Summary Lecture 9 Systems of Particles Systems of Particles 9.8-11 9.8-11 Collisions Collisions 9.12 9.12 Rocket propulsion Rocket propulsion Rotational Motion Rotational Motion 10.1 10.1 Rotation of Rigid Body Rotation of Rigid Body 10.2 10.2 Rotational variables Rotational variables 10.4 10.4 Rotation with constant accelerat Rotation with constant accelerat Problems:Chap. 9: 27, 40, 71, 73, 78 Chap. 10: 6, 11, 16, 20, 21, 28, Friday March 24 20-minute test on material in lectures 1-7 during lecture
Friday March 24 20-minute test on material in lectures 1-7 during lecture. Systems of Particles 9.8-11Collisions 9.12Rocket propulsion Rotational Motion 10.1Rotation of Rigid Body 10.2Rotational variables 10.4Rotation with constant acceleration. Summary Lecture 9. - PowerPoint PPT Presentation
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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
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.