Mechanics Assignment 5 Solved

Question 1. Points A and B are constrained to move in the plane on lines intersecting at angle θ, Figure 1. Describe precisely the fixed and moving centrodes realizing the planar motion of body AB starting from a configuration like the one showed in the figure. Provide a rigorous mathematical proof that the instantaneous center traces the curves described by you. 

 

  

Figure 1. 

 

 

 

Question 2. Points A and B are constrained to move in the plane on circles with centers OA and OB with |OAA| = |OBB|, |OAOB| = |AB|, and |OAA| > |AB|, Figure 2. Describe precisely the fixed and moving centrodes realizing the planar motion of body AB starting from a configuration like the one showed in the figure. Provide a rigorous mathematical proof that the instantaneous center traces the curves described by you. Plot the centrodes for the link-length parameters using a table of value given below. 

 

Question 3. In the conditions of Question 2, let points OA and A be fixed, while points OB and B move on circles. Describe precisely the fixed and moving centrodes realizing the planar motion of the body OBB starting from a configuration like the one showed in the figure. Provide a rigorous mathematical proof that the instantaneous center traces the curves described by you. Make a drawing (or a computer simulation) illustrating and explaining the rolling of the centrodes. Plot the centrodes for the link-length parameters using a table of value given below. 

 

 

  

 

Figure 2. 

 

 

 

Question 4 (bonus).For each of Questions 1, 2 and 3: Does the rolling of the identified centrodes yield all possible configurations of the moving body allowed by the linkage? Prove your answer. 

 

Hint. In all cases, the centrodes are relatively simple, familiar curves. The proofs required in Questions 1-3 can be made using simple planar geometry. 

 

Note. You can get up to 100% of the mark for this homework by answering the first there questions, and up to 110% by answering all four. 

Simulations can be performed in Matlab/Mathematica/Maple or CAD. For Question 2 and 3,  

|𝐴𝐵| = 𝑙𝑎 

The value of 𝑙𝑎 is given by first digit of your student ID/Matricola. 

|𝑂𝐴𝐴| = 𝑙𝑏 

𝑘 = 0.1 + (0.012)𝑛 

𝑙𝑏 = 1/𝑘 Value of n is given below:

 

Nome 
Cognome 
n 
Akshith 
Sirigiri 
1 
Aliya 
Arystanbek 
2 
Andrea 
Pitto 
3 
Andrea 
Tiranti 
4 
Aurora 
Bertino 
5 
Azay 
Karimli 
6 
Chetan Chand 
Chilakapati 
7 
Chiara 
Terrile 
8 
chiara 
saporetti 
9 
Cristina 
Naso Rappis 
10 
Daniel 
Nieto 

Rodriguez 
11 
Daulet 
Babakhan 
12 
Davide 
Piccinini 
13 
Ege Doruk 
Sayin 
14 
Elena 
Merlo 
15 
Emanuele-riccardo 
rosi 
16 
Filip 
Hesse 
17 
Francesco 
Porta 
18 
Francesco 
Testa 
19 
Gabriele 
Reverberi 
20 
Gerald 
Xhaferaj 
21 
Geraldo 
Margjini 
22 
Giulia 
Scorza Azzarà 
23 
isabella-sole 
bisio 
24 
JIHAD 
KHALIL 
25 
Josep 
Rueda I Collell 
26 
Justin Lawrence 
Lee 
27 
Kamali  
Babu 
28 
Laiba 
Zahid 
29 
Latif 
Xeka 
30 
Luca 
Tarasi 
31 
Luca 
Covizzi 
32 
manoj 
kunapalli 
33 
Marco 
Demutti 
34 
Matteo 
Dicenzi 
35 
Matteo 
Palmas 
36 
Mohamed Emad Lotfy Fahmy 
Qaoud 
37 
Mohan Krishna 
Dasari 
38 
Muhammad 
Tahir 
39 
Muhammad Raza 
Rizvi 
40 
Muhammad Sayum 
Ahmed 
41 
Muhammad Talha 
Siddiqui 
42 
MUHAMMAD USMAN 
ASHRAF 
43 
Paul Toussaint 
Ndjomo Ngah 
44 
riccardo 
lastrico 
45 
roberta 
alessi 
46 
Roberto 
Albanese 
47 
Roberto 
Canale 
48 
saivinay  
manda 
49 
Sandeep 
Soleti 
50 
Sara 
Romano 
51 
Sathish kumar 
subramani 
52 
Sebastiano 
Viarengo 
53 
Serena 
Roncagliolo 
54 
Silvana Andrea 
Civiletto 
55 
Simone 
Voto 
56 
soundarya 
pallanti 
57 
Srikanth Gopichand 
Koppula 
58 
srinivasan 
anbarasan 
59 
STANLEY 
MUGISHA 
60 
Steven 
Palma  

