linear momentum. center of mass. - umass...

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


Chapter 9

2-D collisions Systems of particles (extended objects) Center of mass

Outline


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:


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.


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


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


Center of Mass (CM)


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


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.


How to find the center of mass?

A CM point depends only on the mass distribution of an object.


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


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


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


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


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


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


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) ?


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)


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


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


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


Thank youSee you on Monday


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.

linear momentum. center of mass. - umass...

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