Chapter 7 Systems of particles

7-5 Conservation of momentum in a system of particles

7-4 Center of mass of solid objects

7-3 Many-particle system

7-2 Two-particle system

7-1 The motion of a complex object

When can the object studied be regarded as a mass point? Doing only translational motion

Translational + Rotational motion

Translational motion Rotational motion

7-1 The motion of a complex object

When we projectile a rigid body, the motion of the body looks very complicated.

We can consider the motion of the rigid body to be a combination of a parabolic trajectory of a “center of mass” (rotational motion is not considered) plus a rotation about “center of mass” (translational motion is not considered).

How to find the center of mass (CM) of a particle system?

7-2 Two-particle system

As an example, we consider a baton consists of two particles , at its ends A and B, connected by a thin rigid rod of fixed length and negligible mass.

12 2mm

1m 2m

1m

2m

A

B

Do an experiment to find CM in a two-particle system.

Clearly both and are accelerated, however, one point in the rod (point C) moves with constant velocity.

2m1m

We give the rod a push along the frictionless horizontal surface and examine its motion.

Snapshots of the locations of points A and B at successive intervals of time.

If point c is regarded as a reference, A and B points rotate with a constant rotational speed.

View the motion from the reference of point C.

So point c is actually the center of mass.

Fig 7-5

1m

2m1r

x

y

O21

2211

mm

rmrmrcm

21

2211

mm

xmxmxcm

21

2211

mm

ymymycm

cmr

2r

'cmr

C C’

By building a Cartesian coordinate, position of point c is found at :

(7-2)

(7-1)

or written as:

1m

2m

A

BC

From Eq. (7-1), the velocity and acceleration of the CM are: (7-4) (7-6)

21

2211

mm

vmvm

dt

rdv cmcm

21

2211

mm

amam

dt

vda cmcm

Can we also find is zero from Eq. (7-6)?cma

rr FFamam 212211

0

ca

rr FF 12

How about the motion if the system has net external forces?

Suppose there is an external force on each particle in above expt., then

This looks very like a particle of mass located at the center of mass.

rextrext FFFFFFamam 2211212211

021

rr FF

extextext FFF 21, and if write

cmext ammF )( 21

extFamam

2211

21

2211

mm

amamacm

21 mm

Newton’s second law for systems of particles

7-3 Many-particle system

Consider a system consisting of N particles of masses …… . The total mass is

(7-10) Each particle can be represented by its

location , velocity and its acceleration . The CM of the system can be defined by

logical extension of Eq(7-1):

1m 2m Nm

nmM

nvnr

na

In terms of components, Eq(7-11) can be written as

Taking the derivative of Eq(7-11) (7-13)

Differentiating once again: (7-14)

nncm xmM

x1 nncm ym

My

1 nncm zm

Mz

1

nncm vmM

v1

nncm amM

a1

Nnncm FFFamaM 21Or (7-15)

nnN

NNcm rm

Mmmm

rmrmrmr

1

21

2211

(7-11)

(7-12), , .

By Newton’s third law, in Eq(7-15) the vector sum of all the internal forces is cancelled, and Eq(7-15) reduces to

(7-16)

cmext aMF

cma

Eq(7-16) is just the Newton’s second law for the system of N particles treated as a single particle of mass M located at the center of mass( ), experiencing .

cmr

nncm amM

a1

nncm rmM

r1

We can summarize this important result as follow:

“The overall translational motion of a system of particles can be analyze using Newton’s law as if the mass were concentrated at the center of mass and the total external force were applied at that point.”

These are general results that apply equally well to a solid object.

Sample problem 7-3 A projectile( 射弹 ) of mass 9.8kg is launched from the

ground with initial velocity of 12.4m/s at an angle of above the horizontal (Fig 7-11). At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass 6.5kg, is observed at 1.42s after the launch at a height of 5.9m and a horizontal distance of 13.6m from the launch point. Find the location of the second fragment at that same time.

54

0v

0

CM

1m

2m

Solution: If the projectile had not exploded, the

location of the projectile at t=1.42s should have been

It is the location of the CM.

mssmtvx x 4.1042.1)/3.7(0

m

ssmssm

gttvy y

3.4

)42.1()/80.9(2

1)42.1()/0.10(

2

1

22

20

21

2211

mm

xmxmxcm

21

2211

mm

ymymycm

? ?

By Eq(7-12)

Fig 7-11

mm

ymMyy cm 9.0

2

112

m

kg

mkgmkg

m

xmMxx cm

7.3

1.3

)6.13()5.6()4.10()6.9(2

112

0v

0

Cm

1m

2m

7-4 Center of mass of solid objects 1) If an object has symmetry, the CM must lie at the geom

etrical symmetrical center of the object. Suppose the mass is uniformly distributed.

cm

2) If the object has no symmetry, sometimes it is also easy to find its cm position:

(a) (b)

cm

cm

Sample problem 7-4 Fig 7-13 shows a

circular metal plate of radius 2R from which a disk of radius R has been removed. Find the cm (x) of the plate.

Fig 7-13

x

y

C xDR

Solution: Due to the mirror symmetry about the x axis,

the cm must lie along the x axis. If the hole is filled with a disk of the same

material of radius R, the cm of composite disk is at the origin of the coordinate system.

RRRR

Rx

m

mx D

x

Dx 3

1)(

)2()(

22

2

xD

xxDDc mm

xmxmx

0 CM for the big circular plate

?

3) If we encounter solid irregular objects, we can divide infinite small elements. And the sums of Eqs(7-12) transform into integrals:

xdmM

mxM

x nnm

cm

1lim

10

ydmM

myM

y nnm

cm

1lim

10

zdmM

mzM

z nnm

cm

1lim

10

dmrM

rcm1 (7-19)In vector form are

(7-18)

x

O

y

z

Sample problem 7-5 A thin strip of

material is bent into the shape of a semicircle of radius R. Find its center of mass.

Fig 7-14

x

y

0

d

Mdm cmy

Solution: The strip has symmetry about the y axis.

So

( the small element of mass dm subtends an angle . The location of the dm is )

0cx

RR

dR

dM

RM

ydmM

ycm

637.02

sin

sin11

0

0

dsinRy

7-5 Conservation of momentum in a system of particles 1) For a system containing N particles, the total

momentum is (7-21) Here (7-22)

If the net external force acting on a system is zero, then and so the total linear momentum of the system remains constant.

cm

N

n

nnN

nnn

N

nn vM

M

vmMvmPP

111

extcmcm FaMdt

vdM

dt

Pd

P

)( imM

0dtPd

P

2) If we view the system from the cm frame, the velocity of a particle in this frame is

(7-24) Then in this cm frame, the total momentum is

(7-25)

0

''111

cmcm

cm

N

nn

N

nnn

N

nnn

vMvM

vmvmvmP

)(' cmnn vvv

'nv

Sample problem 7-8 As Fig 7-17 shows a cannon whose mass M is 1

300kg fire a 72kg ball in a horizontal direction with a speed vbc of 55m/s relative to the cannon. The cannon is mounted and can recoil ( 后退 ) freely.

(a) what is the velocity vcE of the recoiling cannon with respected to the Earth?

(b) what is the initial velocity vbE of the ball with respected to the Earth?

bcv

MvcEm

Fig 7-17

0

fxix PP

bcv

MvcEm

Fig 7-17

Solution:

0 bEcEf mvMvP

cEbEbc vvv

Momentum in horizontal direction is conserved.