Morera 
61 
Surishoba surendra 
Reddy polaka 
62 
Syed Hani Hussain 
Kazmi 
63 
Vincenzo 
Di Pentima 
64 
Vishruth 
 
65 
Vivek Vijaykumar 
Ingle 
66 
Zaid 
Zaid 
67 
Yasmin 
El Sayed 
68 
Zere 
Gumar 
69 
muhamed  
irfan 
70 
 

 

Chiara Saporetti s4798994 Assignment 5 

Exericise 1:  

In the figure there are the bar AB and the rails on which it moves. Theta is the angle that the rails form when they intersect. The lines a and b are the ones passing through A and B respectively and parallel to the rails.  

  

To find the IC:  

∈ 𝑙𝑖𝑛𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐵, 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑏

                                                     𝐼𝐶 ≡ 𝐶 {                                                                                

∈ 𝑙𝑖𝑛𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐴, 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑎

The point is represented in the figure below:

  

 

 

 

 

We need to understand how C moves wrt AB. To do so we analyze the relationship between O, A,

B,C.  

  

Firstly we consider the congruence between some angles:

𝑂𝐶̂𝐵 ≡ 𝑂𝐴̂𝐵 = 𝛼 because they insist on the same chord of circumpherence γ. 𝑂𝐵̂𝐶 ≡ 𝑂𝐴̂𝐶 = 𝜋   for construction

2 We can now consider the theorem of sinus:

  

In our case:

                                                                                                            𝑂𝐵                 𝑂𝐶

From triangle OBC: 𝑠𝑖𝑛𝛼 = sin(𝜋)  

2

 𝑂𝐵                     𝐴𝐵 From triangle OBA:            =            

                                                                                                             𝑠𝑖𝑛𝛼           sin𝜃 

                                                                                    𝑂𝐵                 𝑂𝐶                  𝐴𝐵

So: 𝑠𝑖𝑛𝛼 = 𝜋 = sin𝜃    sin( ) 

2

                                                                                            𝑂𝐶         𝐴𝐵

                                                                                      →           =            

                                                                                              1         sin𝜃 

Where: AB is constant (it’s the length of the bar) and theta is constant too, since it’s the physical angle that the rails form when intersecting in O.

ð OC=const: the fixed centrode is a circle with center in O and radius equal to OC. So it has equation

 𝑥2 + 𝑦2 = sin 𝐴𝐵2(2𝜃) {𝑖𝑓𝑖𝑓  𝜃𝜃 == 0𝜋 ++ 𝑘𝜋𝑘𝜋 
 

2

In the first case we’d have a pure translation (superimposed rails). In the second case we’d have a circle with diameter coincident with the lenght of the bar: OC=AB.

 We can also see this circle as a locus of points:

𝐹 =  {𝑃 ∈ 𝑅2   𝑠. 𝑡.   𝑃𝑂 = 𝑐𝑜𝑛𝑠𝑡}  
 

ð For any configuration of AB, given a reference frame fixed on AB, C must move as previously described, with OC=const. This means that A, B, C, O always describe a circle with diameter OC and passing through the four points. This circle, which represents the moving centrode, is γ.

                 


Exercise 2: 

In the figure we can see the antiparallelogram ABOAOB, with OAOB fixed.  

  

|𝑂𝐴𝐴| = |𝑂𝐵𝐵| 

|𝑂𝐴𝑂𝐵| = |𝐴𝐵| 

|𝑂𝐴𝐴| > |𝐴𝐵| 

We look for the relations between the points like before.

