Summary Lecture 9
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Transcript of Summary Lecture 9
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