Summary Lecture 9Systems of ParticlesSystems of Particles9.8-119.8-11 CollisionsCollisions9.129.12 Rocket propulsionRocket propulsion

Rotational MotionRotational Motion

10.110.1 Rotation of Rigid BodyRotation of Rigid Body

10.210.2 Rotational variablesRotational variables

10.410.4 Rotation with constant accelerationRotation with constant accelerationProblems:Chap. 9: 27, 40, 71, 73, 78

Chap. 10: 6, 11, 16, 20, 21, 28,

Friday March 24

20-minute teston material in

lectures 1-7

during lecture

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

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)

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 fifi

f vmmv

mvmvm

v 21

21

1

22111

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

if vmmmv 1

21

12 )2(

if v

mmmmv 1

21

211 )(

v2f(1 + m2/m1) = 2v1i

v2f = v1i + v1f …………….…(4)

if vmmmv 1

21

12 )2(

if v

mmmmv 1

21

211 )(

If m1>> m2 v2f 2v1i

If m1= m2

If m2>>m1

v1f 0

v2f 0

v2f v1i

v1f v1i

v1f -v1i

Analyze the equations

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

mmmV

Collision viewed from Lab. Ref. frame

m1

v1i

m2

v2i =0

vcm

CM

Collision viewed from Lab. Ref. frame

Note that the CM moves at constant vel

Because there is no EXTERNAL force acting on the system

Collision viewed from Lab. Ref. frame

Note that the CM moves at constant vel

Because there is no EXTERNAL force acting on the system

Collision viewed from Lab. Ref. frame

Let’s observe the elastic collision from the view point

of the centre of mass

Note that the CM is at rest

Collision viewed from CM Ref. frame

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

mmmV

m1

v1i

m2

v2ivcm

CM

Completely inelastic collision

Observing from the Lab. reference frame

Note that the CM moves at constant vel

Because there is no EXTERNAL force acting on the system

Observing from the Lab. reference frame

Note that the CM moves at constant vel

Because there is no EXTERNAL force acting on the system

Observing from the Lab. reference frame

Let’s observe the inelastic collision from the view point of

the centre of mass

Observing from the CM reference frame

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

22f2

21f1

21i1 vm

21vm

21vm

21

7 variables! 3 equations

Elastic collisions in 2-D

m1v1i

before

21

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

P fy=

p f sin

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

P fy=

p f sin

=

mAvA

Divide equ (2) by (1)

AA

BB

vmvmtan

____________________ 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

P fy=

p f sin

=

mAvA

= 39.80 Use equ 2 to find Vf

sin)mm(vmvBA

BBf

Gives

Vf = 48.6 kph

AA

BB

vmvmtan

Can the investigators determine who was speeding?

sin)mm(vmvBA

BBf

AA

BB

vmvmtan

mA= 830 kgmA= 830 kgmA= 830 kg

mB = 550 kgmB = 550 kg

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

Conservation of Energy

½ mvf2 = f.d

d

f = N = mA + mB) g

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

V = U - v

v m

U = Vel. of gas rel. to rocket

Burns fuel at a rate

dtdm

Mom. of gas = mV = m(U - v)

F dt = v dm - U dm

v+v

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

Impulse is mom. transfer (p)

So since F = dp/dt, p = Fdt

= - change in mom. of rocket (impulse or p)

Force on Rocket

An example of an isolated system where momentum is conserved!

Note:

since m is not constant dtdvm

Now the force pushing the rocket is F = dt

dprocket

(mv)dtdF i.e.

mdtdvv

dtdmF

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

dv = -U dmm

F dt = v dm - U dm

F dt = v dm + m dv

This means: If I throw out a mass dm of gas with a velocity U, when the rocket has a mass m, the velocity of the rocket will

increase by an amount 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 lnf

i

mm

mf

mi

f

i

dmm1U

v

vdvThus

dv = -U dmm

mfmi

vfvi ]m[U]v[ ln

)mm(Uvv ifif lnln

)mm(U fi lnln

f

ifi m

mUv0vif ln

= logex = 1/x dx

e = 2.718281828…

This means: If I throw out a mass dm of gas with a velocity U, when the rocket has a mass m, the velocity of the rocket

will increase by an amount dv.

