Summary Lecture 9 Systems of Particles 9.8-11Collisions 9.12Rocket propulsion Rotational Motion...
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Transcript of Summary Lecture 9 Systems of Particles 9.8-11Collisions 9.12Rocket propulsion Rotational Motion...
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 acceleration
Systems 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,
Problems: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
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???
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 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 vmmm
v 1
21
12 )
2(
if v
mmmm
v 1
21
211 )(
v2f(1 + m2/m1) = 2v1i
v2f = v1i + v1f …………….…(4)
if vmmm
v 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
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
mm
mV
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
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
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
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
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
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 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: 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
m
m
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
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.)
X
Y
Rotation axis (Z)
tttav
12
12
is a vectordtd
ttinst
0limit
Angular Velocity
At time t1
At time t 2
tttav
12
12
is a vector
direction of change in .
Units of -- rad s-2
is the analogue of a
Angular Acceleration
dt
d
ttinst
0
limit
0= 33¹/³ RPM
sec/rad602π
x3
100ω0
= -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
2
2
1 atuts 2
21
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
s
s = 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
dt
dsv
vr
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
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)(ωdt
d
dt
dva x
rdt
dω
dt
drωa 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