Have you ever wondered what the sky really is? Have you ever asked why the sky acts like the largest canvas you have ever seen, a canvas that stretches from the horizon to the zenith (the point directly overhead) which consistently carries varying pictures at various times? If you have evolved to have become a human being, you must have. We humans have always wondered at the enormity of the sky. Not just wondered, we have shed our sheer animal instincts and have evolved as humans by relentlessly trying to look at the sky and understand it. No other animal than Humans have acquired the physiological vantage point of the human head that is the best suited to watch the sky. Or should we say, the heavens, as commonly perceived.

So, what is the Sky and where does it those colours that it flaunts?

The post Per Aspera Ad Astra (Part 1) appeared first on ThePhysicist.

]]>This is a series of articles intended to educate aspiring astronomers about observational and theoretical Astronomy. Our discussion will range from the basic unaided techniques to the frontiers of ongoing Astronomical research which will include fun Mathematics and Physics involved in Astronomy. Advanced technologies, and engineering methods prevalent in modern Astronomy will also be covered in the forthcoming articles of this series. A dedicated post shall also look into the various career options related to Astronomy.

**Introduction**

Have you ever wondered what the sky really is? Have you ever asked why the sky acts like the largest canvas you have ever seen, a canvas that stretches from the horizon to the zenith (the point directly overhead) which consistently carries varying pictures at various times? If you have evolved to have become a human being, you must have. We humans have always wondered at the enormity of the sky. Not just wondered, we have shed our sheer animal instincts and have evolved as humans by relentlessly trying to look at the sky and understand it. No other animal than Humans have acquired the physiological vantage point of the human head that is the best suited to watch the sky. Or should we say, the heavens, as commonly perceived.

So, what is the Sky and where does it those colours that it flaunts?

To aid your imagination to fully grasp the answers, you need to recall all that you have seen of what you think is the sky, such as, a cloudy sky, a sunrise, a sunset, a night sky, a foggy sky, etc. Now take a look at the following image.

Every dot of light you see in this image is a galaxy. This image contains an estimated 10,000 galaxies of which some are as distant as 13 billion lightyears, for example GN-z11 (the oldest and farthest observed galaxy) which is 13.4 billion light years away. So, the light from them has taken almost as long as the age of our universe. And now as we see this image, some of these ultra old galaxies might not even be existing anymore and we can’t know what happened to them because as the universe is expanding faster than light (yes, space-time is expanding and it can expand faster than light can cross the increasing separation).

The Sky is everything that lies above the surface of the Earth. Yes, by the literal definition, Sky contains the atmosphere and the outer space, including all the Sun, the Moon, the planets of our solar system, the asteroids, planetoids, gas, dust, exoplanets, stars, galaxies, dark matter and what not. This also answers our other question as to what gives the Sky its various colours. You see, there are unimaginably huge sources of objects and processes in the sky above us that produce the effect we see as colours. Just as the scattering of the white light from Sun by our atmosphere gives the sky a blue colour when there are no clouds. The same scattering also gives the sky the reddish orange hue during the sunrise or sunset. The interaction of solar storms carrying high speed electrically charged particles with the Earth’s magnetic field, produces the magical effects we see are aurora. The sky is literally a natural canvas on which nature paints many plays for us to see.

**Why and How**

Life in the modern era has given us many advantages over those of our ancestors who started as apes. We, today, have so much of technology, engineered conveniences, scientific lifestyle, and power over nature, that if a thinking ape sees us, it would be lost in awe. Think about it. As apes, we once used to live in caves and forests, hunting and gathering during days and lying under the sky at nights. But today, it hardly matters what time of the day it is. We have virtually acquired the power to turn the day into a night and vice versa. But progress has come at a cost.

To put the why into perspective here is an incident. In a 1994 night, there was a blackout in the city of Los Angeles for a few hours. When the residents did not have artificial lights blocking the night sky, they got scared of the night sky as they were witnessing the view of the sky filled with stars and the edge of the Milky Way galaxy. Although in ideal natural atmospheric conditions the night sky would appear so, these residents had their very first live and direct experience of the celestial beauty to the extent that they got frightened thinking something must be wrong. They called the emergency and the Griffith Observatory to report the night sky.

With advancement of technology we humans have been living an increasingly easier lifestyle than before. There is no doubt that in a practically individual’s point of view, life is hard. But when we take the average survival into account, it’s becoming increasingly less in demanding effort and attention. Most of us are so much in deep in artificial lights that we hardly notice what’s it like outside for real for days. This may not sound much alarming, but it is actually the worst thing that must be happening to the human evolution. Humans were literally carved out of their ancestors by the Sky. The art and skill of recognizing the patterns that exist between the weather, the climate and the seasons and the sky were the intellectual tools that helped humans to grow as a civilization that started producing food instead of hunting and gathering. The understanding of the concept of time is rooted in the sky. Astronomy is the oldest known science and the Sky is the oldest known laboratory known to mankind. When we forget to respect and appreciate the sky, we forget to respect and appreciate ourselves.

When it comes to the how, that is, the questions pertaining the methods which have been used, are currently being used and will possibly be used in the future of Astronomy, this series, is going to answer such questions over great elaborate lengths in the future episodes. To give a glimpse of all that, we can say that the art and science of Astronomy begins with your mind and your eyes. Yes, the unaided eyes you have and that intelligent mind you are, are all that are needed to start walking the path of becoming an Astronomer. Depending upon what ambitions and what intentions you possess, there would come the requirement of more and more advanced tools and thinking techniques. Rest assured, by the time you take a break and ponder over yourself, you’d be amazed to find the manifold expansion you would have achieved as a human mind.

**A Conceptual Glimpse to the Sky**

Professionally, Astronomers see the Sky as a celestial sphere. Although we will cover the concept of celestial sphere in great lengths in the upcoming episodes, let’s just know a little enough to have the feel of it.

The Celestial Sphere (will be used as CS in this series), is an imaginary sphere centred at the centre of the Earth and has an abstract radius, i.e., the radius can be as low as the orbit of a fly to as large as the orbit of the farthest galaxy that can be observed. Although theoretically it can be infinite but to be scientifically realistic, it is the distance to the most distant point of the observable universe, which is currently as of December 2018, almost 46.6 billion light years.

The idea is to obtain the two spatial dimensions of anything that we see in the sky. Barring the distance of the object, as in, the depth of the object into the sky, we can mention the position of the object considering the celestial sphere as a momentary spherical shell. Regarding the depth, we have to use other means which we will discuss later. There are mainly two rectangular coordinate systems used to define the coordinates of objects in the CS, Equatorial and Azimuthal. To understand it better, let’s take a look at the following screenshots of simulation (Stellarium) one by one.

Let’s say that the observer is at a location (mentioned by geographic latitude and longitude), say, in the city of Raipur in the state of Chhattisgarh, India. Now depending upon his location and time of observation there would be a specific configuration of the Sky observable to him. If the observer looks to North, he will see something similar to this:

Note the star labelled as Polaris. This is commonly called the pole star or as Dhruv in Hindi. Polaris always stays north as it is directly above the North Pole of the Earth, independent of Earth’s west to east rotation. Thus in the Northern hemisphere, the position of Polaris varies depending upon the location of the observer. As the observer walks down from the North Pole toward the equator, Polaris will appear to go down from the point directly overhead (called Zenith) towards the Horizon. At the equator, Polaris will appear near the horizon. Because in our present case the observer is less than 45 degrees of the equator, the pole star appears less than 45 degrees of the horizon. As you may have already guessed it, Polaris never appears in the sky observed from the Southern hemisphere as it stays below the horizon there and we cannot see anything below the horizon as Earth itself, being opaque, blocks the view. That’s why you actually can view a celestial hemisphere and not the entire sphere. In the simulation however, you can make the Earth as transparent and see the celestial sphere as a whole. Now, how would you express the location of Polaris on the celestial sphere? Using coordinate systems, right? Well, for that now we are going to use the two types of coordinate systems mentioned above, starting with the Equatorial Grid System.

