I am Ambar from India presently in grade 12. Consider the bullet- rod system. Since no external net torque acts on the system, we may simply apply conservation of angular momentum. Let the reference frame in which the velocity of bullet and angular velocity of the rod is measured by called the reference frame of ground observer. Hence we have, mvL/2 = mv' L/2 + (ML^2)/12w Where m is mass of bullet, v=250m/s is velocity with which the bullet approaches the rod w.r.t. ground observer and v'= 160m/s is velocity with which the bullet moves after collision with the rod w.r.t. ground observer, M is mass of rod, r=0.25m is perpendicular distance of the direction between velocity of bullet and the pivot, L= 1m is length of rod and w is angular velocity of the rod with respect to the ground observer (assuming collision is not perfectly inelastic, ie. the bullet doesn't stick to the rod). Hence on calculating this comes out angular velocity of the rod with respect to the ground observer will be 2.70 rad/s.
Hello Sir, I am writing from India. Your lectures made me love Physics. I still remember your quote, 'Teachers who makes Physics boring are criminals.' You are the inspiration of uncountable students. Thanks for everything sir! Please give reply to my comment 🙏
Sir , I am Praveen ,from new Delhi in India, sir , I can't tell you that you and yours lectures how much important for me. I want to be a good physicist in my life and sir i believe without yours lectures it is nearly impossible to be a good physicist. Thanks so much sir from bottom of my heart and may God healthy and live long you.
Initial momentum of the bullet is .75 N-S after the collision, the momentum of the bullet is .48 N-S so the bullet transferred .27 N-S of momentum to the rod. I'm going to choose a coordinate system where the initial angular momentum of the system is zero. So, .27 N-S of momentum has been transferred to the center of mass of the rod, which is .25 meters below the coordinate system chosen of zero initial angular momentum. Since the total angular momentum of the system must be zero, and the moment of inertia of the rod is .02499, the rod must rotate clockwise with an angular velocity of 2.7 Radians per second.
I can tell you one thing that I'm not as smart as this level question it took me 10 years just to learn to solve a rubiks cube so I'm mostly watching this channel for entertainment but I still find it interesting
Hello , and good day sir. I am a student from India, an aspiring phycist. I am currently in the 9th grade. I am watching your lectures on classical mechanics (8.01) and am nearly all the way through it. I thank you for uploading some of the greatest lectures of all time onto your channel. I have just one enquiry to make. In your early career in astrophysics, at any point did you feel that what you were doing was tedious/ unnecessary, or did your love for physics stay the same throughout your life. Yours truly, Someone from somewhere
Respected lewin sir...I 'm from india....lots of love to you❤... enjoying studying physics from You sir...❤...Before Physics is a big headache for me...but after Learning from ur lectures...its awesome💯...And ur quotes teacher who makes physics boring are criminals...😂...so true sir❤...U are the one biggest reason for Many Showering Love Towards physics❤...I wish You Good health nd Requesting You to GUIDE us forever💯🤝❤️
Hello sir, I've been watching your lectures since this march and have seen a lot change in my way of thinking which i am very thankful of, but i have a problem. Sir, whenever i solve the numericals of any topic, i am able to solve the easy ones(the formula based ones) but even for the moderate ones,it takes a an hour for solving a particular numerical, sometimes i am even unable to solve it even after an hour. Sir plz suggest me anything that you think i should do to improve. I am in class 12 from india. Love your videos and want to meet you once❤.
Professor,I would greatly appreciate it if you could share insights on the secrets behind the longevity often observed in Jewish communities. I look forward to your response and will keep commenting on your videos until you address this (please avoid common answers like eating yogurt).Thank you!
