Problem 1 Solution

8/14/2019 Problem 1 Solution

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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

9

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.

Problem 1 Solution

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