The Equatorial grid is just the projection of the geographical grid we follow in determining the coordinates of a point on Earth, to the celestial sphere. This system is global (independent of the location of the observer on Earth) and it is centred at the centre of the Earth. It has the poles above the geographical North and South poles respectively. So, for our observer in Raipur, this is how the gridded sky look like:

Note that Polaris in this system appears at the North Pole of the grid and will always appear there no matter where the observer goes on Earth, because the grid system is global.

Now let’s turn towards the other coordinate system, the Azimuthal Grid System. This grid system also has the centre at the centre of Earth. But it differs from the Equatorial Grid System in terms of poles. Here, the poles are dependent on the location of the observer. In this system, the point directly above the observer, viz, Zenith is taken as the primary pole. Obviously the other pole or the secondary pole is sphero-diametrically opposite to the Zenith. However, we will only need the primary pole in this system. As you can deduce, when the observer changes his location on Earth, his Zenith will also move with him. Thus this system is not global but rather local. For our observer based in Raipur, this is how the gridded sky appears:

You may also note that these two coordinate systems are interconvertible through a bit of complicated Mathematics which we will learn later. Use of a coordinate system to grid the celestial sphere or Sky as you know it, combined with the calculated depth using Physics and Mathematics, makes it super effective to pin point the position of an object and thus its trajectory.

**Conclusion**

For better understanding of the series and a better coherence of perception, here is the map of the series. As we move further in the series, we would cover each bullet point and the subpoints therein.

**Overview****Naked Astronomy (the Astronomy you can do with the unaided eyes)****Optical Astronomy (the Astronomy you can do with the optical telescopes)****EM Astronomy (the Astronomy you can do with the electromagnetic radiation coming from the Sky)****Frontiers of Astronomy (the advanced and possibly future technologies in Astronomy)****Astronomers (the master Astronomers who we owe our existence to and why)****Careers (various career options based on Astronomy that you can pursue)**

Know that there is no limit when you are convinced that the Sky is the limit. And the only way to the stars is through hardship. Thus the title, ‘Ad Astra Per Aspera’.

Thank You for reading.

Stay in tune with the Cosmos.

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]]>The post The Summers that inspired Physics appeared first on ThePhysicist.

]]>When I was a kid, during the summer vacations, I would go to my granny's place and enjoy life in the countryside for about a month. The warmth of the love I received at my granny's house, made me long for the summer every year. My lovely Granny, maternal uncles, aunts, and even the village folks used to shower love and affection on me. There used to be two Temples at the two ends of the village. The one at the entrance was of Shiva (the personification of deep thinking that brings order out of chaos) and the one at the end was of Narayani (the personification pure and prime energy). Every morning and every evening I used to hear the bells and the conches during the aartis. The evenings were the most crucial time for me since, they hosted clear night skies full of planets, stars, and cosmic bands of haze. Lying down on a cot in the courtyard, being caressed by the cool evening summer breeze, looking up at the sky that appeared as a perfectly crafted sheet of silk laden with gems of different colours and radiance, made me wonder about the true nature of Nature.

It always was a surreal feeling, especially from the dusk till dinner: me looking at the starry night, asking questions about the universe, with the background music from the temples, and my most beloved aunt (now alive only in memories) answering my questions the best way she could. Her name was "Sharmishtha", an ancient Hindu astronomical name for the beautiful northern sky constellation of Cassiopeia. Call it a coincidence, or maybe my biased appreciation, but she was indeed the most beautiful woman I have ever seen. It was she who had taught me the very first letters. It was she who had taught me how to think, behave, and live. She had a very weak heart (a congenital hole in it) but when it came to daring, she was a braveheart who could always fight for what's right. It was she who taught me how to be brave and not be afraid of ghosts. In fact, she had told me that not ghosts or animals but humans are the ones we must watch out for any real danger.

She had given many answers to my many questions. When I had asked her about the beautiful sky, she would tell me that it is infinite. She would tell me that the sky isn't just a two-dimensional sheet. She told me that the sky is just an appearance of what is otherwise a glimpse of the immense universe. She told me that the airplanes that I so eagerly watch flying are only a few kilometres high. She told me about astronauts who frequent to the outer space. She told me about the scientists who devote their lives trying to understand the universe. And she told me one day, that it is the scientists who are next to God who actually answer our questions. She was merely a graduate in Arts.

Those summers subconsciously infused a strong layer of rationale and an attitude of courageous thinking in my core. To be honest, while growing up, I had lost my aspirations to pursue a career in Science many times. It happens to all of us. We keep changing our aspirations depending on our situations. With new information, our minds evolve and our decisions change. Even our beliefs are mended. Inspiration and motivation are mere emotional functions that need to be supported by our intelligence and rationale. That 'support' means hard work. For me, hard work was something I had always avoided until I realized my 'passion'.

Passion is a very elusive thing. It's hard to pin down with precision. One day you might find your passion in one thing and the next day it could be something else. The only way to recognize your passion for something is through the amount of hard work can it support indefinitely. We human beings never do anything difficult. We always break down things into small steps. Then, we take one step at a time. That converts what was initially difficult into a series of easy steps. But to walk a path is very different from knowing a path because to walk needs much more energy to be invested. Investing energy means draining yourself mentally and physically. Now, to overcome the primordial human tendency of avoiding energy loss, you need passion. It is the passion that keeps you well enough supplied by drawing energy for you from your own life so that you could invest it in the indefinitely long series of energy draining process which your aspiration calls for.

I am not talking about success. Success is relative. Attempt and gain are what actually matter. In the turbulent flow of life, I had seen both success and failure from my own standards. In many things I had instant success without much attempt. And eventually nothing much was gained. But in a few things, I have failed despite passionate attempts. And I have always gained from them. When I switched from the successful job of an IT Advisor in a prominent international banking corporation, I was fairly successful but without much gain in terms of life satisfying experiences. What I switched to, was sitting at my study desk, opening up old Physics and Mathematics books that I had left long ago. When I started solving the problems in them, I failed many times. But I was gaining all the time. Each failure gave me a life satisfying experience. After the umpteenth attempt, when I understood my mistake as well as the Physics behind the question, I was rewarded by goosebumps. Those tiny bumps in my failures were missing in the successes I used to get in my previous job.

Now that I have come back to where my life had started, trying to understand the universe, it felt very natural to me that I should maintain a log of my learning process and make it public so that someone else could learn from me and help me learn. This weblog (or famously called Blog) is what ThePhysicist is all about. It is going to be a witness of my attempts, towards a life of research in Cosmology, starting from the summers that inspired Physics in me.

For any requests, suggestions, and complaints, please use the comments section below or use the contact page.

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]]>The post Richard Feynman appeared first on ThePhysicist.

]]>**Happy Birthday to Richard P. Feynman.** (आर.पी. फाइनमेन)

अनोखा व्यक्तित्व, हँसमुख स्वभाव,अद्भुत मस्तिष्क के धनी, गणनाओं के महारथी, प्रखर आधुनिक भौतिक विज्ञानी R.P. Feynman.