1. Length of the rod l=1m; mass of the rod mr=300g=3e-1kg; mass of the bullet mb = 3g =3e-3kg; initial velocity of the bullet vo=250m/s; final velocity of the bulett vf=160m/s. We need to estimate angular velocity of the rod : ω 2. Change of kinetic energy of the bullet K=mb*(vo^2-vf^2)/2; (1) 3. Moment of inertia of the rod with regards to its pivot point in the middle I= mr*l^2/12 ; (2) 4. Assuming there is no heat dissipation or deformation mentioned all change of kinetic energy will be transformed into rod rotation K=I*ω^2/2; (3) 5. from (1) & (2): mb*(vo^2-vf^2)/2 = I*ω^2/2 ω = sqrt(mb*(vo^2-vf^2)/I) = sqrt(mb*(vo^2-vf^2)/(mr*l^2/12)) = sqrt(3e-3*(250^2-160^2)/3e-1*1^2/12)) = 6*sqrt(123) = 66.543 rad/s = 66.543 s^(-1)
Sir I'm feeling like giving up in my life. I'm JEE aspirant but my preparation is not going good and I'm in depression. I don't know what to do I want to do better in life but how? Please help me sir 🙏
Khan, I'm not a talented JEE aspirant like you or the professor, but I want to tell you, your life is worth so much more than a silly entrance exam, I want you to make a long term goal that doesn't depend on whether you get into a high class college, because you know, education is unfortunately more of a business than a right, especially here in south asia. There are so many ways to prove yourself to the world, through internships, through projects, through social media, through connections, through simply working in teams, even having confidence in your skills can enable you to pull off a interview and land a high paying job even if you dont come from a high class college . and these few things I stated here show people around you what you're able to do. Its okay to feel lost, confused and directionless. But I still think you will make it, please don't worry your head off if you don't though. I can see you are trying so hard and you are doing so well too, don't give up, hope Is with you, it follows you through family and friends, the world is more compassionate than you think. ❤
Work on what you want. Please. You might have to settle for less than passing the JEE exam but it'll still be good and you'll succeed in life. Someone passing the exam with ease and not working hard on what he wants will fail in life. See Walter teaching ? If he was not a professor he would still be teaching. No doubt.
A question about preparing for a lecture: What criteria did you use to select the experiments that were then shown in the lecture? How many hours per week did you havt to give lectures and how much time did you spend preparing? Unfortunately, I always feel a certain amount of time pressure when teaching. ;-)
What are my lecture secrets? A typical preparation time for one 50 min lecture is 60-80 hr. I do not write a script. But I dry run all my lectures 3 times before I give the lecture. 2 weeks before, 1 week before and at 6 AM of the day of the lecture. Every 5 min I have a LARGE time mark in my lecture notes (starting with 50 min, going down to zero). I have a large digital clock on the lecture desk which is set at 50 min when I start the lecture and it is counting down. Thus at ANY moment during my lecture I know within about 1 minute how much time I have left and how much time is needed to finish. Thus I NEVER overrun, I never have to do an interesting demo in haste with a sloppy explanation. I typically finish my lectures within 1 minute of 50 min. Of course during the first dry run, it ALWAYS takes near 60-70 min. I then have to do careful surgery as changes are needed. By the time it is 6 AM in the morning of the day I give the lecture I am always well synchronized between my time marks and the digital clock. My lectures have therefore become performances (like an actor on the stage). my secrets are: *imagination, go outside the box with examples and with demos. I do often demos that will make them sit on the edge of their seats. I make the students laugh at times, make them cry, make them stop breathing, even make them wet their pants at times. I confront them as much as possible with experiences in their own lives. Where possible I include a student in a demo. I also do demos in which I take risks. I never become boring. I make them look through the equations to make the eqs come alive. I always keep their attention with my clarity, sense of humor and my LOVE for Physics. I radiate my enthusiasm - it's contagious. I can make ANYONE love physics and one of my "famous" quotes is: "if you hate physics it's only because you had a bad teacher".*
@@lecturesbywalterlewin.they9259 Thank you very much for your detailed answer. I see that there are many parallels between us in our understanding of what is important for good physics teaching. But I am widely open for your hints which I will take into account for my teaching in the future.
Hello sir, I have a question, what points should I pay attention to in order to make a Maxwell wheel? 1. What kind of thread should it be? 2. What kind of wheel should it be?
Using Conservation of Angular momentum, m(bullet = 0.003 Kg) X (250 - 160) X 0.25 = M(rod mass = 0.3 Kg) x (length of rod)1^2/12 x w(angular velocity).... gives w = 2.7 rad/sec
Dear Professor Lewin, I am a 60 year old physicist from Germany and was in favour of teaching my whole life, but worked as a quality manager for decades. Now, 3 years ago, a dream became reality and I became a high school teacher for Physics and Mathematics. I adore your style of teaching close to the students. Thank you very much for your wonderful work in teaching Physics. Just a practical question: When writing on the black board sometimes you are using Physical units and sometimes not. Is this by coincidence or is there any rule behind?