The triangles ABOB and OAOBA are congruent:

𝐴𝐵 = 𝑂𝐴𝑂𝐵 for construction

𝑂𝐵𝐵 = 𝐴𝑂𝐴 for construction

𝐴𝑂𝐵 in common

ð 𝐴𝐵̂𝑃 = 𝑃𝑂̂𝐴𝑂𝐵 
The triangles APB and OAOBP are congruent:

𝐴𝑃̂𝐵 = 𝑂𝐴𝑃̂𝑂𝐵 = 𝛼 because they are opposite angles wrt P

𝐴𝐵 = 𝑂𝐴𝑂𝐵 for contruction

𝐴𝐵̂𝑃 = 𝑃𝑂̂𝐴𝑂𝐵 because of the previous demostrated congruence

Since A rotates around 𝑂𝐴 and B rotates around 𝑂𝐵, IC must be in the intersection between line b and line a  

ð 𝐼𝐶 𝑃 
 

To find the fixed centrode we must consider P moving around 𝑂𝐴𝑂𝐵: to understand how it moves we then need to analyze 𝑂𝐴𝑃 𝑎𝑛𝑑 𝑃𝑂𝐵, using the relations found before.  

𝑂𝐴𝑃 + 𝑃𝑂𝐵 = 𝐵𝑃 + 𝑃𝑂𝐵 = 𝐵𝑂𝐵 = 𝑐𝑜𝑛𝑠𝑡 → 𝑂𝐴𝑃 + 𝑃𝑂𝐵 = 𝑐𝑜𝑛𝑠𝑡 
So we need to have a trajectory curve that respects this relation: it is an ellipse with foci

𝑂𝐴 𝑎𝑛𝑑 𝑂𝐵 and center O, middle point of segment 𝑂𝐴𝑂𝐵. We can then describe the moving centrode as a locus of points:

𝐹 = {𝑃 ∈ 𝑅2  𝑠. 𝑡.   |𝑃𝑂𝐴| + |𝑃𝑂𝐵| = 2𝑎 }

 

Demostration of fixed centrode (similar for the moving one): 

  

𝑂𝐴𝑃 + 𝑃𝑂𝐴 = 𝐿

Where L is the length of the longest side of the antiparallelogram: 𝐵𝑂𝐵.

 

If I consider a reference frame in O, I can write the previous equivalence as:

 

√𝑦2 + (𝑥 + 𝑐)2 + √𝑦2 + (𝑥 − 𝑐)2 = 𝐿

Where (x,y) are the coordinates of P, while c = 𝑂𝐴𝑂𝐵

2 By elevating to square the equivalence I obtain:

 

√𝑦2 + (𝑥 + 𝑐)2 = −√𝑦2 + (𝑥 − 𝑐)2 + 𝐿

 

                                                              2                                          2

                                                      (√𝑦2 + (𝑥 + 𝑐)2)       = (−√𝑦2 + (𝑥 − 𝑐)2 + 𝐿)

 

 

𝑦2 + 𝑥2 + 𝑐2 + 2𝑥𝑐 = 𝑦2 + 𝑥2 + 𝑐2 − 2𝑥𝑐 + 𝐿2 − 2𝐿√𝑦2 + (𝑥 − 𝑐)2

 

 

4𝑥𝑐 = −2𝐿√𝑦2 + (𝑥 − 𝑐)2 + 𝐿2

 

𝐿2 − 4𝑥𝑐
 = √𝑦2 + (𝑥 − 𝑐)2

2𝐿 By again squaring:

(4𝑥2𝑐2 + 𝐿2 − 2𝑥𝑐) = 𝑦2 + 𝑥2 + 𝑐2 − 2𝑥𝑐

                                                               𝑙2                 4
 

                                                                             𝐿2 − 4𝑐2                                𝐿2 − 4𝑐2

                                                                     𝑥2 (  𝐿2 ) + 𝑦2 =  4        

Keeping into account that 2c = a:

                                                                                   𝑥2                  𝑦2

                                                                             2 + 𝐿2                   𝑎2 = 1

𝐿

                                                                                    4          4 − 4

 

I obtained the equation of an ellipse with semiaxis:

                                                                                       𝐿2                            𝐿2         𝑎2

                                                                             𝑎1 =   ;   𝑏1 =         −      

                                                                                        4                     4        4

 

To find the moving centrode we fix AB: P must move around it. To understand how it moves we then need to analyze AP, PB, using the relations found before.  

𝐴𝑃 + 𝑃𝐵 = 𝐵𝑃 + 𝑃𝑂𝐵 = 𝐵𝑂𝐵 = 𝑐𝑜𝑛𝑠𝑡 → 𝐴𝑃 + 𝑃𝐵 = 𝑐𝑜𝑛𝑠𝑡

The curve that respects this relation is an ellipse with foci A, B and center in the middle point of AB.

