tuesday, june 28, 2011phys 1443-001, summer 2011 dr. jaehoon yu 1 phys 1443 – section 001 lecture...
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PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
1Tuesday, June 28, 2011
PHYS 1443 – Section 001Lecture #13
Tuesday, June 28, 2011Dr. Jaehoon Yu
• Collisions• Center of Mass• Rotational Motion• Rotational Kinematics• Relationship Between Angular and
Linear quantities
Tuesday, June 28, 2011 2
Announcements• Reading Assignment
– CH9.10• Quiz #3 tomorrow, Wednesday, June 29
– Beginning of the class– Covers CH8.1 through CH9.9
• Planetarium Show extra credit– Must obtain the signature of the “Star Instructor” AFTER
watching the show on the ticket stub– Tape one side of the ticket stubs on a sheet of paper with
your name on it– Submit it on the last class Thursday, July 7
• Late submissions will not be accepted!!!PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Valid Planetarium Shows• Regular shows
– TX star gazing; Nanocam; Ice Worlds• Private shows for a group of 15 or more
– Bad Astronomy; Black Holes; IBEX; Magnificent Sun– Microcosm; Stars of the Pharaohs; Time Space– Two Small Pieces of Glass; SOFIA– Violent Universe; Wonders of the Universe
• Please watch the show and obtain the signature on the back of the ticket stub
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Extra-Credit Special Project• Derive the formula for the final velocity of two objects which
underwent an elastic collision as a function of known quantities m1, m2, v01 and v02 in page 8 of this lecture note in a far greater detail than in the note.– 20 points extra credit
• Show mathematically what happens to the final velocities if m1=m2 and explain in detail in words the resulting motion.– 5 point extra credit
• NO Credit will be given if the process is too close to the note!• Due: Start of the class Tuesday, July 5
PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Extra Credit: Two Dimensional Collisions
• Proton #1 with a speed 5.0x106 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ. This must be done in much more detail than the book or on page 13 of this lecture note.• 10 points• Due beginning of the class Wednesday, July 6
Tuesday, June 28, 2011
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Collisions
Consider a case of a collision between a proton on a helium ion.
The collisions of these ions never involve physical contact because the electromagnetic repulsive force between these two become great as they get closer causing a collision.
Generalized collisions must cover not only the physical contact but also the collisions without physical contact such as that of electromagnetic ones on a microscopic scale.
1 21dp F dtrr
t
F F12
F21
Assuming no external forces, the force exerted on particle 1 by particle 2, F21, changes the momentum of particle 1 by
Likewise for particle 2 by particle 1 2 12dp F dt
rr
Using Newton’s 3rd law we obtain So the momentum change of the system in a collision is 0, and the momentum is conserved
dpur
2
dpur12F dt
r21F dt
r −dp
ur1
=dpur1 + dp
ur2
pursystem =p
ur1 + p
ur2
constant0
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Elastic and Inelastic Collisions
Collisions are classified as elastic or inelastic based on whether the total kinetic energy is conserved, meaning whether it is the same before and after the collision.
A collision in which the total kinetic energy and momentum are the same before and after the collision.
Momentum is conserved in any collisions as long as external forces are negligible.
Elastic Collision
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision, moving together with the same velocity.Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost.
Inelastic Collision
A collision in which the momentum is the same before and after the collision but not the total kinetic energy .
Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Elastic and Perfectly Inelastic Collisions In perfectly inelastic collisions, the objects stick together after the collision, moving together. Momentum is conserved in this collision, so the final velocity of the stuck system isHow about elastic collisions?
1 21 2i im v m vr r
In elastic collisions, both the momentum and the kinetic energy are conserved. Therefore, the final speeds in an elastic collision can be obtained in terms of initial speeds as
1 21 2i im v m vr r
21
211 fi vvm
m1 v1i −v1 f( ) =m2 v2i −v2 f( )
v1 f m1 −m2
m1 m2
⎛
⎝⎜⎞
⎠⎟v1i
2m2
m1 m2
⎛
⎝⎜⎞
⎠⎟v2i
1 2( ) fm m v r
1 21 2
1 2( )i i
fm v m v
vm m
r rr
1 21 2f fm v m v r r
222
211 2
1
2
1ii vmvm 2
22211 2
1
2
1ff vmvm
22
222 fi vvm
m1 v1i −v1 f v1i v1 f m2 v2i − v2 f( ) v2i + v2 f( )
From momentum conservation above
v2 f 2m1
m1 m2
⎛
⎝⎜⎞
⎠⎟v1i
m1 −m2
m1 m2
⎛
⎝⎜⎞
⎠⎟v2i
What happens when the two masses are the same?
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Example for CollisionsA car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the two become entangled. If the lighter car was moving at 20.0m/s before the collision what is the velocity of the entangled cars after the collision?
pi
The momenta before and after the collision are
What can we learn from these equations on the direction and magnitude of the velocity before and after the collision?
m1
20.0m/s
m2
vf
m1
m2
Since momentum of the system must be conserved
p
i=pf
The cars are moving in the same direction as the lighter car’s original direction to conserve momentum. The magnitude is inversely proportional to its own mass.
Before collision
After collision
=m1v1i + m2v2i =0 + m2v2i
p
f =m1v1 f + m2v2 f
= m1 + m2( )vf
m
1+ m2( )vf =m2v2i
v
f
=m2v2i
m1 + m2( ) =900×20.0900 +1800
=6.67m/ s
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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The mass of a block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet.
Ex.9 – 11: A Ballistic Pendulum
m
1v
f 1+ m2vf 2 =
m
1+ m2( )
v01
What kind of collision?No net external force momentum conserved
Perfectly inelastic collision
m1v
01+ m2v02
m1v
01fv
Solve for V01
1 2
1
fm m v
m
What do we not know? The final speed!!How can we get it? Using the mechanical
energy conservation!
