problem 1 solution
TRANSCRIPT
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SOLUTION
PROBLEM PHYSICS
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By
Pristiadi Utomo
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SOLUTION
Soluti on of the theoretical problem 1
Back -and- Forth Rolling of a Liquid-Filled Sphere
1.1Let1
I and 2Idenote the rotational inertia of the spherical shell and W in
solid state respectively, while Ibe the sum of1
I and 2I. The surface mass density of
the spherical shell is24
m
r
= . Cut a narrow zone from the spherical shell
perpendicular to its diameter, which spans a small angle d with respect to the center
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of the sphere C, while the spherical zone makes an angle with the diameter of the
spherical shell, which is called Caxes hereafter, as shown in Fig. 1. The rotational
inertia of the narrow zone about the C axis is 22 sin ( d ) ( sin )r r r , therefore
integral over the whole spherical shell gives
2 2
10
2
2 sin ( d ) ( sin ) .3 I r r r mr
= =(1A.1)
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dC
Figure 1
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r
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The volume density of W is3
4 / 3
M
r
= . By using above result for the spherical
zone it can be seen that the rotational inertia of the solid W about the Caxis is
2 2 2
20
2 2' 4 ' d ' .
3 5
r
I r r r Mr = = (1A.2)
Then, 2 21 2
2 2
.3 5 I I I mr Mr = + = +(1A.3)
2According to the Newtons second law we can derive the translational motion
equation of the center of mass for the sphere along the tangent of the bowl,
( )( ) ( ) ,m M R r m M g f + = + +&& (1A.4)
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D(m+M)gCCOER
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f
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Figure 2
where ( 1)
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Equations (1A.4)-(1A.6) lead to
5 7( )( ) ( ) .3 5
m M R r m M g + = +&&
This is a motion equation of the type of simple harmonic oscillator. Therefore, we
obtain the angular frequency and period of the sphere rolling right and left:
rR
g
Mm
Mm
++
= 5/73/51 (1A.7)
Mm
Mm
g
rRT
++
=5/735
21 . (1A.8)
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2. This case can be treated similarly, except taking that the ideal liquid does not rotate
into consideration. Therefore Eqs. (1A.4) and (1A.6) are still applicable, while Eq.
(1A.5) needs to be modified as
2
1
2.
3 fr I mr = = && & (1A.9)
Equations (1A.4), (1A.6), and (1A.9) result in
(5 / 3 )( ) ( ) .m M R r m M g + = +&&
Then, the angular frequency and period of the sphere rolling back-and-forth are
obtained respectively.
2 ,5 / 3
m M g
m M R r
+=
+ (1A.10)
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2
5 /32 .
R r m M T
g m M
+=
+(1A.11)
3. The time taken by the sphere from position A0 to equilibrium position O is2 / 4T ,
1/ 4T from O to '
0A, and
2 / 4Tfrom '
0Ato O,
1 / 4Tfrom O to
1A . Although the angular
amplitude decreases step by step (see below) during the rolling process of the sphere
right and left, the period keeps unchanged. This means
3 1 2
1 5 / 3 7 / 5 5 / 3( ) .
2
R r m M m M T T T
g m M m M
+ += + = + + +
(1A.12)
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Next, we calculate the change of the angular amplitude. When the sphere passes
through the equilibrium position O after it rolled down from the initial position0
A ,
the velocity of its center is
2 0 0( ) ( ) .5 / 3
C
m Mv R r g R r
m M
+= =
+(1A.13)
Now the angular velocity of the spherical shell rotating about the Caxis is
0 ( ).5 / 3
Cv m M g R r r r m M
+ = =
+(1A.14)
where Caxis is the axis of rotation through the center of the sphere and perpendicular
to the paper plane of Fig.2. When Wbehaves as liquid (before it changes into solid
state), the angular momentum of the sphere relative to point O is
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1( ) .
CL m M v r I = + + (1A.15)
When W changes suddenly into solid state, due to the fact that both gravitational and
frictional force pass through point O, the angular momentum of the sphere relative to
O is conserved, we have
2
1( ) [ ( ) ] '.CL m M v r I I m M r = + + = + + (1A.16)
where and ' represent the angular velocity of the sphere immediately before and
after passing through point O. Therefore
12( ) 5 / 3' ,
( ) 5 / 3 7 / 5C Cm M v r I v m M
I m M r r m M + + + = =+ + + (1A.17)
while after passing through point O the velocity of the center of the sphere becomes
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' 5 / 3 .5 / 3 7 / 5
C C
m Mv r v
m M
+= =+
(1A.18)
Once the sphere reaches the left highest position '0A corresponding to the left angular
amplitude '0
we have
1 0' ( ) '.Cv R r =
However,2 0( ) .
Cv R r =
From above two expressions we obtain
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'' 20 0 0
1
5 / 3.
5 / 3 7 / 5
c
C
v m M
v m M
+= =
+(1A.19)
Similarly we can treat the process that the sphere rolls from position '0A back toA
1 ,
the second highest position on the right, corresponding to the second right angular
amplitude 1, and obtain
01
0 0
' .'
=
Then,' 2
01 0
0
5 / 3.
5 / 3 7 / 5
m M
m M
+= =
+
Following the similar procedure repeatedly we finally obtain:
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0
5 / 3.
