Summary Lecture 9Systems of ParticlesSystems of Particles9.129.12 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 acceleration

Systems of ParticlesSystems of Particles9.129.12 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 acceleration

Problems:Chap. 9: 27, 40, 71, 73, 78

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

Problems:Chap. 9: 27, 40, 71, 73, 78

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

This Friday

20-minute teston material in

lectures 1-7

during lecture

This Friday

20-minute teston material in

lectures 1-7

during lecture

Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

http://www.amazon.com/gp/product/images/B000002KO7/ref=dp_image_0/104-9783589-2464723?%5Fencoding=UTF8&n=5174&s=music

Principle of Rocket propulsion:

In an ISOLATED System (no external forces) Momentum is conserved

Momentum = zero

We found that the impulse (p = Fdt) given to the rocket by the gas thrown out the back was

v m

U = Vel. of gas rel. to rocket

Burns fuel at a rate

dtdm

F dt = v dm - U dm

v+v

Force on Rocket

An example of an isolated system where momentum is conserved!

The impulse driving the rocket, due to the momentum, of the gas is given by

Note:

since m is not constant dt

dvm

Now the force pushing the rocket is F = dt

dprocket

(mv)dt

dF i.e.

mdt

dvv

dt

dmF

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

dv = -U dm

m

F dt = v dm - U dm

F dt = v dm + m dv

This means: Every time 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.

mf

mi

f

i

dmm

1U

v

vdvThus

dv = -U dm

m

mf

mi

vf

vi]m[U]v[ ln

)mm(Uvvifif

lnln

)mm(Ufi

lnln

f

i

fi 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

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

=2.3 x 2800

= 6400 N

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

= 6400/850 = 7.6 m s-2

Final vel.

1

f

i

f

sm4300180

8502800

m

mUv

ln

ln

F = ma

Thrust –mg = ma

6400 – 8500 = ma

a = -2100/850

= -2.5 m s-2

= U dm/dt

Thrust = 6400 N

mg = 8500 N

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

is a vector

is the rotational analogue of v

Angular Velocity

is rate of change of units of …rad s-1

How do we specify its angular velocity?

right hand

tttav

12

12

is a vector

direction: same as .

Units of -- rad s-2

is the analogue of a

Angular Acceleration

dt

d

ttinst

0

limit

angular acceleration

Consider an object rotating according to:

= -1 – 0.6t + 0.25 t2

= d/dt

e.g at t = 0 = -1 rad

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

= - .6 + .5t

Angular motion with constant accelerationAngular motion with constant acceleration

0= 33¹/³ RPM

sec/rad60

2πx

3

100ω0

= -0.4 rad s-2

Q1 How long to come to rest?

Q2 How many revolutions does it

take?

= 3.49 rad s-1

= 8.7 s

2

2

1 atuts 2

21

0 tt

= 15.3 rad

= 15.3/22.43 rev.

atuv

0

0

t

t

An example where is constant

0

-

++

Relating Linear and Angular variables

r

s = r

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

and s

s

s = r

dt

dsv

v

r

and v

ωrv

r)(dt

dv

rdt

dθv

V, r, and are all vectors.

Although magnitude of v = r.

The true relation is v = x r

Not quite true.

s

Relating Linear and Angular variables

v = x r

r v

Grab first vector () with right hand.

Direction of screw is direction of third vector (v).

Turn to second vector (r) .

Direction of vectors

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 = 0 + k AxBy - kAyBx + 0 = 0 - k ABsin

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

Is a vector?

However is a vector!

Rule for adding vectors:

The sum of the vectors must not depend on the order in which they were added.

Rule for adding vectors:

The sum of the vectors must not depend on the order in which they were added.

This term is the

tangential accel atan.

(or the rate of increase of v)

The centripetal acceleration of circular motion.

Direction to centre

a and rv x

)rω(dt

d

dt

dva x

rdt

ωd

dt

rdωa xx

r

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

rαvωa xx

vv

Relating Linear and Angular variables

Total linear acceleration a

The acceleration “a” of a point distance “r” from axis

consists of 2 terms:

Tangential acceleration

(how fast v is changing)

a and

a = r & v2/r

Central accelerationPresent even when is zero!

r

Relating Linear and Angular variables

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

The tangential acceleration at the end is twice this, but the acceleration due to gravity of any mass point is only g

cosThe rod only falls as a body because it is rigid

The Falling Chimney

L

and at the end: atan = L

gcos

………..the chimney is NOT.………..the chimney is NOT.