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Ex. A Ballistic Pendulum, cnt’d
v01=
Using the solution obtained previously, we obtain
m1 m2 vf
m1
Now using the mechanical energy conservation
=mgh
m
1+ m2( )ghf =
gh
f=
fv
12
mv2
12
m1+ m2( )vf
2
12v
f2
2gh
f 22 9.80m s 0.650 m
Solve for Vf
m1 m2 2ghf
m1
=
0.0100 kg+ 2.50 kg0.0100 kg
⎛
⎝⎜⎞
⎠⎟2 9.80m s2( ) 0.650 m( )
=+896m s
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Two dimensional Collisions In two dimension, one needs to use components of momentum and apply momentum conservation to solve physical problems.
1 21 2i im v m v r r
21 12
1i
vm
m2
m1
v1i
m1
v1f
θ
m2
v2f
φ
Consider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us:
1 21 2i im v m vr r
And for the elastic collisions, the kinetic energy is conserved:
What do you think we can learn from these relationships?
fxfx vmvm 2211 θ coscos 2211 ff vmvm
iyvm 11 0 fyfy vmvm 2211 =m1v1 f sinθ − m2v2 f sinφ
11 im vr
ixvm 11
222
211 2
1
2
1ff vmvm
1 1 2 2ix ixm v m v
1 1 2 2iy iym v m v
x-comp.
y-comp.
1 21 2f fm v m vr r
1 1 2 2fx fxm v m v
1 1 2 2fy fym v m v
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Ex. 9 – 13: Two Dimensional CollisionsProton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ.
Since both the particles are protons m1=m2=mp.Using momentum conservation, one obtainsm2
m1
v1i
m1
v1f
θ
m2
v2f
φ
ipvm 1
Canceling mp and putting in all known quantities, one obtains
smv f /1080.2 51
From kinetic energy conservation:
3.50 ×105( )2=v1 f
2 + v2 f2 (3)
Solving Eqs. 1-3 equations, one gets
(1) 1050.3cos37cos 521 ff vv
Do this at home
θ coscos 21 fpfp vmvm
θ sinsin 21 fpfp vmvm 0
x-comp.
y-comp.
(2) sin37sin 21 ff vv
smv f /1011.2 52
0.53
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Center of MassWe’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situations objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and
represents the motion of the system as if all the mass is on the point.
Consider a massless rod with two balls attached at either end.
CMx
The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass.
ar= F
ur∑ / M
What does above statement tell you concerning the forces being exerted on the system?
m1 m2x1 x2
The position of the center of mass of this system is the mass averaged position of the systemxCM CM is closer to the
heavier object1 1 2 2m x m x
1 2
m m
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Motion of a Diver and the Center of Mass
Diver performs a simple dive.The motion of the center of mass follows a parabola since it is a projectile motion.
Diver performs a complicated dive.The motion of the center of mass still follows the same parabola since it still is a projectile motion.
The motion of the center of mass of the diver is always the same.
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Example 9 – 14 Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and x3=6.0m. Find the position of CM.
Using the formula for CM
ii
iii
CM m
xmx
1.0M 12.0
3
M
M
M M M 4.0( )m
5.0M 6.0M
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Ex9 – 15: Center of Mass in 2-DA system consists of three particles as shown in the figure. Find the position of the center of mass of this system.
Using the formula for CM for each position vector component
ii
iii
CM m
xmx
One obtains
CMx
CMrr
If kgmmkgm 1;2 321
3 40.75
4CM
i jr i j
r rr r r
m1y=2 (0,2)
m2
x=1
(1,0)m3
x=2
(2,0)
(0.75,1)rCM
ii
iii
CM m
ymy
ii
iii
m
xm
321
332211
mmm
xmxmxm
321
32 2
mmm
mm
CMy
ii
iii
m
ym
321
332211
mmm
ymymym
321
12
mmm
m
CMx ir 2 3 1
1 2 3
2 2m m i m j
m m m
r r
CMy jr
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Center of Mass of a Rigid ObjectThe formula for CM can be extended to a system of many particles or a Rigid Object
A rigid body – an object with shape and size with mass spread throughout the body, ordinary objects – can be considered as a group of particles with mass mi densely spread throughout the given shape of the object
CMx
ii
iii
CM m
ymy
The position vector of the center of mass of a many particle system is
CMrr
xCM ≈Δmixi
i∑
M
CMx
1CMr rdm
M
r r
Δmi
ri
rCM
ii
iii
CM m
zmz
CM CM CMx i y j z k r r r i i i i i i
i i i
ii
m x i m y j m z k
m
r r r
iii
CM
m rr
M
rr
M
xmi
ii
mi
Δ
Δ 0lim =
1
Mx dm∫
1 1 2 2 n nm x m x m x i ii
m x i
i
m1 2
nm m m
Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu
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Ex 9 – 16: CM of a thin rod
The formula for CM of a continuous object is
Lx
xCM xdmM
x0
1
Therefore
L
x dxdm=λdx
Since the density of the rod (λ) is constant;
CMx
Show that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length.
Find the CM when the density of the rod non-uniform but varies linearly as a function of x, λ=αx
CMxM
dxdm LM /
The mass of a small segment
Lx
xxdx
M 0
1 Lx
x
xM
0
2
2
11
2
2
11L
M
ML
M 2
11
2
L
Lx
xdx
0
Lx
xxdx
0α
Lx
x
x
0
2
2
1α 2
2
1Lα
Lx
xxdx
M 0
1
Lx
xdxx
M 0
21 αLx
x
xM
0
3
3
11 α
3
3
11L
Mα
ML
M 3
21
3
2LCMx