EYP

              This post is not about science either. Is about a great activity for teens all around Europe. It is called The European Youth Parliament. About a week ago,I participated at the National Selection Session Romania 2012, where my country chooses its representatives for the International Session. This project is founded by the European Union and its purpose is to encourage teenagers to be active and involve on the social plan.
              During EYP sessions, various committees are formed, using the model of the EU and each is assigned to create a resolution on a specific theme. For example, I was in the Transport and Tourism Committee and our theme was urban mobility. After three days of exhausting committee work, our team managed to create a resolution I am quite proud of(it could be better, but  the time was really short). This is how it looked like:

Here are some photos taken during the committee work and team-building( when we do games to know each other better)

During and after the General Assembly( when we debate the resolutions):

 About a month ago, I discovered an amazing Youtube science channel, called Vsauce. A friend recommended me a video....and my GOD!!! ....it was so f...ing awesome!!! I immediately started to watch the Vsauce videos  like a maniac. I'll stop talking and let you enjoy their fantastic work. Thank you Vsauce !
          These are among my favourites:

Me at TV

        Realitatea TV, an important TV channel from Romania started a campaign called" Cine, cum si cand SCHIMBA ROMANIA?""( in translation it means" When, how and who CHANGES ROMANIA?"). During the campaign, important businessmen, scientists, writers and poets from Romania or even from other states are invited to tell their opinions regarding the future of our country, that has many economical and social issues.

        I am also very pleased to see that the Romanian media takes a break from exposing the boobs of a so called" celebrity" or spreading trivial rumors about public figures and finally starts to promote the real values of our country, people who made and can make significant good changes. This kind of campaigns offer Romanians the necessary models  they need to follow in their struggle to overcome the the mentality flaws caused by communism( although Romania is not a communist country since 1989, the harm of the regime on people's mentality can still be felt)

           I am very proud to say that I am among the interviewed people, as the representant of the young generation of Romania. I always had concerns regarding social development and a genuine interest in political science, so that was a great chance to share my view and why not, to inspire students to involve in the public life, by showing them that a teenager's opinions do matter among a lot more experienced persons.

         This video is with me. Unfortunately, I speak in Romanian and I can't put in subtitles,but you can still enjoy it!

     

That's me!

Here are some other videos from the campaign:

Maia Morgenstein, Romanian actress( Maia Morgenstein played the role of Mary in "Passion of the Christ" movie, directed by Mel Gibson):

Francois Rantrua, the president of the World Bank:

Laurentiu Damian, Roumanian theatre director:

Steven van Groningen, the president of the Romania Comertial Bank

Ilinca Dumitrescu, Romanian Pianist:

Radu Gologan, Mathematics Teacher, Coordinator of The Romanian Mathematics Olympic Team

   

Charge images

      While working a problem given at IPHO 2010, which can be found at this link:http://ipho.phy.ntnu.edu.tw/problems-and-solutions/2010/IPhO_2010_Theo_Problem%201.pdf

I gained a great interest in the charge image method.

The problem asks what's the attraction force between the charge q and the sphere connected to a 0 potential. Following the relatively laborious solution, I started wondering what happened if the sphere was connected to a V potential, or at no potential at all. This is exactly what I will study in this post.

First, let's see what happens if we have two charges, q and -q at a distance d from each other, in an infinite empty space. The equipotential lines and field lines look like this :

You can observe that the equipotential ares have the shapes of " elongated spheres" with one exception (right in the middle is an infinite plane of 0 potential)
The trick is that by replacing charge q with a conductor charged with q as well that has the exact shape of any of the equipotential areas, the exterior field and potential distribution is maintained. O...the conductor must be either initially charged with q, or connected to a potential equal to the potential of the equipotential surface it replaced. q is thus called the "image charge" of the conductor.
For example I'll give you the example when I replace the infinite plane with a conductor. As I said, there are two options: either to charge the conductor with q from the beginning, either to connect it to a zero potential( which is the potential of the area it replaces). In either case, the final result is the same: the conductor is charged with q, has 0 potential and the exterior distribution of field is the same. The attraction force between the plane (or any other conductor) is equal to the attraction force between the initial charges,since the field distribution is kept constant.

So, in order to solve the case with the sphere and the charge q, we must find the charge q', that, in the presence of q, creates an equipotential sphere of 0 potential and radius R.

If you write the equations for the potential in P and solve the equations in the ZOY plane, you will obtain that the equipotential zone of 0 potential is a circle, with the equation  Z^2+ (Y-OO')^2=R^2  (that means that in XYZO space, the zone is a sphere)

You can solve the IPHO problem by finding  the equation of the circle I mentioned previously. Thus you will have the radius and OO' distance. R=-q'd/q; OO'=d q'^2/q; this distribution is possible only if q'/q<0!!!, so pay attention to the signs. In the problem , the radius R and charge q are given, so you can easily find q' and OO'. Since the field distribution is the same outside the sphere( the field is 0 inside), the attraction force between the charge and the sphere is equal to the attraction force between the 2 charges.