**प्रारंभिक जीवन (Early life)**

**जन्म**

11 May 1918 (Queens, New York, USA)

**शिक्षा (Education**)-

Feynman प्रखर मस्तिष्क के धनी थे। उच्च तार्किक क्षमता के बल पर विद्यालयी समय में ही उन्होंने Advanced Algebra , Trigonometry को आत्मसात कर लिया।

B.Sc. - 1939 (MIT)

M.Sc. - 1942 ( Princeton University )

**कार्य और उपलब्धियाँ ( work and achievements)**

Research Assistant - Princeton University (1940-41)

Professor in Theoretical Physics - Cornell University ( 1945 - 1950)

Professor in Theoretical Physics - California Institite of Technology ( 1950 - 1959)

गणित में कार्य करते करते इन्हें लगा की गणित अमूर्त (Abstract) है तब इन्होंने विद्युत अभियांत्रिकी (Electrical Engineering) को अपना कार्य क्षेत्र चुना और भौतिकी में इनकी रुचि जागृत हुई । इसके बाद ये एक सैद्धांतिक भौतिक विज्ञानी बनकर उभरे। ये Bongo बजाने के भी शौकीन थे।

इनको पढ़ाना बहुत पसंद था । इनका मानना था कि यदि आपको किसी विषय पर महारथ हासिल करनी है तो आपको इसे पढ़ाना प्रारम्भ कर देना चाहिए। किसी भी कठिन से कठिन सिद्धांत को आसान शब्दों में व्याख्या(Explain) करने की क्षमता अद्वितीय थी इनमें। इनकी पुस्तक Feynman Lecture of Physics भी भौतिक विज्ञान की प्रारंभिक समझ विकसित करने के लिए अद्वितीय (Unique) पुस्तक है। Feynman एक अच्छे आर्टिस्ट भी थे। उन्होंने विद्यार्थियों के लिए किसी भी Concept को आसानी से समझने और याद रखने का एक अनोखा तरीका सुझाया जिसे Feynman Technique के नाम से जाना जाता है। इसमें रेखाचित्रों (sketch) के माध्यम से किसी भी विषयवस्तु को याद रखा जा सकता है। क्वांटम भौतिकी (Quantum Physics)और आण्विक भौतिकी(Particle Physics) में कणों के मध्य होने वाली अंतः क्रिया को समझाने के लिए Feynman Diagram दिए जो आज भी प्रासंगिक हैं। इस कार्य के लिए इन्हें 1965 में नोबल पुरस्कार मिला।

**पुरस्कार(Prizes)**

1. Nobel Prize (1965, In 1/3)

2.Albert Einstein Award (1954, Princeton Uni.)

3. Einstein Award (1962, Albert Einstien institute of advanced Medicine)

4. Lawrence Award (1962)

**Nobel Winning Work**

क्वांटम भौतिकी और विद्युतगतिकी के स्थापित होने के पश्चात मूल कणों के मध्य की अंतःक्रिया को समझने हेतु Feynman Diagrams दिए। Feynman diagram अवपरमाणुक(Subatomic Particles) कणों के मध्य अन्तः क्रिया के गणितीय रूप को रेखाचित्रों के रूप में प्रदर्शित करने में सहायक होते हैं|

**Nobel lecture theme-**

About the development of the space-time view of Quantum Electrodynamics ( The theory of interaction between Light and Matter)

**The Manhattan project-**

द्वितीय विश्व युद्ध के समय अमेरिका का बम विकास कार्यक्रम( Bomb Developement Programme) में महत्वपूर्ण भूमिका निभाई। Feynman गणितीय गणना करने में बहुत तेज़ थे इसलिए इन्हें इस प्रोजेक्ट में विशेष जिम्मेरदारी दी गयी।

**QUOTES OF FEYNMAN**

इस महान वैज्ञानिक का जीवन यूँ तो बहुत समस्याओं से ग्रस्त रहा परंतु ये हमेशा हंसते मुस्कुराते भौतिक विज्ञान में नए आयाम स्थापित करके इस जिंदगी से 15 February 1988 को विदा ले गए।

इनका नोबेल व्याख्यान पढ़ने हेतु आप नीचे दिए गए लिंक पर क्लिक करिए।

ThePhysicist की तरफ से आपका अपना Mohit Tiwari.

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]]>The post Preparation Status Report appeared first on ThePhysicist.

]]>**What is PSR?**

PSR or Preparation Status Report is a status reporting system designed for the preparation group (common for the paying and free) community members of ThePhysicist. The system generates the spreadsheet containing data on an individual member's educational qualification, target competitive exam, current self-study topic as well as the topics already covered. This helps the students in the following ways:

- Becoming conscious of their own progress
- Developing a stronger determination and commitment
- Becoming aware of which member stands where and who to approach in the community for quicker resolution of subject related doubts

**What are preparation groups?**

ThePhysicist has dedicated preparation groups to aid students of Physics preparing for competitive examinations. All such students are community members on ThePhysicist. The ones who require active group support have to only request to be added to a suitable group. The groups offer total flexibility to the members to progress as per their own speed and style. The groups are there to help them out wherever they get stuck during their preparation.

**Do I have to pay to become a member?**

Not necessarily. Although there are paying members who get the benefits of priority support, it is not at all mandatory to pay a fee to become a community member or a preparation group member. Moreover, ThePhysicist is currently allowing only free community members and preparation group members because of the unavailability of new slots for paying members.

**How do I become a member of a preparation support group?**

In order to become a member of a preparation support group, you must upgrade yourself as a registered community member on ThePhysicist (click here) by creating a community User ID. Once you have activated your community membership account, you have to log in to your community account and then contact the support team on WhatsApp with a request to be added to a suitable preparation group.

Once added to a preparation group, you will be given a PSR ID that you shall keep as your private key for reporting your individual preparation status in the PSR system. You can access the PSR system, either from the main menu > community > PSR or you can click here.

For any requests, suggestions, and complaints, please use the comments section below or use the contact page.

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]]>The post Everything a science student should know about coronavirus appeared first on ThePhysicist.

]]>**Introduction :**

An infection with SARS-CoV-2 is called COVID-19 diseases (Coronavirus disease in 2019). Genesis has been tracked to (Possible but not confirmed ) in a food market in Wuhan, China,( or lab-engineered by China) in December 2019, to countries as far-reaching as the United States. Regardless of the global fear in the news, its improbable to contract SARS-COV-2 unless came in contact with an infected area or living thing.

**Causes:**

Coronaviruses are a family of viruses. Coronaviruses are zoonotic (first develop in animals before developing in humans). Virus transmission from animals to humans due to close contact with an animal that carries the infection. Coronaviruses can be spread from person to person through respiratory droplets (wet stuff that moves through the air when you cough or sneeze). The viral material hangs out in these droplets and can be breathed into the respiratory tract (your windpipe and lungs)leading to infection. Researchers believe that possibly the virus may have been transmitted to humans via bats to another animal — either snakes or pangolins. This transmission likely occurred in the open food market in Wuhan, China, or may have been engineered by a lab in Wuhan.

**Symptoms:**

Different things are learned every day by doctors. COVID-19 may not initially cause any symptoms as people may carry the virus for 2 days or up to 2 weeks before you notice symptoms.

Some common symptoms

• shortness of breath

• having a dry cough that gets more severe over time

• a low-grade fever that gradually increases in temperature

• fatigue

Less common symptoms include:

• chills

• repeated shaking with chills

• sore throat

• headache

• muscle pain and aches

• loss of taste

• loss of smell

These symptoms may become more severe

• trouble breathing

• blue lips or face

• persistent pain or pressure in the chest

• confusion

• excessive drowsiness

**Who is at increased risk?**

High risk for contracting SARS-CoV-2 (Covid19)if it came into contact with someone who’s carrying it, specifically if exposed to their saliva or cough or sneeze. Without taking proper preventative measures, you’re also at high risk if you:

- living with infected (with Covid19) person.
- are providing home care to infected (with Covid19) person.
- have an intimate partner who was infected (with Covid19).