5 / 3 7 / 5
n
n
m M
m M
+ = + (1A.20)
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Solution of theoretical problem 2
2A . Optical pr operties of an unusual material
1. 1
i
r
A
B
C
D
iair
medium E
ProveAssume E-D shows one of the wavefronts of the refracted light. According to
the Huygens principle the phase accumulation from A to E should be equal to that
from B via C to D:
AE BC CD = + 2A.1
With the Hints given these phase differences can be calculated respectively, then
r r AE BC r r C kd kd kd = 2A..2
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Simplification of2A.2gives
( )r r AE CD BC d d d = 2A.3
Because 0BCd > and 0
r r > we obtain
AE CD
d d< 2A.4
Therefore the schematic ray diagram of the refracted light shown in above figure is
reasonable.
2From the above figurethe refraction anglerand incidence angle
isatisfy
sin , sin BC AC i CD AE AC r d d d d d = = 2A.5
respectively. Substitution of2A.5into2A.3results in
sin sinr r r i
= 2A.6
( 3
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i
r
D
Ci
B
Amedium
air E
ProveAssume E-D shows one of the wavefronts of refracted light.
According to the Huygens principle the phase accumulation from A to E should be
equal to that from B via C to D:
AE BC CD = + 2A.7
With the Hints given these phase differences can be calculated respectively, then
AE r r BC CDkd kd kd = +
2A.8
Simplification of2A.8gives
AE CD r r BC d d d =
2A.9
Because 0BCd > and 0
r r > we obtain
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AE CD
d d< 2A.10
Therefore the schematic ray diagram of the refracted light shown in the above figure
is reasonable.
(4) From the above figure the refraction angler
and incidence anglei
satisfy
sin , sin BC AC i CD AE AC r d d d d d = = 2A.11
respectively. Substitution of2A.11into2A.9results in :
sin sinr r r i = 2A.12
2. The ray diagram is shown below.
Illustration:
The light is negatively refracted at both interfaces, and the refraction angle equals to
incidence angle. Meanwhile according to the Hints provided there is no reflected light
from each interface. Therefore within the medium light rays converge strictly at a
point symmetric to the source about the left side of the medium, and on the other side
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of the medium the rays converge strictly at a point which is symmetric to the image of
the source within the medium about the right side of the medium.
d
r=
r=-1
d
1
2
3
3
4
3. The phase difference between the two waves transmitting through the right side of
the medium in succession is
2 ( 0.4 ) 2 0.5 0.4 2k d d k d = + 2A.13
On the right side of above equation the first term shows the phase difference of the
light wave accumulated during its propagation in air, the second term shows the phase
difference of the light wave accumulated during its propagation in the unusual
medium, while the third term accounts for the phase difference of the light wave
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accumulated due to the two reflections in succession from the interface between air
and the medium. Taking 2k
= 2A.13changes into
20.8 2d
= + 2A.14
Resonant condition means
20.8 2 2d m
+ = 2A.15
Thus 0.8 , 2,3,4,...1
dm
m = =
2A.16
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4.
r'
i
'
r
receiving plane
o
2R
X
Y
i
x
From the given conditions the ray diagram can be accordingly constructed. Above
figure shows schematically the ray diagram for 0x > .Because for the unusual
medium 1r r = = from the solution of the first question we have
' 'i r i r
= = = .Therefore the direction of the final out-going light deviates from
that of the incident light by 4i
.Because the direction of the incident light is given in
the y direction, only if the condition
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/ 2 4 3 / 2 / 8 3 / 8i i
2A.17
is satisfied the light signal can not reach the receiving plane. Notice
sin /i
x R = 2A.18
and the similarity of the monotonicity of sinto that ofin the range of[0, / 2] ,
we find that2A.17goes to
sin( / 8) sin(3 / 8)x
R . 2A.19
Further taking the symmetry about the y axis into consideration we obtain that if the
following condition
sin( / 8) sin(3 /8) R x R 2A.20
is satisfiedthe light emitted from a light source located on the x axis can not reach
the receiving plane.
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2B. Dielectric spheres inside an external electric field
1. (1) Adopting the polar coordinates, the z component of the electric field produced
by a dipole located at the origin with its axis parallel to the z axis is:
2
3
0
1 3cos( , )
4z
pE
r
=
(2B.1)
where ris the length of the relative position vector of the two dipoles. In the
external electric fieldE, the energy of a dipole with its axis parallel to the z axis
is:
zU p E pE = =
r Therefore we obtain that the
interaction energy between two contacting small dielectric spheres is
2 2
12 3
0
1 3cos4 (2 )pU
a
= (2B.2)
(2) Based on Eq. (2B.2) for the configuration (a) we obtain:
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2
2
3 3
0 0
1 1 3 1
4 (2 ) 4 4a
pU p
a a
= = (2B.3)
For configuration (b)
2
2
3 3
0 0
1 1 0 1
4 (2 ) 4 8b
pU p
a a
= =
1B.4
For configuration (c),
2 2
2
3 3
0 0
1 1 3cos / 4 1
4 (2 ) 4 16c
pU p
a a
= = 1B.5
(3) Comparison between (2B.3)(2B.4) and (2B.5) shows that configuration (a)
has the lowest energy, corresponding to the ground state of the system.
2. With the similar approach to question 1 the interaction energies for the three
different configurations can also be calculated.
For configuration (d),
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2 2 2
3 3 3
0 0
1 1 32
4 4 32 4 16d
p p pU
a a a
= + =
2B.6
For configuration (e),
2 2 2
3 3 3
0 0
1 1 172
4 4 32 4 32e
p p pU
a a a
= =
(2B.7)
For configuration (f),
2 2 2
3 3 3
0 0
1 1 172
4 8 64 4 64f
p p pU
a a a
= + =
(2B.8)
Comparison shows that configuration (e) of the lowest energy is most stable,
while configuration (f) of the highest energy is most unstable.