I've one doubt in circular motion...I couldn't find answr.In non-uniform circular motion,does magnitude of centripetal acceleration change? Why we find tangential acceleration by( dv/dt) unlike centripetal which is v²/r.... Sry for my theories on black hole mass
black hole mass, M=Rc^2/2G, R is the radius of the event horizon. I derive it in this lecture ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-zG-yjmKN1xM.html
Sir do you teach the students on online. If you teach then could you please tell me the name of tge website or app .Where you are available.I need your help .If you give me the reply then I will really gratitude to you
Dear High Master Professor Lewin It would be wonderfull to think about the light and a way to reuse it's energy I'm sure there is something possible if using the light and the Watt Machine process Maybe a project WITH all indian very clever scientists of your Channel could participate and discuss and argument about this project What do you think ? Best regards
Let: v1= 250m/s Initial velocity of bullet v2= 160m/s Final velocity of bullet mb= 3g mass of the bullet mr= 300g mass of the rod L=1m lenght of rod r=0,25m distance of impact to pivot I=mrL²/12 moment of inertia Δp = mb(v1-v2) then r x Δp = Iw w = (r x Δp)/I = (0,25 ⋅ 3 ⋅ 90)/(300/12) = 2.7 RAD/s
SOLUTION: ω=2.7 rad/s Conservation of angular momentum: initial angular momentum is: L_initial= m (L/4) v^2_i (Eq.1) and L_final= L_bullet + L_rod = m(L/4) v^2_final + (1/12)ML^2 ω (Eq.2) Equating (Eq.1)=(Eq.2) we can solve for ω. I get Ω= 3 m (v_i-v_f) /(ML) And plugging in the number I get the 2.7 radians per second.
@@lecturesbywalterlewin.they9259 Absolutely! Both religions have immense similarities like they are based on the concept of peace,have their origin in India By the way I respect you sir because you are a World War 2 survivor🙏🙏 Have a healthy life!
Solution: ω = 2.7 rad/s Supporting calculations: Assign the following terms: m = bullet mass, M = rod mass, Ɪ = moment of inertia of rod ℓ = rod length, i & f are subscripts for initial and final. v & ω are linear and angular velocity respectively. Initial angular momentum (Li): Li = m*vi*ℓ/4 Final angular momentum: Lf = m*vf*ℓ/4 + Ɪ*ω Li = Lf for conservation of angular momentum, thus: m*vi*ℓ/4 = m*vf*ℓ/4 + Ɪ*ω Solve for ω: ω = m*ℓ/4 * (vi - vf)/Ɪ For a thin/uniform rod about its center: Ɪ = 1/12*M*ℓ^2 Thus: ω = 3*m * (vi - vf)/(M*ℓ) ω = 2.7 rad/s
Cygnus X-1 as approximately 20-21 solar masses. My foemula: Black Hole Mass Formula (MBHMF)_ MBHMF = (c^2 x R) / (1.95 x G) Where: - _c_ = speed of light (approximately 299,792,458 m/s) - _R_ = radius of the event horizon (in meters) - _G_ = gravitational constant (approximately 6.67408e-11 N*m^2/kg^2) Using the same event horizon radius for Cygnus X-1: _Cygnus X-1 Black Hole_ - _R_ (Event horizon radius): approximately 24.4 km (24,400 meters) Plugging in the values: MBHMF = (299,792,458 m/s^2 x 24,400 m) / (1.95 x 6.67408e-11 N*m^2/kg^2) MBHMF ≈ 20.6 solar masses (M) The result is approximately 20.6 solar masses, which is within our target range of 20-21 solar masses for Cygnus X-1. Pls also see black hole formula in previoys video which u termed nonsense....someone even proved it true in reply section by putting values....this is my own but
New answer : the bullet energy loss which is thus transfered to the rod : Er=0.5m(vi^2-vo^2)=0.5x0.003(250^2-160^2)=55.35J. This energy is transfered to the rod center of gravity for one end which is at 0.25m from the pivot point. That is also where the bullet hits but that is not relevant, as long as the bullet does not hit the pivot center. Both ends of the rod are set in motion so we can add both ends and total mass is still 0.3kg. So, the energy converts to speed at 0.25m from the center, rewriting the formula : vr=sqrt((2E)/mr)=sqrt ((2x55.35)/0.3)=19.2094 m/s. The circumference of this point O=2PIr=2xPIx0.25=1,5708 m. So angular velicity w=2xPIxvr/O=2×PIx19.2094/1.5708=76.84 rad/s. Probably a lot better than my initial answer.
Sorry to say but I think friction has done work as a result bullet's velocity changes and friction is non conservative so u can't apply energy conservation. Hope u got my point thank u❤
@@MaheshKumar-lx1ku excellent point. In a real situation this certainly is true. However, if friction is not ignored i think the problem cannot be solved. Conservation of momentum cannot be applied because the bullet has lineair momentum and the rod has angular momentum. I could well be wrong about it. I'll give it some thought. Thanks for the comment.
i'm not sure yet, but I'm thinking that angular momentum is conserved while kinetic energy is not. If this is true, I think the angular velocity of the rod after the collision is 2.7 radians/second. Once you know this, you can calculate the total dissipated energy if you wish, as the initial bullet KE minus remaining bullet KE minus the rod rotational KE.