People with a health conditions and old people have a higher risk of severe complications if are infected .These health conditions include:

- lung conditions, like COPD and asthma
- certain heart conditions
- immune system issues like HIV.
- cancer requiring treatment
- severe obesity
- other health conditions, if not well-managed, such as diabetes, kidney disease, or liver disease

Pregnant women have a higher risk of other viral infections, but the same isn't confirmed with Covid19.

**How are Coronaviruses diagnosed?**

COVID-19 can be diagnosed using a blood, saliva, or tissue sample. Though, most tests use a cotton swab to retrieve a sample from your nostrils.

The emergency use of diagnostic kit is authorized for use by people who healthcare professionals have identified as having suspected COVID-19.

If you notice any COVID-19 symptoms, talk to your doctor. Your doctor will advise you about the action you have to take like to stay home, quarantine, to come doctor's office for evaluation, and to visit the hospital in an emergency .

**What treatments are available?**

There’s currently no specific treatment approved for COVID-19, treatments and vaccines are currently under study. Treatment is based on managing symptoms as the virus has no complete medicine. Other coronaviruses (SARS AND MERS) are also treated by managing symptoms.

**Therapies** used for these illnesses are :

- antiviral medications
- breathing support, like mechanical ventilation
- steroids to reduce lung swelling
- blood plasma transfusions

** Possible complications from COVID-19** **:**

The most serious complication of COVID-19 is a type of pneumonia that’s been called 2019 **novel corona virus-infected pneumonia (NCIP).** Into hospitals in Wuhan, China 138 people were admitted with NCIP, 26 percent of those admitted had severe cases and needed to be treated in the intensive care unit (ICU). Around 4.3 percent of these people died in ICU from this type of pneumonia. People who were admitted to the ICU were **on average **older and had more underlying health conditions than **people that **didn’t **attend **the ICU.. Till now, NCIP is the only complication linked to the COVID-19. Researchers have seen **the subsequent **complications in **people that **have developed COVID-19 like

- acute respiratory distress syndrome (ARDS)
- irregular heart rate (arrhythmia)
- cardiovascular shock
- severe muscle pain (myalgia)
- fatigue
- heart damage or heart attack

**Methods of prevention:**

The best way to prevent the spread of infection is to avoid or limit contact with people who are showing symptoms of COVID-19 or any respiratory infection. The next best thing you can do is practice good hygiene and social physical distancing to prevent bacteria and viruses from spreading.

**Prevention tips:**

- Wash your hands frequently for
**a minimum of**20 seconds at a time with warm water and soap. How long are 20 seconds? About as long**because it**takes to sing your “ABCs.” - when your hands are dirty, don’t touch your face, eyes, nose, or mouth.
- Don’t
**leave**if you’re feeling sick or have any cold or flu symptoms - Stay at 2 meters far aw
**ay from**people. - Cover your mouth with a tissue or
**the within**of your elbow whenever you sneeze or cough. Throw away any tissues**you employ directly**. - Clean any objects you touch
**tons**. Use disinfectants on objects like phones, computers, utensils, dishes, and doorknobs.

**Should you wear a mask?**

- If you call
**at**a public setting where it’s difficult to follow physical distancing guidelines, recommends**that you simply**wear a cloth**mask**that covers your mouth and nose. - When worn correctly, and by large percentages of the public, these masks can help to slow the spread of SARS-CoV-2.
- That’s because they can block the respiratory droplets of people who may be asymptomatic or people who have the virus but have gone undiagnosed. Respiratory droplets get into the air when you exhale, talk, cough, or sneeze.
- Cloth masks are preferred for
**the overall**public since other**sorts of**masks should be reserved for healthcare workers. - It’s critical
**to stay**the mask clean. Wash it**whenever you employ**it. Avoid touching the front of it**together with your**hands. Also,**attempt to**avoid touching your mouth, nose, and eyes**once you**remove it. This prevents you from possibly transferring the virus from a mask to your hands and from your hands to your face. - Keep in mind that wearing a mask isn’t a replacement for other preventive measures,
**like**frequent hand washing and practicing physical distancing. All of them are important. - Certain people shouldn’t wear face masks. They include children under 2 years old,
**people that**have trouble breathing,**and other people**who are unable**to get rid of**their masks.

**In the next post, I am going to share a few details about the Coronavirus family virus of MERS, SARS, and basic difference.**

For any requests, suggestions, and complaints, please use the comments section below or the contact page.

The post Everything a science student should know about coronavirus appeared first on ThePhysicist.

]]>The post Combined E&M fields – Tutorial 1 appeared first on ThePhysicist.

]]>**Question:**

**A uniform electric field of strength $E$ acts parallel to the positive Y-axis**. **A uniform magnetic field of strength $B$ acts parallel to the positive Z-axis. A particle of charge $q$ with mass $m$ placed at the origin O is released from rest. If at a certain time, the velocity of the particle is given as ($\vec{v}=3\hat{i}+4\hat{j}$), find out the following:**

**the Y coordinate of the particle at that instant****the ratio $\frac{E}{B}$**

**If you prefer watching over reading, you can also watch the video lecture below.**

This problem is a typical JEE Main as well NEET level Physics question. However, in this post, the second question is added to ensure that the reader understands the basics very well and is comfortable with not only the Physics tricks but also the Mathematics involved. The detailed analysis of the background has been covered in the previous post. In this post we shall illustrate how and why the principle of conservation of energy should be used to save time, instead of taking the lengthier approach that has been already discussed.

**Diagram 1:**

Refer to the original normalized plot.

We have already obtained the trajectory curve of the particle in the lecture. There, however, are a few things we should note very carefully.

- $x$ starts from $0$ and keeps increasing with $t$ without bound.
- $y$ starts from $0$, rises to $y_{max}$ and then falls back to $0$.
- $\vec{F_l}$ does no work. Only $\vec{F_c}$ does the work in moving the particle through $y$.
- At every point on $x$, there is only 1 corresponding $y$ but for any $y$ there are infinitely many values of $x$.
- The $x$ and $y$ components of the velocity $\vec{v}$, $v_x$ and $v_y$ are always $=0$ on the
**X-axis**. - $v_x$ is never negative but $v_y$ can be both positive and negative.
- For every $x$, there is only one $\vec{v}$.
- Except on the
**X-axis**and at $y_{max}$, for every $y$ there are two values of $\vec{v}$ that are same in magnitude but different in direction. In both of them $v_x$ is the same but if in one $v_y$ is positive, in the other $v_y$ is negative. - When $y=y_{max}$, $v_y=0$ and $|\vec{v}|=v_x$.
- $v_x=0\Rightarrow v_y=0$ but $v_y=0 \nRightarrow v_x=0$

Now, we can solve the $1^{st}$ part of the question by using the principle of conservation of energy.

At $y$, $\vec{v}=v_x\:\hat{i}+v_y\:\hat{j}=3\hat{i}+4\hat{j}$

Work done by $\vec{F_c}$ in moving the particle of charge $q$ through $y$ $=qEy$.

At $y$, the speed $v=\sqrt{3^2 + 4^2}=5$

Starting from $0$, the change in kinetic energy when the particle is at $y$,

$= \frac{1}{2}mv^2=\frac{1}{2}m(5^2)=\frac{25m}{2}$

Thus $qEy=\frac{25m}{2}$

$\Rightarrow \boxed{y=\frac{25m}{2qE}}$---(Ans 1)

To solve the $2^{nd}$ part of the question, we have to consider the $v_x$ and $v_y$ as functions of time as in $Eq^n\:(5)$ and $Eq^n\:(6)$ of the lecture post.

$v_x=\frac{E}{B}(1-cos\:\omega t)$ and $v_y=\frac{E}{B}sin\:\omega t$.