Fraction of mass burnt as fuel

Spee

d in

uni

ts o

f gas

vel

ocity

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

=2.3 x 2800

= 6400 N

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

= 6400/850 = 7.6 m s-2

Final vel.

1

f

if

sm43001808502800

mmUv

ln

ln

F = ma

Thrust –mg = ma

6400 – 8500 = ma

a = -2100/850

= -2.5 m s-2

= U dm/dt

Rotation of a body about an axis

RIGID n FIXED

Every point of body

moves in a circle

Not fluids,. Every point is

constrained and fixed relative to

all others

The axis is not translating.

We are not yet considering

rolling motion

reference line fixed in body

X

Y

Rotation axis (Z)

The orientation of the rigid

body is defined by .(For linear motion position is

defined by displacement r.)

The unit of is radian (rad)

There are 2 radian in a circle

2 radian = 3600

1 radian = 57.30

X

Y

Rotation axis (Z)

tttav

12

12

is a vectordtd

ttinst

0limit

Angular Velocity

At time t1

At time t 2

Angular

velocity

is a vector

is rate of change of

units of …rad s-1

is the rotational analogue of v

tttav

12

12

is a vector

direction of change in .

Units of -- rad s-2

is the analogue of a

Angular Acceleration

dtd

ttinst

0limit

= -1 – 0.6t + .25 t2

= d/dt = - .6 + .5t

e.g at t = 0 = -1 rad

e.g. at t=0 = -0.6 rad s-1

Rotation at constant acceleration

0= 33¹/³ RPM

sec/rad602πx

3100ω0

= -0.4 rad s-2

How long to come to rest?

How many revolutions does it

take?

=3.49 rad s-2

= 8.7 s

221 atuts 221

0 tt

= 45.5 rad

= 45.5/27.24 rev.

atuv

0

0

0

t

t

An example where is constant

Relating Linear and Angular variables

r

ss = r

Need to relate the linear motion of a point in the rotating body with the angular variables

and s

Relating Linear and Angular variables

s = r

dtdsv

vr

and v

ωrv

r)(dtdv

rdtdθv

V, r, and are all vectors.

Although magnitude of v = r.

The true relation is v = x r

Not quite true.

s

v = x r

vr

So C = (iAx + jAy) x (iBx + jBy)

= iAx x (iBx + jBy) + jAy x (iBx + jBy)

= ixi AxBx + ixj AxBy + jxi AyBx + jxj AyBy

Ay = Asin

Ax = Acos

A

B

C = A x B

Vector Product

A = iAx + jAy B = iBx + jBy

C= ABsin So C = k AxBy - kAyBx

= 0 - k ABsin

now ixi = 0 jxj = 0ixj = k jxi = -k

This term is the

tangential acceln atan.

(or the rate of increase of v)

Since = v/r this term = v2/r (or 2r)

rαvωa xx

The centripital acceln of circular motion.Direction to centre

r

a and Relating Linear and Angular variables

rωv x

r)(ωdtd

dtdva x

rdtdω

dtdrωa xx

Total linear acceleration a

Thus the magnitude of “a”

a = r - v2/r

Tangential acceleration

(how fast V is changing)

Central acceleration

r

Relating Linear and Angular variables

a and

CMg

The whole rigid body has an angular acceleration

The tangential acceleration atan distance r from the base is

atanr

at the CM, atanL/2, and at end atanL

Yet at CM, atan= g cos (determined by gravity)

gcos

At the end, the tangential acceleration is twice this, yet the maximum tangential acceleration of any mass point is g cosThe rod only falls as a body because it is rigid…the chimney is NOT.

The Falling Chimney

L