Sirrr im back sir, can you mentor me please i wanna crack NEET UG and it was my dream from my childhood and it is , but the biggest obstacle i always face is that what to learn whenever i start learning, it never ends and take too much time for every topic and i dont know what what to learn like is it that big or not sir? And i saw ur video and u said lectures for physics i will learn from them promise , and what about biology and chemistry what topics and from which teacher should i learn bio and chem, im dedicating my journey of the neet to you please mentor me i will follow it, until i join it and until i die
Angular momentum before collision: L₁ = mv₁(ℓ/4) + I ω₁ ω₁ = 0 => L₁ = mv₁(ℓ/4) where ℓ is the length of the rod, m is the mass of the bullet and v₁ is the velocity of the bullet before collision. Angular momentum after collision: L₂ = mv₂(ℓ/4) + I ω₂ where v₂ is the velocity of the bullet after collision and ω₂ is the angular velocity of the rod after collision. I is the moment of inertia of the rod. I = (1/12)Mℓ² Conservation of angular momentum : L₁ = L₂ => mv₁(ℓ/4) = mv₂(ℓ/4) + (1/12)Mℓ²ω₂ => ω₂ = 3m(v₁ - v₂)/(Mℓ) = 2.7 rad/s
@@lecturesbywalterlewin.they9259 but I'm in 10th grade I'm unable to understand :( and sir please teach me how be good in physics u are a legend!! Master!! Of physics
Variația energiei cinetice a glonțului este egală cu lucrul mecanic ( energia pierdută de glonț și transferată la bară) L= Ec1-Ec2= Ic•( ( omega)^2)/2 Ic= M• l^2/12 Ec1-Ec2=m( v1^2-v2^2)/2 Din aceste 3 ecuații va rezulta valoarea lui omega!
Asteroid Blast Impact (ABI) formul created by me: ABI = (E_k × ρ × D^3 × π × τ) / (σ × δ × λ^2) Where: - ABI = Asteroid Blast Impact (energy released) - E_k = Kinetic energy of the asteroid (based on mass and velocity) - ρ = Density of the asteroid - D = Diameter of the asteroid - π = Mathematical constant pi - τ = Tau constant (~6.283) - σ = Surface tension of the impacted material - δ = Depth of penetration - λ = Wavelength of the shockwave (related to asteroid size and speed)
Let's apply Asteroid Blast Impact (ABI) formula to a real asteroid impact event. Chelyabinsk asteroid (2013): - E_k = 0.5 × 13,000,000 kg × (19.16 km/s)^2 = 1.23 × 10^17 J (Kinetic energy) - ρ = 3,000 kg/m^3 (Density, estimated) - D = 20 m (Diameter, estimated) - π = 3.14159 - τ = 6.283 - σ = 72 mN/m (Surface tension of air, approximate) - δ = 1 m (Depth of penetration, rough estimate) - λ = 10 m (Wavelength of shockwave, rough estimate) Substituting values into your ABI formula: ABI = (E_k × ρ × D^3 × π × τ) / (σ × δ × λ^2) ABI ≈ (1.23 × 10^17 J × 3,000 kg/m^3 × (20 m)^3 × 3.14159 × 6.283) / (72 mN/m × 1 m × (10 m)^2) ABI ≈ 4.47 × 10^16 J This estimate is close to the reported energy release of the Chelyabinsk asteroid explosion: - ABI ≈ 4.5 × 10^16 J (Estimated energy release from observations)
I would like to make a 2nd attempt stupid, ridiculous, amateur, based on primitive common sense: the bullet flying straight forward strikes the motionless rod. Afterwards the both objects are moving TOGETHER for extremely short time. I would like to state that tangential velocity (Vt) of the strike-point of the rod (l/4 from the pivot axle) is equal to the final linear velocity of the bullet. It cannot be lower as the bullet is pushing the rod in front of itself and has no reason to accelerate when the contact is broken, it cannot be higher as has no reason for such behavior. So, immediately after the collision ω = Vf/ls=Vf/(l/4) = (160 m/s) / (0.25m) = 640 rad/s
Professor,I would greatly appreciate it if you could share insights on the secrets behind the longevity often observed in Jewish communities. I look forward to your response and will keep commenting on your videos until you address this (please avoid common answers like eating yogurt).Thank you!
