Summary Lecture 9

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Summary Lecture 9
Systems of Particles 9.8-11 Collisions 9.12 Rocke
t propulsion Rotational Motion 10.1 Rotation of
Rigid Body 10.2 Rotational variables 10.4 Rotation
with constant acceleration
ProblemsChap. 9 27, 40, 71, 73, 78 Chap. 10
6, 11, 16, 20, 21, 28,
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coll
isions
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Collisions
Elastic collisions Energy and momentum are
conserved
Inelastic collisions Only momentum is conserved
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Elastic collision
In 1 dimension
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Elastic Collision 1D
Mom. Cons. m1v1i m1v1f m2v2f(1)
? m2v2f m1(v1i- v1f)(2)
Energy Cons ½ m1v1f2 ½ m2v2f2 ½ m1v1i2
? ½ m2v2f2 ½ m1(v1i2 - v1f2)
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Mom. Cons. m1v1i m1v1f m2v2f(1)
? m2v2f m1(v1i- v1f)(2)
Energy Cons ½ m1v1f2 ½ m2v2f2 ½ m1v1i2
? ½ m2v2f2 ½ m1(v1i2 - v1f2)
Mult. by 2 and factorise
? m2v2f2 m1(v1i- v1f) (v1i v1f) (3)
Divide equ. (3) by (2)
? v2f v1i v1f .(4)
V1i is usually given, so to find v2f we need to
find an expression for v1f. Get this from equ.
(1).
m1v1f m1v1i - m2v2f ?
Substitute this form of v1f into equ 4
? v2f v1i v1i m2/m1 v2f
? v2f(1 m2/m1) 2v1i
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Analyze the equations
If m1gtgt m2
v2f ? 2v1i
v1f ? v1i
If m2gtgtm1
v2f ? 0
v1f ? -v1i
If m1 m2
v1f ? 0
v2f ? v1i
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Collision viewed from Lab. Ref. frame
m1 v1i
m2 v2i 0
vcm
What is Vcm? Mom of CM mom of m1 mom of
m2 (m1 m2 ) Vcm m1v1i m2v2i
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Collision viewed from Lab. Ref. frame
m1 v1i
m2 v2i 0
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Collision viewed from Lab. Ref. frame
Note that the CM moves at constant vel Because
there is no EXTERNAL force acting on the system
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Collision viewed from Lab. Ref. frame
Note that the CM moves at constant vel Because
there is no EXTERNAL force acting on the system
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Lets observe the elastic collision from the view
point of the centre of mass
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Collision viewed from CM Ref. frame
Note that the CM is at rest
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Collision viewed from CM Ref. frame
Note that the CM is at rest
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inelastic collision
In 1 dimension
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m1 v1i
m2 v2i 0
What is Vcm? Mom of CM mom of m1 mom of
m2 (m1 m2 ) Vcm m1v1i m2v2i
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Completely inelastic collision
m1 v1i
m2 v2i
Observing from the Lab. reference frame
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Note that the CM moves at constant vel Because
there is no EXTERNAL force acting on the system
Observing from the Lab. reference frame
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Note that the CM moves at constant vel Because
there is no EXTERNAL force acting on the system
Observing from the Lab. reference frame
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Lets observe the inelastic collision from the
view point of the centre of mass
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Observing from the CM reference frame
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Collisions in 2 dimensions
Elastic billiard balls comets a-particle
scattering
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Elastic collisions in 2-D
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
7 variables!
3 equations
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Collisions in 2 dimensions
Inelastic Almost any real collision! An example
Automobile collision
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Cons. Momentum gt pA pB pf X component PA
Pf cosq mAvA (mA mB) vf cosq.(1) Y
component PB Pf sinq mBvB (mA mB) vf
sinq.(2)
____________________ mAvA (mA mB) vf cosq
Divide equ (2) by (1)
Gives q 39.80
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Gives Vf 48.6 kph
Use equ 2 to find Vf
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Can the investigators determine who was speeding?
Conservation of Energy ½ mvf2 f.d
f mN m(mA mB) g
http//www.physics.ubc.ca/outreach/phys420/p420_9
6/danny/danweb.htm
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momentum conservation and
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That's all folks
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Dm
An example of an isolated system where momentum
is conserved!
IN THE EARTH REF. FRAME Vel of gas rel me vel
of gas rel. rocket - vel of rocket rel me V
U - v
Mom. of gas ?mV ?m(U - v)
- change in mom. of rocket (impulse or ?p)
Impulse is mom. transfer (?p) So since F
dp/dt, ?p Fdt
i.e. F dt ?m(v - U)
F dt v dm - U dm
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F dt v dm - U dm
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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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.
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An example
Mi 850 kg mf 180 kg U 2800 m s-1 dm/dt
2.3 kg s-1
Thrust dp/dt of gas
U dm/dt
2.3 x 2800 6400 N
?F ma Thrust mg ma 6400 8500 ma a
-2100/850 -2.5 m s-2
Initial acceleration F ma gt a F/m
6400/850 7.6 m s-2
Final vel.
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Rotation
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n FIXED
Rotation of a body about an
axis
RIGID
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The orientation of the rigid body is defined by
q. (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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Angular Velocity
At time t2
At time t1
w is a vector
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Angular velocity w
w is a vector w is rate of change of q units of
wrad s-1
w is the rotational analogue of v
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Angular Acceleration
a is a vector direction of change in w. Units of
a -- rad s-2 a is the analogue of a
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?

• -1 0.6t .25 t2

e.g at t 0 ? -1 rad
? d?/dt ? - .6 .5t
e.g. at t0 ? -0.6 rad s-1
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Rotation at constant acceleration
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An example where ? is constant
3.49 rad s-2
8.7 s
? -0.4 rad s-2 How long to come to rest? How
many revolutions does it take?
45.5 rad 45.5/2??7.24 rev.
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Relating Linear and Angular variables
q and s
Need to relate the linear motion of a point in
the rotating body with the angular variables
s qr
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Relating Linear and Angular variables
w and v
s qr

Not quite true.
V, r, and w are all vectors. Although magnitude
of v wr. The true relation is v w x r
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v ? x r
w
r
v
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Vector Product
C A x B
A iAx jAy B iBx jBy
So C (iAx jAy) x (iBx jBy) iAx x
(iBx jBy) jAy x (iBx jBy) ixi
AxBx ixj AxBy jxi AyBx jxj AyBy
now ixi 0 jxj 0 ixj k jxi -k
C ABsin?
So C k AxBy - kAyBx
0 - k ABsin?
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Relating Linear and Angular variables
a and a
The centripital acceln of circular
motion. Direction to centre
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Thus the magnitude of a a ar - v2/r
Total linear acceleration a
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The Falling Chimney
The whole rigid body has an angular acceleration a
The tangential acceleration atan distance r from
the base is atan ar
q
at the CM, atan aL/2, and at end atan aL
Yet at CM, atan g cosq (determined by gravity)
At the end, the tangential acceleration is twice
this, yet the maximum tangential acceleration of
any mass point is g cosq.
The rod only falls as a body because it is
rigidthe chimney is NOT.