. Then the center of mass is at x/2 because A and B have the same mass. If v = x/t tells us how fast the position of B is changing, then the position of the center of mass must be changing like (x/2)/t, which is simply v.
ConcepTest 1 Motion of CM
12
12
12
Xm m vv/2CM ?
2/2
)()0( vm
vmmvCM
MvmvmvCM
2211
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
CM of a solid objectLet’s find CM of an extended body:
ir im
xCM 1M
x dm yCM 1M
ydm zCM 1M
z dm
i
iiCM rmM
r 1
n
iiiCM rm
Mr
1
1 Before, for many particles we had
Now, let’s divide mass into smaller sections ∆mi
i
iimCM rmM
ri
0
lim1
dmrM
rCM 1
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
CM of solid symmetrical objects The easiest trick is to use symmetry
CMCM CM
If we break a symmetry, the CM will be shifted
CM Old CMCM
Old CM
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
CM of Solid Objects (nice trick) How to deal with objects like this?
m1
CMm2
CMCM of the original object
m1
m2
Divide it into symmetrical objects
ConcepTest 2 Center of Mass
(1)
XCM
(2)
A) higherB) lowerC) at the same placeD) there is no definable
CM in this case
CM
The disk shown below in (1) clearly has its center of mass at the center.Suppose a smaller disk is cut out as shown in (2).Where is the center of mass of (2) as compared to (1) ?
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Before we used Newton’s 2nd law for points (which had masses but not sizes)
Now,Newton’s 2nd law for a system of particles
(or extended bodies)
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Newton’s 2nd law for a system of particlesM m1 m2
n
iiiCM rm
Mr
1
1 Position vector of the CM:
CMr1r
2r
3r
Since , then CM acceleration is :td
rda CM2
2
n
iiiCM am
Ma
1
1
n
iiiCM amaM
1
extnet
n
i
extiCM FFaM
1
ernali
externalii FFF int
iparticleonactingforcesFWhere i
)3.(int lawrdNothereachcancelforcesernalHowever
The CM of a system (total mass M) moves like a single particle of mass M acted upon by the same net external force.
n
iiF
1
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
CM point describes translational motion of a system
F
F
F
F
CM
It doesn’t matter where you applied an ext. force, transl. motion of the system will be the same
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Newton’s 2nd law for a system of particles (II)
In the absence of external forces, the motion of the center of mass of a system of particles (or an extended object) is unchanged.
extnetCM FaM
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Thank youSee you on Monday
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 16
Review: Ballistic PendulumA device used to measure the speed of a bullet.
hvom
M v1M+m
Collision:
Mech. Energy is not conserved (wood is crushed).
Lin. Momentum is conserved(ext. forces cancel each other)
FT FT
mg mg
FT FT
Swinging:
Mech. Energy is conserved .
Lin. Momentum is not conserved(ext. forces don’t cancel each other)
ConcepTest 1 Nuclear Fission IA uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?
A) the heavy one
B) the light one
C) both have the same momentum
D) impossible to say
1 2
The initial momentum of the uranium was zero, so the final total momentum of the two fragments must also be zero. Thus the individual momenta are equal in magnitude and opposite in direction.