Solutions

A12-1
Official (English)

A Mechanical Model for Phase Transitions
General Grading Guidelines
When student’s solutions are correct and s/he also show how solutions were obtained, the stduent gets
full credit. The scheme oulined below is helpful if the student’s answers are partially correct. Attention will be paid to the detailed solution so, if the final answer is correct but it is obtained by incorrect
method(s) then no credit will be given. Alternative solutions may exist and will be given due credit.
Partial or full outcomes obtained for later sections in the problem which are incorrect solely because oferrors being carried forward from previous sections, but are otherwise reasonable, will not be furtherpenalized. For example a dimensioanlly wrong answer when carried forward will not get any credit in thesubsequent sections. A numerically wrong evaluation when carried forward will get credit in subsequent
sections unless the numerical answer is patently wrong (e.g. the value of g is 981 m/sec2 ! )Incorrect or no labeling of an axis is penalized by -0.1 points
The numerical answer (i) must be correct to +/- 10% AND (ii) must respect significant figures.
It maybe noted that NO micro-marking scheme takes care of all contingencies. A certain amountof discretion rests with and a certain level of judgement is invested in the academic committee.
A.1 (0.5 pt)
Equations of motion
The radial component 𝐹𝑟 yields:
𝑚𝑅𝜃2̇

2

= 𝑁 − 𝑚𝑔 cos(𝜃) − 𝑚𝑅 sin (𝜃)𝜔2

The tangential component 𝐹𝜃𝑚𝑅𝜃 ̈ = 𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − sgn (𝜃)̇ 𝑘𝑁(1)[0.2](2)OR𝑚𝑅𝜃 ̈ = 𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − 𝑓𝑘𝑁 (𝑓 = 1)

(3)
[0.3]

Solutions

A12-2
Official (English)

B.1 (1.0 pt)
Equilibrium angle(s)We set 𝑘 = 0 in the equation for the tangential component of the force. Thus𝑚𝑅𝜃 ̈ =𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃)(4)

For equilibrium we set 𝜃0̈ = 0 in the above equation. Then 𝜃0 = 0 is an equilibrium angle for all valuesof 𝜔[0.4]
The other values are given by𝑔𝜔2cos 𝜃0 = 2 = 𝑐2(5)
𝜔 𝑅
𝜔
OR
𝜔2(6)
𝜃0 = ± ∣cos−1 𝑐2 ∣𝜔[0.6]
with values of 𝜃0 between -𝜋/2 to 𝜋/2. The ± indicates that there are two equivalent positions. The
bead could rise on either side of the axis shown in the figure depicted in the problem. Note that for𝜔 < 𝜔𝑐 , Eq. (5) implies cos 𝜃0 > 1. This is clearly unphysical. A little reflection will convince us that𝜃0 = 0 for 𝜔 < 𝜔𝑐 .
B.2 (0.5 pt)
Sketch of 𝜃0 .

𝜃0𝜋20− 𝜋21𝜔/𝜔𝑐[shape correct: 0.2]
[only if both branches: 0.3]

Solutions

A12-3Official (English)𝑁

B.3 (0.5 pt)
Sketch of the magnitude of the normal reaction

𝑚𝑔1𝜔/𝜔𝑐[𝜔 < 𝜔𝑐 ∶ 0.2]
[𝜔 > 𝜔𝑐 ∶ 0.3]

Solutions

A12-4
Official (English)

B.4 (1.0 pt)
The potential 𝑉 (𝜃)Given that𝐹𝜃 = −

1 𝑑𝑉 (𝜃)𝑅 𝑑𝜃

(7)% and taking 𝑉 (𝜃 = 0) = 0, we obtain on
integrating
Eq. (4) that𝑉𝜃
−𝑅 ∫ 𝐹𝜃 𝑑𝜃 = ∫ 𝑑𝑉 = 𝑉 − 000(8)[0.3]

the left hand side is𝜃𝜃−𝑅 ∫ 𝐹𝜃 𝑑𝜃𝜃=−𝑚𝜔2 𝑅2
∫ sin(2𝜃) + 𝑚𝑔𝑅 ∫ sin(𝜃)𝑑𝜃200=𝑚𝜔2 𝑅2 (cos(2𝜃) − 1)
− 𝑚𝑔𝑅(cos(𝜃) − 1)402