Let $\frac{E}{B}=\alpha$. Thus,

$v_x=\alpha\:(1-cos\:\omega t)$ $\:Eq^n\:(9)$

$v_y=\alpha\:sin\:\omega t$

$\Rightarrow {v_x}^2+{v_y}^2={\alpha}^2[(1-cos\:\omega t)^2+sin^2\:\omega t]$

$\Rightarrow v^2={\alpha}^2(1+cos^2\:\omega t+sin^2\:\omega t-2\:cos\:\omega t)$

$\Rightarrow v^2={\alpha}^2(2-2\:cos\:\omega t)$

$\Rightarrow v^2=2{\alpha}^2(1-cos\:\omega t)$ $\:Eq^n\:(10)$

Dividing each side of $\:Eq^n\:(10)$ with the same side of $\:Eq^n\:(9)$ gives,

$\frac{v^2}{v_x}=2\alpha$

$\Rightarrow \alpha=\frac{v^2}{2v_x}$

Plugging in the numerical values of $v^2$ and $v_x$ gives,

$\boxed{\frac{E}{B}=\frac{25}{6}}$---Ans(2)

If you prefer watching a lecture over reading, you can watch the video lecture below.

This concludes the tutorial. If you have any requests, suggestions, or complaints regarding this post, please mention in your comment below. For new questions, please use the General Physics Forum and I will respond there with the help you require.

The post Combined E&M fields – Tutorial 1 appeared first on ThePhysicist.

]]>The post Charged Particle in Coexisting Electric and Magnetic Fields appeared first on ThePhysicist.

]]>This post is categorized as 'Lecture' because, even if it is based on a single problem, some important concepts of Physics are illustrated in this.

**Question:**

**A uniform electric field of strength $E$ acts parallel to the positive Y-axis**. **A uniform magnetic field of strength $B$ acts parallel to the positive Z-axis. A particle of charge $q$ with mass $m$ placed at the origin O is released from rest. Find out the following:**

**the position vector of the particle as a function of time****if there will be any double point in the trajectory curve; if yes, then will it be a node or a cusp****the shortest time interval between points where the particle comes to rest****the displacement of the particle along the Y-axis when the velocity of the particle becomes parallel to the X-axis for the first time****the velocity of the particle when it becomes parallel to X-axis****if the particle will have a velocity purely along the Y-axis**

**If you prefer watching over reading, you can also watch the video lecture below.**

This is a key problem for students preparing for JEE Advanced, JEE Main, and NEET (undergraduate). Many questions of varying standards and toughness can be formed on its basis, some of which have been put here in the question statement. For example, the magnetic field working perpendicular to the velocity vector all the time is not going to produce any work. Thus the question can be a little modified to be easily solved by using the Work-Energy principle so that the work done by the electric field can be equated to the kinetic energy of the particle. However, in this particular problem, that technique is useless. That's why we must understand the Physics and Mathematics involved in this question. A short tutorial post explaining how a sample variant of this question can be asked has also been published.

In the diagram below, the distribution of the electric field $\vec{E}$ and the magnetic field $\vec{B}$ is shown along the Y and Z axes respectively. The particle of charge $q$ and mass $m$ is placed at the origin **O**. Starting from the time $t=0$, the path of the particle at any instant $t$ is shown along with its velocity, tangential to the path, making an angle $\theta$ with the Y-axis. The corresponding $x$ and $y$ components of the velocity, the Coulomb force $\vec{F_c}$ (electric force), and the Lorentz force $\vec{F_l}$ (magnetic force) along with its components are also shown.

**Diagram 1:**

It is clear that:

The Coulomb force, $\vec{F_c}=qE\:\hat{j}$

The Lorentz force $\vec{F_l}=q{\vec{v}}\times \vec{B}$

Magnitude of $\vec{F_c}=F_c=qvB\:\:\:(\because \vec{v}\perp\vec{B})$

If at any time $t$, the velocity vector $\vec{v}$ of the particle makes an angle $\theta$ with the Y-axis, then:

$v_x=v sin \theta$ & $v_y=v cos \theta$

Net force along X, $F_x=F_{l_x}=F_lcos\theta=qBvcos\theta=qBv_y$

Net force along Y, $F_y=F_c-F_{l_y}=F_c-F_lsin\theta=qE-qB v sin\theta=qE-qBv_x$

Now, let's try to understand what really happens when the particle at the origin **O** is released from rest. This is shown in the diagram below.

**Diagram 2:**

At the time $t=0$, the particle is at **O** and is at rest. At this instant, there is only one force acting on it. This is the Coulomb force, $\vec{F_c}$ along the +Y direction. From this, the acceleration along the Y-axis, $a_y$ can be found as:

$F_c=qE\implies a_y=\frac{qE}{m}$

This vertically upward acceleration will cause the particle to acquire a differential vertical velocity $dv_{y_1}$ in a differential time interval of $dt_1$. The only possible force along X is the Lorentz force ($\vec{F_l}$). Since $\vec{F_l}$ depends on the net velocity ($\vec{v}=v_x\:\hat{i}+v_y\:\hat{j}$) of the particle, till $t$ reaches the value of $dt_1$, $F_l$ remains negligibly 0. Thus far, the particle moves mostly along Y. In the next time interval $dt_2$, at $t=dt_1 + dt_2$, the vertical velocity $dv_{y_2}$ of the particle causes a horizontal differential Lorentz force ($dF_l$), thus, a horizontal differential acceleration ($da_x$) and hence a horizontal differential velocity ($dv_{x_2}$). At this instant, the particle begins to deviate from moving purely along Y into a curve in the XY plane or Z plane. Let's summarize all the data we have for the initial moment, which can also be called boundary conditions, i.e., at $t=0$

$x=0$ |

$y=0$ |

$v_x=0$ |

$v_y=0$ |

$a_x=0$ |

$a_y=\frac{qE}{m}$ |

Now, we are ready to answer the first question, i.e., the position vector $\vec{r}$ of the particle as a function of time:

$$\boxed{\vec{r}=x(t)\:\hat{i}+y(t)\:\hat{j}}$$

To find $x(t)$ and $y(t)$, we have to begin with $a_x$ and $a_y$ at any time $t$ where the velocity of the particle $\vec{v}$ makes an angle $\theta$ with the **+Y** and proceed as follows:

$a_x=\frac{F_l cos\theta}{m}=\frac{qB}{m}v_y$ $\:\:\:Eq^n\:(1)$

$a_y=\frac{F_c}{m}-\frac{F_l sin\theta}{m}=\frac{qE}{m}-\frac{qB}{m}v_x$ $\:\:\:Eq^n\:(2)$

Now, we can clearly see that $a_x$ depends on $v_y$ and $a_y$ depends on $v_x$. So, these two quantities are entangled. In order to solve this situation, we are going to use a few mathematical techniques. Let's differentiate $a_y$ once WRT $t$ and get rid of the constant term. This will also differentiate $v_x$ and give us $a_x$ which depends only on $v_y$. Thus, we can have an equation purely along **Y **and then we'll see what's next. Okay? Let's do that.

$\frac{d}{dt}(a_y)=-\frac{qB}{m}a_x=-(\frac{qB}{m})^2v_y $ $\:\:\:Eq^n\:(3)$

$\because a_y={v_y}^{'} \implies \frac{d}{dt}(a_y)={v_y}^{''}$

Thus $Eq^n\:(3)$ becomes: ${v_y}^{''}=-(\frac{qB}{m})^2v_y$ $\:\:\:Eq^n\:(4)$

One way of solving $Eq^n\:(4)$, is to use the technique of solving the simple harmonic motion equation.

If we get the equation

$r^{''}=-\omega ^2 r$

then the general solution is

$r=r_{max} sin (\omega t+\delta)$

in which we can use boundary conditions to obtain the particular solution.

