Summary Lecture 8

Systems of ParticlesSystems of Particles

9.29.2 Centre of massCentre of mass

9.39.3 Newton 2 for system of particlesNewton 2 for system of particles

9.4-79.4-7 Conservation of momentumConservation of momentum

9.8-119.8-11 CollisionsCollisions

9.129.12 Rocket propulsionRocket propulsion

Systems of ParticlesSystems of Particles

9.29.2 Centre of massCentre of mass

9.39.3 Newton 2 for system of particlesNewton 2 for system of particles

9.4-79.4-7 Conservation of momentumConservation of momentum

9.8-119.8-11 CollisionsCollisions

9.129.12 Rocket propulsionRocket propulsion

Problems:Chap. 9: 1, 6, 10 , 15, 20,

27, 40, 71, 73, 78

Problems:Chap. 9: 1, 6, 10 , 15, 20,

27, 40, 71, 73, 78

Tomorrow 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

Tomorrow 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

Week March 20 – 2420-minute teston material in

lectures 1-7

Week March 20 – 2420-minute teston material in

lectures 1-7

So far we have considered the motion of

POINT PARTICLES

FINITE OBJECTS

can move as a whole

(translational motion)

and also

rotate about the “Centre of Mass”

What happens depends on where we apply the force

The “Centre of Mass” is that point where if we apply a force, the object will not rotate.

The motion of the Centre of Mass is a simple parabola.

(just like a point particle)

The motion of the entire object is complicated.

This motion resolves to

•motion of the CM

The Centre of Mass

•motion of points around the CM

Lizzie Borden took an axeAnd gave her mother forty whacks.And when she saw what she had done,

She gave her father forty-one

m1 m2

m1

m2

m1 m2

m 1

m 2

m1 m2

m1g m2g

d1 d2

M = m1 + m2

The Centre of Mass

CMM

m1g x d1 = m2g x d2

Centre of Mass (1D)

0x1 x2

xcm

m1M m2

M = m1 + m2

M xcm = m1 x1 + m2 x2

Mxmxm

x 2211cm

In general iicm xmM1

x

moment of M = moment of individual masses

Centre of Mass (3D)

iicm rmM1

r

For a collection of masses in 3D

cmr1

r3r2

Centre of Mass (3D)

iicm rmM1

r

For a collection of masses in 3D

cm

rcm = ixcm +jycm

iicm ymM

y1

So in a solid body we can find the CM by finding xcm and ycm

iicm xmM

x1

y

x0

2

21

1

3 kg

8 kg

4 kg

iicm rmM1

r

iicm ymM1

y

)1x42x80x3(15

1ycm

Xcm = 16/15 = 1.07 m

ycm = 20/15 = 1.33 m

15 kg

rCM

1.07

1.33

iicm xmM

1x

)2x41x80x3(15

1xcm

becomes an integral: dmrM

1rcm

For odd shaped objects this probably needs to be determined experimentally

For symmetric objects this can often be calculated

1. Look for a symmetry axis

2. Then carry out the integral to find the position of xcm along the axis.

Finding the Centre of Mass

For solid bodies the sum: mrM

1rcm

dx

h

Rr

dmxM

xcm 1

dm = r2 dx

but r = (R/h)x

dm = (R/h)2x2 dx

x

dxxMhR

x hcm 0

3

2

2

4

4

2

2 hMhR

xcm

Mass of cone M = 1/3 R2 h4

3 hMM

xcm

xcm = ¾ h

xcm

4

2 hM

hRxcm

Symmetry linex

Solid cone

cmMaF

cm

Sum of all EXTERNAL forces acting on system

The total mass of the system

The acceleration of the CM of the system

zcmzext MaF

ycmyext MaF

xcmxext MaF

For a system of particles, the dynamics of the Centre of Mass obeys Newton 2.

cmCM

cmMaF

zcmzext MaF

ycmyext MaF

xcmxext MaF

For a system of particles, the dynamics of the Centre of Mass obeys Newton 2.