Noting that 𝑐𝑜𝑠(2(𝜃) − 1) = −2 sin (𝜃) and 𝜔𝑐2 = 𝑔/𝑅 we obtain
𝑉 (𝜃) = 𝑚𝑔𝑅 [(1 − cos 𝜃) −

𝜔22sin 𝜃]2𝜔𝑐2

We can also verify the above equation by substitution into Eq. (7)𝑃 = 𝑚𝑔𝑅
𝑄 = −𝑚𝑔𝑅
𝜔2 𝑚𝑔𝑅
𝑆=−2𝜔𝑐2

(9)
[0.4]

(10)
[0.3]

Solutions

A12-5
Official (English)

B.5 (1.0 pt)
The coefficients
We use the expansions for the trigonmetric functions sin(𝜃) and cos(𝜃) in Eq. (10). We shall keep
terms upto and inculding order 𝜃4 . Thus
𝑉 (𝜃)

≈≈

𝜔2(𝜃 − 𝜃3 /6)2 ]2𝜔𝑐2
𝜔2𝑚𝑔𝑅 𝜔2𝑚𝑔𝑅
1[1 − 2 ] 𝜃2 +[ 2 − ] 𝜃42𝜔𝑐6𝜔𝑐4𝑚𝑔𝑅 [1 − 1 + 𝜃2 /2 − 𝜃4 /24 −Thus𝑎(𝜔) =𝑚𝑔𝑅
𝜔2(1 − 2 )
2
𝜔𝑐
[0.5]

𝑏(𝜔) =𝑚𝑔𝑅 𝜔21(− )
6 𝜔𝑐2
4
[0.5]

Note: no penalty if the 1/4 term is missed.
B.6 (1.0 pt)
Representative plots of the potential𝑉 (𝜃)𝜔 < 𝜔𝑐𝜔 > 𝜔𝑐0−𝜃0− 𝜋2

𝜃00𝜃

𝜋2

[𝜔 < 𝜔𝑐 ∶ 0.5]
[𝜔 > 𝜔𝑐 ∶ 0.5]

Solutions

A12-6
Official (English)

B.7 (1.0 pt)
Bead analogues
For 𝜔 → 𝜔𝑐+ , 𝜃0 is close to zero. Hence on expanding the cosine term in Eq. (5),
1−

𝜃02
2=

𝜃0=

𝜔𝑐2
𝜔21/2
√
𝜔2± 2 [1 − 𝑐2 ]𝜔(11)Also note from Eq. (5) that as 𝜔 → ∞, 𝜃0 → ±𝜋/2. This plot also has an analogue in phase transition.
The magnetization ℳ goes to zero as 𝑇 goes to 𝑇𝑐 in a similar fashion. Thus the role of ℳ is playedby 𝜃0 and temperature is inversely related to 𝜔. Increasing temperature is equivalent to decreasing
𝜔. Summarizing,ℳ⟶𝜃[0.4]
𝑇𝑐 ⟶ 1/𝜔𝑐2𝑇 /𝑇𝑐 ⟶ 𝜔𝑐2 /𝜔2
Equivalent value of 𝛽 for bead is = 1/2.

[0.4]

[0.2]
[Note: The critical exponent is 1/2 in our case and also in Landau theory. However experimentallyand in more elaborate theories the exponent of vanishing magnetization is 1/3].