For our case, the general solution is

$v_y=v_{y_{max}} sin\: (\omega t+\delta)$ where $\omega=\frac{qB}{m}$

From the boundary conditions, we know that, at $t=0$, $v_y=0$ and $v_y^{'}=\frac{qE}{m}$.

This gives us the following:

$0=v_{y_{max}} sin\:\delta \implies \delta=0\implies v_y=v_{y_{max}} sin\: \omega t$

$\frac{qE}{m}=\omega v_{y_{max}}\implies v_{y_{max}}=\frac{E}{B}$

Thus, the particular solution we need and now we have is:

$$\boxed{v_y=\frac{E}{B}\:sin\:\omega t}$$ $\:\:\:Eq^n\:(5)$

The other and mathematically more rigorous way of solving $Eq^n\:(4)$, is to use the standard method of solving linear homogeneous differential equations of second order:

For an equation of the form $r^{''}+ar^{'}+br=0$ where $a$ and $b$ are constants, the quadratic characteristic equation is:

$\lambda^2+a\lambda+b=0$

Then we need to find the roots of this equation and proceed to the general solution depending upon whether the roots are (i) real and distinct, (ii) real and repeated or (iii) complex conjugates. In our case, if we rewrite our $Eq^n\:(4)$ as:

${v_y}^{''}+(\frac{qB}{m})^2v_y=0$,

we shall see that, $a=0$ and $b=(\frac{qB}{m})^2$.

In this form, we can see that the quadratic characteristic equation is going to give us imaginary roots $\pm i\omega$, where $\omega=\frac{qB}{m}$ and $i=\sqrt {-1}$.

Thus, we can obtain the general solution:

$v_y=e^{\frac{-a}{2}t}[C_1\:cos\:(\omega t) + C_2\:sin\:(\omega t)]$ (where $C_1$ and $C_2$ are arbitrary constants)

From the boundary conditions, we know that, at $t=0$, $v_y=0$ and $v_y^{'}=\frac{qE}{m}$.

Thus, our general solution reduces to the particular solution:

$\boxed{v_y=\frac{E}{B}\:sin\:\omega t}$ $\:\:\:Eq^n\:(5)$ (as obtained above)

Now that we have $v_y$, we can write $a_x$ as:

$a_x=\frac{qB}{m}v_y=\frac{qE}{m}\:sin\:\omega t$

Since $a_x=\frac{dv_x}{dt}\implies dv_x=\frac{qE}{m}\:sin\:\omega t\:dt$

By integrating the above equation for $v_x$ varying with limits from $0$ to $v_x$ and time from $0$ to $t$, we can get $v_x$.

$\int_{0}^{v_x} dv_x=\frac{qE}{m}\int_{0}^{t} sin\:\omega t\:dt$

$$\implies\boxed{v_x=\frac{E}{B}(1-cos\:\omega t)}$$ $\:\:\:Eq^n\:(6)$

Now, we can find both $x(t)$ and $y(t)$ by integrating $\:\:\:Eq^n\:(6)$ and $\:\:\:Eq^n\:(5)$ respectively, within proper limits.

$v_x=\frac{E}{B}(1-cos\:\omega t)$

$\implies \frac{dx}{dt}=\frac{E}{B}(1-cos\:\omega t)$

$\implies dx=\frac{E}{B}(1-cos\:\omega t)dt$

$\implies \int_{0}^{x} dx=\frac{E}{B}\int_{0}^{t}(1-cos\:\omega t)dt$

$$\implies \boxed{x=\frac{E}{B}(t-\frac{sin\:\omega t}{\omega})}$$ $\:\:\:Eq^n\:(7)$

Similarly $y$ can be found:

$v_y=\frac{E}{B}\:sin\:\omega t$

$\implies \frac{dy}{dt}=\frac{E}{B}\:sin\:\omega t$

$\implies dy=\frac{E}{B}\:sin\:\omega t\:dt$

$\implies \int_{0}^{y} dy=\frac{E}{B}\int_{0}^{t}\:sin\:\omega t\:dt$

$$\implies \boxed{y=\frac{mE}{qB^2}(1-cos\:\omega t)}$$ $\:\:\:Eq^n\:(8)$

Now, using $\:\:\:Eq^n\:(7)$ and $\:\:\:Eq^n\:(8)$, we can answer

*Part 1: the position vector as a function of time:*

$$\boxed{\vec{r(t)}=\frac{E}{B}(t-\frac{sin\:\omega t}{\omega})\:\hat{i}+\frac{mE}{qB^2}(1-cos\:\omega t)\:\hat{j}}$$

*Part 2: if there will be any double point in the trajectory curve; if yes, then will it be a node or a cusp*

To answer this question, we need to trace the curve of the particle's trajectory and see if there are any double points on it, i.e., points where the curve crosses itself with real and distinct tangents to the branches of the curve at that point (node) or touches itself with a single real tangent to all the branches at that point (cusp).

**Curve Tracing** is an interesting topic in calculus and ThePhysicist will cover that topic through dedicated posts. But here we can find the answer through some simple analysis of $x(t)$, $y(t)$, $v_x (t)$, and $v_y (t)$.

$x(t)$ contains a linear part as $t$ and a negative periodic part as $\frac{sin\:\omega t}{\omega}$. Thus $x(t)$ is going to increase with time without bounds but the rate of increase will fluctuate periodically. This can be seen in $v_x$ as it is also periodic, being $0$ (minimum) when $t=\frac{2n\pi}{\omega}$ and $\frac{2E}{B}$ (maximum) at $t=\frac{(2n+1)\pi}{\omega}$, for $n=0,1,2,3,...$.

$y(t)$ is a purely periodic function being $0$ (minimum) when $t=\frac{2n\pi}{\omega}$ and $\frac{2mE}{qB^2}$ (maximum) at $t=\frac{(2n+1)\pi}{\omega}$, for $n=0,1,2,3,...$. This can be seen in $v_y$ as it is a sine function, being $0$ (mean) when $t=\frac{n\pi}{\omega}$, $\frac{E}{B}$ (maximum) at $t=\frac{\pi}{\omega}(\frac{4n+1}{2})$, $-\frac{E}{B}$ (minimum) at $t=\frac{\pi}{\omega}(\frac{4n+3}{2})$ for $n=0,1,2,3,...$.

Thus, the curve starts from **O**, then moves continuously along **+X**, while rising from $0$ in **Y** to $\frac{2mE}{qB^2}$ and then back to $0$. This can help us plot a qualitative cycle of the curve as shown below.

**Diagram 3:**

If we wish to verify what we have concluded about the shape of the trajectory, we can do it very easily using some easily available CAS tools. There are some which have apps for Android, iOS, Mac OS and Windows. Some also have web apps. The most easily available web apps are desmos and GeoGebra.

For simplicity for plotting the trajectory in desmos, let's normalize all the parameters, $E$, $B$, $q$, $m$, and $\omega$ to unity. This will give us a parametric form of the trajectory $\left(\left(t-\sin t\right),\left(1-\cos t\right)\right)$ which we can put in desmos and obtain the trajectory as below.

**Diagram 4:**

**From this, we can say that, starting from O, at every $\frac{2\pi E}{\omega B}$ units along +X, we shall have a cusp on the trajectory.**

*Part 3:**the shortest time interval between points where the particle comes to rest*

The particle comes to rest when $v_x=v_y=0$. This happens when $sin\:\omega t=0$ and $cos\:\omega t=1$. The minimum period for that is $\omega t=2\pi$. Hence, the minimum time interval between two such consecutive points is

$$\boxed{\delta t=\frac{2\pi}{\omega}}$$.

*Part 4:**the displacement of the particle along the Y-axis when the velocity of the particle becomes parallel to the X-axis for the first time*

The velocity of the particle becomes parallel to the X-axis when $v_y=0$ but $v_x$ is maximum. This happens for the first time when $\omega t=\pi\implies t=\frac{\pi}{\omega}$.