This also applies to a solid body, where the individual particles are rigidly connected. The dynamics of the Centre of Mass obeys Newton 2

You will recall that madt

dpF

Where p=mv is the momentum of each particle

For a system of particles

P = p = Mvcm

Linear Momentum of system of particles

This also applies to extended objects

maFext

cm

dt

dPFext

for system of particles

Conservation of Linear Momentum

If Fext = 0 NO EXTERNAL forces act on the system

0dtPd

P is a constantThat is: Px, Py and Pz remain constant

if Fext-x, Fext-y and Fext-z are zero

In an isolated system, momentum is conserved.

dt

dPFext

Exploding rocket

Why?

No external horizontal forces

so horiz momentum unchanged

C of M

m = 3.8 g, n =12

v = 1100 m s-1

Define systemInitial momentum Pi = n mv + M Vi

= n mv + 0

V

nmMnmv

V

124 smV .

M=12 kg

M= 12 kg

+ 12 m

Final momentum pf = (M + nm) V = Pi = n mv

m = 3.8 g, n =12

v = 1100 m s-1

Initial momentum Pi = n mv + M Vi

= n mv + 0

V

nmMnmv

V

124 smV .

M=12 kg

M= 12 kg

+ 12 m

Final momentum pf = (M + nm) V = Pi = n mv

KE initial

½ n mv2

½ x 12 x 0.0038 x (1100)2

27588 J

KE final

½ (M + 12m)V2

½ x (12.0456) x 4.22

163 J

What is a collision?Collisions

An isolated event involving 2 or more objects

No external forces

Momentum is conserved

Usually interact (often strongly) for short time

Equal and opposite impulses are exerted on each other

p = F dt

Elastic collisionsEnergy and momentum are conserved

Inelastic collisionsOnly momentum is conserved

Collisions

But Energy is always conserved???

In 1 dimension

m1

v1i

m2

v2i = 0Before

m1

v1f

m2

v2f After

Elastic Collision 1D

We want to find V1f and V2f

m2v2f2 = m1(v1i- v1f) (v1i+ v1f) ……(3)

Mom. Cons. m1v1i = m1v1f + m2v2f………………(1) m2v2f = m1(v1i- v1f)…………………(2)

Energy Cons ½ m1v1f2 + ½ m2v2f

2 = ½ m1v1i2

½ m2v2f2 = ½ m1(v1i

2 - v1f

2)Mult. by 2 and factorise

Divide equ. (3) by (2)

V1i is usually given, so to find v2f we need to find an expression for v1f. Get this from equ. (1).

m1v1f = m1v1i - m2v2f fi

fi

f vmm

vm

vmvmv 2

1

21

1

2211

1

Substitute this form of v1f into equ 4 v2f = v1i + v1i – m2/m1 v2f

if vmm

mv 1

21

12 )

2(

if v

mmmm

v 1

21

211 )(

v2f(1 + m2/m1) = 2v1i

v2f = v1i + v1f …………….…(4)

if vmm

mv 1

21

12 )

2(

if v

mmmm

v 1

21

211 )(

If m1>> m2 v2f 2v1i

If m1= m2

If m2>>m1

v1f 0

v2f 0

v2f v1i

v1f v1i

v1f -v1i

m1

v1i

m2

v2i =0

vcm

CM

What is Vcm?

Mom of CM = mom of m1 + mom of m2

(m1 + m2 ) Vcm = m1v1i + m2v2i

i121

1cm v

mm

mV

Motion of the C of M

m1

v1i

m2

v2i =0

vcm

CM

Motion of the C of M

Let’s observe the elastic collision from the view point of the centre of mass

In 1 dimension

m1

v1i

m2

v2i =0

vcm

CM

What is Vcm?