We can then describe the moving centrode as a locus of points:

𝑀 = {𝑃 ∈ 𝑅2  𝑠. 𝑡.   |𝑃𝐴| + |𝑃𝐵| = 2𝑎 } where 2a is the major axis of the ellipse.

With the values given (long side=4.8077, short side=4):

  

Because of the type of mechanism, we can see that the IC are not equally spaced, but the shape of the curve is clearly an ellipse.

Exercise 3: 

  

We have the same mechanism as before, but in this case 𝑂𝐴 and A are fixed, while 𝑂𝐵 and B can move. IC is in the intersection between the lines a and b, so it coincides with P2. We analyze the distances P2𝑂𝐴 and P2A.

Triangles AP2𝑂𝐴 and B𝑂𝐵𝑃2 are congruent:

𝐴𝑂𝐴 = 𝐵𝑂𝐵for construction

𝐵𝐴̂𝑃 = 𝑃𝑂̂𝐵𝑂𝐴 demonstrated before

𝐴𝑂̂𝐴𝑃2 = 𝑃2𝐵̂𝑂𝐵 because they are supplementary to equivalent angles (𝑃𝑂̂𝐴𝑂𝐵 = 𝐴𝐵̂𝑃)

ð From this equivalence we obtain 𝑃2𝐵 = 𝑃2𝑂𝐴

This last equivalence is useful because:

𝑃2𝐴 − 𝑃2𝐵 = 𝐴𝐵 

{𝑃2𝐴 − 𝑃2𝐵 = 𝑃2𝐴 − 𝑃2𝑂𝐴 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑐𝑒

ð 𝑃2𝐴 − 𝑃2𝑂𝐴 = 𝑐𝑜𝑛𝑠𝑡 
The curve that respects this relation is the hyperbola with foci A, 𝑂𝐴. So the fixed centrode can be described as a locus of points s.t.:

𝐹 = {𝑃 ∈ 𝑅2  𝑠. 𝑡.   |𝑃2𝐴| − |𝑃2𝑂𝐴| = 2𝑎 }

 

 

 

 

 

 

Demonstration of fixed centrode (similar for the moving one): 

Similarly as before, if we consider a reference frame fixed in O, middle point of OAA:

 

𝑃2𝐴 − 𝑃2𝑂𝐴 = √𝑥2 + (𝑦 − 𝑐)2 − √𝑥2 + (𝑦 + 𝑐)2 = 𝑙 
 

Where, in this case: l is the length of AB, and 𝑐 = 𝑂𝐴𝐴 2

 𝑦

  

                                                           𝑦  𝑙2        4             𝑥𝑐 
                                                                     𝑥  𝑙2             4             

Since 2c=AOA=L:

                                                                                   𝑥2                 𝑦2

                                                                              2 − 𝐿2                    𝑙2 = 1

𝑙
                                                                                    4          4 − 4

 

And the moving centrode can be described as a locus of points s.t.:

𝑀 = {𝑃 ∈ 𝑅2  𝑠. 𝑡.   |𝑃2𝐵| − |𝑃2𝑂𝐵| = 2𝑎 } 
 

 

 

 

With the given values(long side=4.8077, short side=4):

  

 

 

Exercise 4: 

As a definition of centrodes, according to the book Theory of Machines and Mechanisms of Shingley, we read: “The plane motion of one rigid body relative to another is completely equivalent to the rolling motion of one centrode on the other. The instantaneous point of rolling contact is the instantaneous center”. This means that the rolling of the fixed and moving centrodes yields all possible configurations of the moving body allowed by the linkage.  

 

 

 

 

 

 

 

 

 

 

Note on the matlab code: 

As reference for the construction of the mechanism I created this model, where the long sides are equal to b and the short sides are equal to a:

  

For the cosine theorem:

𝑥 = 𝑎2 + 𝑏2 + 2𝑎𝑏𝑐𝑜𝑠𝜃

For the sine theorem:

                                                                                 𝑠𝑖𝑛𝛼           𝑠𝑖𝑛𝜃

                                                                                         =         

                                                                                   𝑥                  𝑎

For the sum of the internal angles of a triangle:

𝛽 = 𝜋 − 𝛼 − 2𝜃

These are the values used for the angles in the matlab code. The points P1, P2, P3, P4 are calculated with reference to them.