Putting that value of $t$ in $\:\:\:Eq^n\:(8)$, we get the required answer.

$$\boxed{y=\frac{2mE}{qB^2}}$$

*Part 5: the velocity of the particle when it becomes parallel to X-axis*

This can be found very simply as it is related to the previous question. When the velocity of the particle becomes parallel to the X-axis for the first time, $t=\frac{\pi}{\omega}$ for which $\:\:\:Eq^n\:(6)$ shows that $$\boxed{v_x=\frac{2E}{B}}$$.

*Part 6: if the particle will have a velocity purely along the Y-axis*

This is a question based on the sense of trigonometry. It asks if ever, $v_x=0$ would happen when $v_y\neq0$. $\:\:\:Eq^n\:(6)$ shows that $v_x=0\implies\omega t=2n\pi$ for which $\:\:\:Eq^n\:(6)$ shows that $v_y=0$.

*Hence, there never will be a point on the trajectory where the velocity is purely along the Y-axis.*

The differential approach illustrated in diagram 2 is only for illustration purpose.

If you prefer watching a lecture over reading, you can watch the video lecture below.

This concludes the lecture. If you have any request, suggestion, or complaints regarding this lecture, please post your comment below. For new questions, please use the General Physics Forum and I will respond there with the help you require.

The post Charged Particle in Coexisting Electric and Magnetic Fields appeared first on ThePhysicist.

]]>The post Bead embedded to a rotating wire appeared first on ThePhysicist.

]]>This post is categorized as 'Lecture' because, even if it is based on a single problem, some important concepts of Physics are illustrated in this.

**Question:**

**A bead of mass $m$ is embedded to a horizontal long massless thin wire such that the bead is free to move along the wire. The wire is made to rotate around a vertical axis passing through one end $O$ with constant angular velocity $\omega$ at time $t = 0$. The bead at this instant is at a distance $r_0$ from $O$. Find the expression for the angular momentum $L$ of the bead around $O$ as a function of time.**

**Diagram 1: **

**If you prefer watching over reading, you can also watch the video lecture below.**

The quick and easy formula that comes to the mind when asked to find the angular momentum of a particle of mass $m$ with angular velocity $\omega$ at the position $\vec{r}$ relative to the centre of rotation is $L=I\omega$ where $I$ is the moment of inertia of the particle given by $\boxed{I=mr^2}$ or simply $\boxed{L=m\omega r^2}$. Ideally, we should not bother about the radial velocity of the bead, if any, since the radial velocity vector will always pass through $O$ and make the angular momentum null ($\vec{L}=\vec{r}\times\vec{p_r}=0$, where $\vec{p_r}$ is the linear radial momentum ($=m\vec{v_r}$)). But in the problem we have, $r$ is not a constant but a function of time. Thus, the crux is to find $r\equiv r(t)$.

This problem can be solved in at least 3 meaningful ways, **(a) from the ground frame**, **(b) from the rotating frame** and **(c) with Lagrangian**. Let us do it in all of them. But even before we begin to decide whether to use Lagrangian mechanics or not, we need to understand the motion from the ground frame. It is not always advisable to use the ground frame to solve problems in rotational motion. More often than not, the ground frame calls for more variables and thus more equations than the rotational frame. However, using the rotational frame calls for fictitious or pseudo forces and without a firm understanding of the motion through the real forces, there is a big possibility of making mistakes. Hence, ThePhysicist suggests that the students, while practicing, should always try to solve problems of rotational motion from the ground frame first and then from the rotational frame. Now, let us see what happens to the bead of mass $m$ once the wire, to which it is embedded, starts to rotate with constant $\omega$.

**(a) from the ground frame**

From the ground frame, once the wire starts to rotate with the angular velocity $\omega$, ignoring gravity, the bead will experience only one force at any instant. This force is the contact force normal to the wire which is exerted on it by the momentary contact point on the wire. Why so? That's because, since the bead is embedded to the wire, it cannot have a relative velocity with respect to the contact point along the normal to the wire at that point. However, the bead is free to move along the wire. This means there is no radial force on the bead from any point on the wire. Since every point at a radial position $\vec{r}$ on the wire is moving in a circle of radius $r$, the instantaneous tangential velocity of that point on the wire at an angular displacement $\theta$ with respect to the initial line, $\vec{v_{\theta}}$, is given as

$\vec{v_{\theta}}=r\omega\,\hat{\theta}$ ,where $\hat{\theta}$ is the momentary direction of the unit vector perpendicular to $\vec{r}$ at that instant.

This is the instantaneous tangential velocity of the bead at any given radial position at any given instant. Now let's try to understand, what generates the radial motion of the bead and form a clearer picture.

**Diagram 2:**

As shown in the diagram above, the direction of the position vector $\vec{r}$ is shown at 3 instants, viz, $t=0$, $t=t_1$ and $t=t_2$. The angular displacements shown, respectively, are $\theta_0=0$, $\theta_1=d\theta_1$ and $\theta_2=d\theta_1+d\theta_2$ where $d\theta$ is infinitesimally small angular displacement. Also, $d\theta_1=\omega t_1$ and $d\theta_2=\omega (t_2-t_1)$.

At $t=0$, the bead starts moving with the same velocity $\vec{v_{\theta_0}}$ as that of the point of contact. When the wire moves through a rotation of $d\theta_1$, the point of contact at $r_0$ moves in a circular arc as it accelerates due to the centripetal force along the wire. The bead, not getting any force along the radial direction, keeps moving with $\vec{v_{\theta_0}}$. When we look at the bead again at $t=t_1$, we see that the contact point of the bead on the wire has now has moved a distance $dr_1$ radially inward. One perspective is that it's not the bead that has moved radially out, but the previous contact point has moved radially in and a new contact point has now been established at $r=r_0+dr_1$. Again, at $t=t_1$. This new contact point now forces the bead to move with a tangential velocity $\vec{v_{\theta_1}}$. Through a subsequent rotation of $d\theta_2=\omega(t_2-t_1)$, this contact point recedes, once again, by $dr_2$. By now, the bead is in contact with another point on the wire at $r=r_0+dr_1+dr_2$. This goes on and on till the bead falls out of the wire at the free end. Since we are concerned only as long as the bead stays on the wire, let's not worry about the bead falling out. At the end of this section, we shall also talk about the other perspective that is more rigorous and complex but explains the real Physics from the ground frame.

Now that we have got a perspective in which the wire that accelerates radially inward and not that the bead accelerates radially outward, we can frame the equation of motion. But we are solving the motion of the bead and not of the instantaneous point of contact. Therefore we have to frame our equation using the apparent acceleration of the bead relative to its momentary point of contact. This relative acceleration $\vec{a_r}$ is radially outward.

$a_r=\omega^{2}r$ where rhs is in the same sign as lhs

$\implies r^{''}=\omega^{2}r$

$\implies$$\boxed{r^{''}-\omega^{2}r=0}$ $Eq^n (1)$

$Eq^n 1$ is of the form of a linear homogeneous differential equation of second order that we will encounter in the other two methods as well. Hence, let us solve it properly and we shall refer to this solution in the other two methods. The standard solution method for such a differential equation is given below.

Any differential equation of the form $r^{''}+ar^{'}+br=0$ has a characteristic equation,

$\lambda^2+a\lambda+b=0$

Let the roots of the characteristic equation be $\lambda_1$ and $\lambda_2$.

Then, the general solution of the differential equation is given as:

$\boxed{r=C_{1}e^{\lambda_1 t}+C_{2}e^{\lambda_2 t}}$

where $C_1$ and $C_2$ are arbitrary constants whose values must be found using boundary conditions.