Solutions

A12-7
Official (English)

B.8 (1.0 pt)
Oscillation frequency
The frequency of oscillation Ω0 of the bead about the "equilibrium" position 𝜃0 isΩ0 =1 √ 𝑉 ″ (𝜃)𝑅𝑚

We take the second order derivative of the potential as given in Eq. (10)𝑉 ″ (𝜃) = 𝑚𝑔𝑅 cos 𝜃 [1 −𝜔2𝜔22cos 𝜃] + 𝑚𝑔𝑅 2 sin 𝜃2𝜔𝑐𝜔𝑐(12)For 𝜃 = 𝜃0 = ± cos−1 (𝜔𝑐2 /𝜔2 )𝑉 ″ (𝜃0 ) = 𝑚𝑔𝑅

𝜔2
𝜔𝑐4(1−)>0
𝜔𝑐2
𝜔4if 𝜔 > 𝜔𝑐(13)For 𝜔 < 𝜔𝑐 , 𝜃0 = 0, and we obtain from Eq. (12) that(14)Ω0 = (𝜔𝑐2 − 𝜔2 )1/2

[0.5]

Similarly for 𝜔 > 𝜔𝑐 , using Eq. (13) we obtainΩ0=𝜔4𝜔 (1 − 𝑐4 )𝜔

1/2(15)[0.5]

Solutions

A12-8
Official (English)

B.9 (1.0 pt)
Sketch of Ω0Ω

Ω00 =𝜔𝜔 < 𝜔𝑐𝜔 > 𝜔𝑐𝜔𝑐𝜔𝑐𝜔[𝜔 < 𝜔𝑐 ∶ 0.5]
[𝜔 > 𝜔𝑐 ∶ 0.5]

Solutions

A12-9
Official (English)

C.1 (1.0 pt)
Condition for equilibrium angles
We substitute the expression for the normal reaction (Eq.(1)) in the angular part (Eq.(3)) to obtain𝑚𝑅𝜃 ̈ =2𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − 𝑓𝑘(𝑚𝑔 cos(𝜃) + 𝑚𝑅 sin (𝜃)𝜔2 + 𝑚𝑅𝜃2̇ )Noting that 𝜔𝑐2 = 𝑔/𝑅 and rearranging terms we have𝜃̈ =22𝜔𝜃̇2⎡𝜔𝑐 ⎢(sin(𝜃)) (cos(𝜃) − 𝑓𝑘 sin(𝜃)) ( ) − sin(𝜃) − 𝑓𝑘 cos(𝜃) − 𝑓𝑘 ( ) ⎤⎥𝜔𝑐𝜔𝑐

⎣

⎦

[0.2]
At equilibrium, 𝜃 ̇ = 0, 𝜃 ̈ = 0 and 𝑓 = sgn(𝜃)̇ = ±1 depending on how this equilibrium was attained,iė., depending on the value of 𝜃 ̇ just before equilibrium was attained. Thus we obtain the expression
for the equilibrium angle 𝜃0 ,2sin(𝜃0 ) (cos(𝜃0 ) − 𝑓𝑘 sin(𝜃0 )) (𝜔) = sin(𝜃0 ) + 𝑓𝑘 cos(𝜃0 ) with 𝜃0 ∈ (−𝜋/2, 𝜋/2)𝜔𝑐

For 𝑓 = 1 and 𝑘 = tan(𝛼) we may express the above as
𝜔
( )
𝜔𝑐2

==sin(𝜃0 ) + tan(𝛼)𝑐𝑜𝑠(𝜃0 )sin(𝜃0 )(cos(𝜃0 ) − tan(𝛼) sin(𝜃0 ))tan(𝜃0 + 𝛼)sin(𝜃0 )

[0.4]

(16)
[0.4]

Solutions

A12-10
Official (English)

C.2 (0.5 pt)
Representative values for 𝜃0We are given the expansions for the trignometric functions in the problem. We notice that the coefficient of friction 𝑘 is small (=0.05). Thus 𝑘 = 𝛼. We then havesin(𝜃0 ) ≈ 𝜃0tan(𝜃0 + 𝛼) ≈ 𝜃0 + 𝛼[0.2]

Thus2(𝜔)𝜔𝑐≈

1+𝑘𝜃0Simple calculations yield(a) 𝜃0 = -0.07 radians(b) 𝜃0 = -0.1 radians
The plot will no longer be symmetric.

[0.3]