Now,what happens if the sphere is connected to Ve? You must find a distribution of charges that gives you desired equipotential  surface,of Ve potential. I said distribution! Nobody said that there must be just one image charge! There can be more. 

If you place another charge right in the middle of the sphere from the previous drawing, you will obtain the equipotential spherical surface of Ve potential I demanded!

the condition that must be respected is simple:


 The electrostatic force between the exterior charge and the sphere will equal the sum of the electrostatic forces between the exterior charge and the two image charges!

What if we don't connect it to a potential? Well, since the sphere cannot take charge through the contact anymore, it will remain neutral. Every conductor is an equipotential surface( electrons move freely until they find equilibrium), so the sphere remains an equipotential surface, but the value of its potential is not interesting anymore. There are also two image charges, but the condition that must be respected is Q1+Q2=0! so, Q1=-Q2. 
         I hope you understand now what's up with the image charges after all. If you understand their concept, you can play easily with electrical charges  conductors and field distributions.

A slippery pencil

Physics is great, because it studies everything, from the expansion of the Universe, to the chaotic movements of molecules. However, before we go that far, I'd like to show you how physics describes the simple falling of a pencil on a table, from a vertical position(θ is 0 at the beginning)

      First, I want you to do the experiment on different surfaces, like in the image above. If you did it right, you should have observed that on smooth surfaces, the head of the pencil touching the surface goes onward or backwards.

It already seems a little complicated. In the next section I will write the equations of motion at a arbitrarily chosen moment during the falling.I strongly urge you to make the calculation by yourself too, because the way I will wrote might not be clear enough.

From the conservation of energy:

Iω^2/2= Mgl ( 1-cosθ) /2
where I is the moment of inertia of the pencil about  the contact point;I=Ml^2/ 3

ω^2=3g(1-cosθ)/l

 the tangential acceleration,  ɛ=dω/dt, is ɛ=3g sinθ /2l

• the equation of motion on the vertical axis is:

Mg-N = Mω^2*cosθ *l/2  + ɛsinθ *l/2

=> N=Mg((3cosθ-1)/2)^2
  
N becomes 0 at cosθ=1/3=> θlim=70.5; that means that at 70.5 degrees the pencil loses contact with the table(that holds only if the pencil hadn't started to slop till then- the equations of motion would change)

• the equation of motion of the horizontal axis is:

The F force is sketched having a arbitrarily chosen direction, being equal to the contact force between the surface and the pencil. If the direction chosen is wrong, I will obtain a negative value of the force from the equations, so that will be just fine( I'll change the direction if so).

F= Mɛ *cosθ* l/2 - Mω^2 *sinθ *l/2

=> F= 0.75 sinθ( 3 cosθ - 2)

the condition for the slipping to start is F= μN, so at that moment we can write:

μ=F/N= 3sinθ( 3 cosθ -2 )/ (3cosθ -1)^2

In order to study properly the motion of the pencil, we should sketch a graph between θ and μ, and see when does the pencil starts to slide and how. However a representation with μ=f(θ), will do just fine. 


for arbitrarily chosen values of  θ varying between 0 and 70.5 degrees, I obtained:

θ         μ

0         0

10       0.1301

20       0.2539

30       0.3512

40       0.3411

50       -0.191

60       -5.196

70       -4042

From μ=f(θ) we can figure out that μ becomes negative at cosθ=2/3 , θ=48

at θ=36, μ=0.37, which is the maximum value obtained.

Next, I'll  sketch the graph and I'll explain the falling of the pencil using it.

• The first part of the graph( black part) is attained when the pencil falls on a smooth surface.  The values of μ are positive, F=μN is positive( N>0 on this interval), so the direction of the force is well drawn. If the direction of F is to the right( as in the drawing), it means that the pencil starts to slip backwards. The conclusion is that the pencil starts to move backwards only if μ=0.37.
• The second part of the graph( red part) cannot be attained, because the pencil already started to slip, so the solutions are not physical anymore.
• The third part ( blue part) is attained when the pencil falls on a rough surface. The values of μ are negative,  F=μN is negative( N>0 on this interval), so the direction of the force F must be drawn in the other direction. The conclusion is that when 0.37<μ<1 the pencil starts moving onward.

Well, that's pretty much it...I leave you the pleasure to study the motion of the pencil after it started slipping. Maybe I'll study it in my future posts....