summary lecture 9 systems of particles 9.12rocket propulsion rotational motion 10.1rotation of rigid...
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![Page 1: Summary Lecture 9 Systems of Particles 9.12Rocket propulsion Rotational Motion 10.1Rotation of Rigid Body 10.2Rotational variables 10.4Rotation with constant](https://reader030.vdocuments.us/reader030/viewer/2022032703/56649d2b5503460f94a010ef/html5/thumbnails/1.jpg)
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
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Principle of Rocket propulsion:
In an ISOLATED System (no external forces) Momentum is conserved
Momentum = zero
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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
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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.
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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.
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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)
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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
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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
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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.)
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The unit of is radian (rad)
There are 2 radian in a circle
2 radian = 3600
1 radian = 57.30
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X
Y
Rotation axis (Z)
tttav
12
12
is a vectordtd
ttinst
0limit
Angular Velocity
At time t1
At time t 2
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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
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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
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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
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Angular motion with constant accelerationAngular motion with constant acceleration
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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
-
++
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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
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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
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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
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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
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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.
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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
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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
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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.