Mom of CM = mom of m1 + mom of m2

(m1 + m2 ) Vcm = m1v1i + m2v2i

i121

1cm v

mm

mV

m1

v1i

m2

v2i

vcm

CM

Let’s observe the elastic collision from the view point of the centre of mass

Elastic

billiard balls

comets

-particle scattering

Collisions in 2 dimensions

Momentum is conserved

Consider x-components

m1v1i= m1v1f cos 1 + m2v2f cos 2

Consider y-components

0= -m1v1f sin 1 + m2v2f sin 2

Since elastic collision energy is conserved

2

2f2

2

1f1

2

1i1 vm21

vm21

vm21

7 variables! 3 equations

Elastic collisions in 2-D

m1v1i

before

2

1

m 2v 2f

m1 v

1f

after

Impact parameter

Inelastic

Almost any real collision!

an example

Automobile collision

Collisions in 2 dimensions

mB = 550 kg

vB = 78 kph

V f =

pA

pB

pf

Pfy=

pf s

in

Pfx= pf cos

mA= 830 kg

va = 62 kph

Cons. Momentum ==> pA + pB = pf

X component PA = Pf cos

mAvA = (mA+ mB) vf cos………….(1)

Y component PB = Pf sin

mBvB = (mA+ mB) vf sin………….(2)

pA

pB

pf

Pfx= pf cos

Pfy=

pf s

in

=

mAvA

Divide equ (2) by (1)

AA

BB

vm

vmtan

____________________ mAvA = (mA+ mB) vf cos

Gives

= 39.80

Cons. Momentum ==> pA + pB = pf

X component PA = Pf cos

mAvA = (mA+ mB) vf cos………….(1)

Y component PB = Pf sin

mBvB = (mA+ mB) vf sin………….(2)

pA

pB

pf

Pfx= pf cos

Pfy=

pf s

in

=

mAvA

= 39.80Use equ 2 to find Vf

sin)mm(

vmv

BA

BBf

Gives

Vf = 48.6 kph

AA

BB

vm

vmtan

Can the investigators determine who was speeding?

sin)mm(

vmv

BA

BBf

AA

BB

vm

vmtan

mA= 830 kgmA= 830 kgmA= 830 kg

mB = 550 kgmB = 550 kg

http://www.physics.ubc.ca/~outreach/phys420/p420_96/danny/danweb.htm

IN THE EARTH REF. FRAMEVel of gas rel me = vel of gas rel. rocket - vel of rocket rel me

= U - v

v v+vm

U = Vel. of gas rel. to rocket

Burns fuel at a rate

dtdm

Mom. of gas = m(U - v) = -change in mom. of rocket (impulse)

i.e. F dt = m(v - U) = v dm - U dm

Note since m is not constant dt

dvm

Now the force pushing the rocket is F = dt

dprocket

)(mvdt

dF i.e.

mdt

dvv

dt

dm

so that v dm + m dv = v dm - U dm

dv = -U dm

m

This means that if I throw out dm of gas with vel. U,I will increase rocket velocity by dv.

F dt = m(v - U) = v dm - U dm

Fdt = v dm + m dv

If I want to find out the TOTAL effect of throwing out gas, from when the mass was mi and velocity was vi, to the time when the mass is mf and the velocity vf, I must integrate.

vf = U ln m

m

i

fif vi = 0 vf = U ln

f

i

m

m

mi

mdm

mU

v

vdv

i

f

i

1Thus

[] [ln]v Umvivf

mimf note

1

mdmmln

vf -vi = - U (ln mf - ln mi)

= + U (ln mi - ln mf)

dv = -U dm

mThis means that if I throw out dm of gas with vel. U,I will increase rocket velocity by dv.

Fraction of mass burnt as fuel

Sp

eed

in u

nit

s of

gas

vel

ocit

y

1

2

.2 .4 .6 .8 1

Constant mass (v = at)

Reducing mass (mf = 0)

An exampleMi = 850 kg

mf = 180 kg

U = 2800 m s-1

dm/dt = 2.3 kg s-1

Thrust = dp/dt of gas = dm/dt U

=2.3 x 2800

= 6400 N

Initial acceleration F = ma ==> a = F/m

= 6400/850 = 7.6 m s-2

Final vel.

14300180

8502800

sm

m

mUv

f

if

ln

ln