Now coming back to our case, the values of $a$ and $b$ in our characteristic equation are $0$ and $\omega^2$ respectively. Thus the solutions of the characteristic equation are,

$\lambda^2-\omega^2=0$

$\implies \lambda^2=\omega^2$

$\implies\boxed{\lambda_1=\omega;\:\lambda_2=-\omega}$

Thus the general solution in our case is:

$r=C_1e^{\omega t}+C_2e^{-\omega t}$

Since, the boundary condition in our case says that at $t=0$, $r=r_0$

$\implies r_0=C_1+C_2$

Also at $t=0$, the points on the wire have no radial velocity since the centripetal acceleration hasn't had any time to generate any velocity. Thus,

$r^{'}=C_1-C_2=0\:\implies\:C_1=C_2$

$\implies\:r_0=2C_1=2C_2$

$\implies\:C_1=C_2=\frac{r_0}{2}$

$\implies\:r=\frac{r_0}{2}(e^{\omega t}+e^{-\omega t})$

$\implies\:\boxed{r=r_0 cosh\:\omega t}$ $\:$($\because \frac {e^{\omega t}+e^{-\omega t}}{2}=cosh\: \omega t$) $Eq^n (2)$

Now we can get to where we started from and simply write angular momentum as a function of time.

$\boxed{L=m\omega [r_0 cosh\:\omega t]^2}$ (Ans)

The better but more complex perspective is to consider that starting from the beginning of motion, the bead keeps acquiring a resultant velocity that is not completely tangential. Look at diagram 2 again and try to understand that when the bead gets to the radial direction at $t=t_2$, it still retains a part each from $\vec{v_{\theta_0}}$ and $\vec{v_{\theta_1}}$. And the vector sum of the parts of $\vec{v_{\theta_0}}$ and $\vec{v_{\theta_1}}$ lies at an angle to the normal contact force at this instant. It also has a radial component. So, basically, it is the varying normal contact force that feeds the radially outward velocity of the bead. The question doesn't contain any information about the contact force, hence we cannot solve the problem directly using Newton's second law of motion. However, we still can solve the problem using the work-energy principle, since all the forces in this system are conservative. But this requires a different method, called the Lagrangian Mechanics which is basically an advanced reformulation of the Newtonian Mechanics. We shall discuss this in the end.

Now that we understand the problem and have solved it from the ground frame, the other two methods, **(b)** and **(c)** should come easy.

**(b) from the rotating frame**

Since the rotating frame involves a constant $\omega$, we can apply a pseudo acceleration called the centrifugal acceleration opposite to the centripetal acceleration as observed in the ground frame. This will make the rotating frame inertial while we are inside it and make it appear to us as if the bead is accelerating radially outward at $\omega^2 r$. Just like **(a)**, we have the same equation and we can solve it in the same way to find the same answer.

**(c) with Lagrangian**

Finally, let's bring in the tool we call **Lagrangian** to solve the problem again from the ground frame. Applying Lagrangian to this particular problem is neither necessary nor makes it easier to solve the problem. But it demonstrates the validity of the concept that the Lagrangian of any mechanical system contains all the necessary information about the system and by using the proper Lagrange equation we can find any information about the evolution of the system.

To denote the Lagrangian we are going to use a scripted $\mathscr{L}$ so that we don't get confused with angular momentum. The definition of Lagrangian for any mechanical system in simple words is the difference between the kinetic energy ($T$) and the potential energy ($V$). Why this is so is something that we can understand when we discuss Lagrangian in another dedicated post about it. Till then it should suffice to say that for complicated mechanical systems, the total mechanical energy ($T+V$) comes less handy than the Lagrangian ($\mathscr{L}=T-V$).

With all the information we already know, let us write the full expression of the Lagrangian and the Lagrange equation for our problem.

$\mathscr{L}=T-V$

Since at any point in time, the bead will have a radial velocity and a tangential velocity, we write down the kinetic energy as the sum of the two components as:

$$T=\frac{1}{2}m{v_r}^{2}+\frac{1}{2}m{v_\theta}^{2}$$

$$\implies T=\frac{1}{2}mr^{'2}+\frac{1}{2}mr^2\omega^{2}$$

Since we are ignoring gravity and there is no other restoring force in the system, there is no potential energy. Thus $V=0$ for all time. Hence, we could write the Lagrangian for the system as:

$\boxed{\mathscr{L}=T=\frac{1}{2}mr^{'2}+\frac{1}{2}m\omega^{2}r^2}$

Now, we have to formulate the Lagrange equation for our system. Without the Lagrange equation (also called as Euler-Lagrange equation in this context) the Lagrangian is useless. It's like having a chest full of diamonds without the key to open it. So, what is the Lagrange equation for this system?

Before we form the Lagrange equation, we need to understand the role of the wire. The wire constrains the movement of the bead along the normal direction but offers no resistance along the wire. Thus, the velocity of the bead at any time can be uniquely determined separately by both $r$ and $\theta$ (refer diagram 2). Thus there is a relation between the position variables $r$ and $\theta$ and we can express that relation as:

$$\boxed{f(r,\theta)=0}$$

Such a constraint is an example of what is called "Holonomic". Since, $\theta=\omega t$ and $r$ can be shown as a function of $\theta$, time $t$ becomes an explicit variable. This makes the system "Rheonomous". In such a system, we can write the Lagrange equation using any of the position variables $r$ or $\theta$ to find the exact answer. Since it is easier to visualize the change of kinetic energy of the bead with the varying radial positions, we write the Lagrange equation using $r$. If you want, you can also try $\theta$ and post your solution in the comments below or in the General Physics Forum for a review.

Now, let's write the Lagrange equation for our system.

$$\boxed{\frac{d}{dt}(\frac{\partial\mathscr{L}}{\partial r^{'}})-\frac{\partial\mathscr{L}}{\partial r}=0}$$ $Eq^n (3)$

Now, let's solve each term on the lhs.

$$\frac{\partial\mathscr{L}}{\partial r^{'}}=mr^{'}$$ $Eq^n (4)$

$$\implies\frac{d}{dt}(\frac{\partial\mathscr{L}}{\partial r^{'}})=mr^{''}$$ $Eq^n (5)$

$$\frac{\partial\mathscr{L}}{\partial r}=m\omega^{2}r$$ $Eq^n (6)$

From $Eq^n (5)$ and $Eq^n (6)$, after cancelling $m$ from both the terms on lhs, we can write the final differential equation of motion as:

$$\boxed{r^{''}-\omega^{2}r=0}$$

which is $Eq^n (1)$ and we have already solved it in **(a)**.

**This concludes the lecture. As you might have already sensed, there is a fourth method too. We can use the Lagrangian in the rotating frame. But to formulate the Lagrangian and the Lagrange equation for the rotating frame, we have to apply proper transformation and that is a topic for another post dedicated to the Lagrangian mechanics.**

If you prefer watching over reading, you can also watch the video lecture below.

If you still have any doubts, or if you find any mistakes, please mention in the comment section below. You can also use the General Physics Forum on ThePhysicist to post your questions and rest assured, ThePhysicist is always hungry for problems to solve.** **

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]]>ThePhysicist was initially planned to be a Physics blog that aimed to provide everything in Physics to everyone in Physics. To achieve the vision, ThePhysicist has been expanded from the idea of being only a blog to a full fledged website with content and features for those who are interested in Physics, viz, enthusiasts, students and researchers. ThePhysicist aims to establish continuity among the levels of its audience through well crafted posts, materials, classes, news and research tools. The aspiration here is to provide a single platform to educators, learners and researchers in Science in general and Physics in specificity. Not only that the content on ThePhysicist shall be unique and empowering for the readers, subscribers and visitors, but also that if some new, interesting and useful content is available elsewhere on the internet, it finds a special mention on ThePhysicist.

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