 वल्क्म डूस्तो, मैं यह आपे दिरज संटम एकट्मी की तरफ से आपका स्ववगत करता हूँ आजका वीडियो जो है आंगलर इंपल्स के बारे में होने वाला है ये एक अचा तोपिक है डूस्तो, जो की इगनोर हो जाता है जब हम और बीडी चाप्टर कर रहे होते हैं, और ये हम बाद में फाँईंडाउड करते हैं दूस्तो जब कुछ एसे सवाल जो की कनजरवेशन अप लीनेर मोमेंटम, आंगलर मोमेंटम, तोरक एकवेशन कीज़े भी नहीं होते हैं जो की ईंटिग्रल आप डवीटी लोग की मुमेंटम, उसी भी चीच बदे, झो की वंठीगरल अप ड़ावाईंटाउड की मेंगाई मुमेंटम के भराभर होती है। आप कुछ चेजे और हैं जो में दियान में रहाएगी जैसे कि आगर अप खुज आप लीन्ँर मूमेंटम लगाते हैं तो जो जो आप को ये छींज दियानदनी परती है कि एक स्तरनल फोर्स ना लगे ये प्राइट तो यहां पर इस तरागा कोई रुस्ट्चन नहीं आप � तो यहांप पर अंगलर इंपलस एकवेशन मे भी इस तरेगा खोई रिस्टिक्छन नहीं है लेकिन दोस तो जब आप लीनेर मोमेंटम यह आंगलर मोमेंटम की जो एकवेशन है खंजरवेख जब आप खंजर परत हैं, लीनेर मोमेंटम और अंगलर मोंटम को तब आप एक से ज़ादा ब्टीग के बारे मे बात कर रहे होते हैं, जैसे की एक आप जडग का लीनेर मोमेंटम पलस दुसरे आपजग का लीनेर मोमेंटम बहुर दह कोलीजन is equal to sum of the linear momentum of both the objects after the collision तो उसी तरह से हम लोग आंगला मुमेंटम के बारे में भी कनजर्ट करते हैं तो before the collision, जो भी आंगला मुमेंटम है, system का, after the collision से हम equate कर देते हैं तो in both theorums में, हम लोग एक से जादा बोडी का एक साथ equation लिकते हैं अप आप आप आप पह यहता लग conjugation of an object लिकते हैं शहागा। तो आप सर्फ आप सर्फ एक अब्ध्यक लिकते हैं At a time. उस अब्ध्यक में जो लीनियर इमपलस रहा थे एक आब अबज्यक के चेंझें मुमेंटम के लोगखा। change in momentum के बरावर होगा similarly उस अबजेक में जो angular impulse है वो उस अबजेक के change in angular momentum के बरावर होगा alright अब दूस तो एक चीज शाएड आप को परेशान कर रही होगी कि बही ये सब तो थीख है equation भी बडया है लेकिन ये integral हम निकालेंगे कैसे integral fdT अब अब अब निकालेंगे कैसे निकालेंगे तो दोस तो इसका समवादान ये है कि जब आप equation लिखेंगे definitely जब impulse लग रहा होगा तो कोई ना कोई impulse लगा ही रहा होगा तो ये जो impulse equation अब ये आप एक से जादा body के बारे में लिखेंगे एक एक करके और ये integral term डो तीन equation में एक साथ हैगा और जब आप उन डो तीन equation को सुल करेंगे इस integral term को you can get rid of अब for example मान लिखें की आपके पास एक equation इस तरह से आती है integral fdT आपके पास मान लिखें आती है 5 और दूस्रा equation मैं एक रैंटम एक साथ लेगा आपे 2 एक साथ मालेजे 6- integral fdT is equal to some... some thing... some constant letter से A तो आप फॉस्ट एक साथ एक अप शुप्शूर कर सकते है और आपको value of A यहापे मिल जाएगी तो इसी तरे से हम लोग रिलाए गरते हैं, की एक से जादा एकवेशन में यें इंटिगल तम आएगा, और हम इस इंटिगल तम को जो है एलिमनेट कर देंगेग. अके, तो ये तो हो गए दूस्तो, how to get rid of this इंटिगल term. और लास्त बट नोट दे लीस्त अगरी बात ये है दूस्तो, की जब हम लिन्यर इंपल्स की एकवेशन लिकते हैं, वो हम किसी भी डारेक्षन में लिक सकते हैं. All right. अगर भी बट लोग उस डारेक्षन में लिखेंगे, जिस दारेक्षन के बारे में, या तो हमें बट स़ी चीजे पता हैं, या तो हमें उस डारेक्षन के बारे में चीजें निकालनी हैं. Right. And, इस को लिखें गे. So, this equation, writing this equation, is most comfortable about the center of mass excess. अभ आप इस यह एक सेठाए, अब उब वह थे आप आप लग भी में साचाछ लग़ भी ख़ब में रवाथ पुदतना वेंँ, अब आप प किसाचक बग़ूँ after discussing this theory, we will try to ask a couple of questions here, and we will know in which way we will be able to solve the problem and how we will be able to solve the problem and how we will be able to solve the problem after discussing this theory, we will try to ask a couple of questions here, and we will know in which way we will be able to use this concept in multiple ways let's go let's go, friends, let's apply what we have read so far here is the first question in front of you here I will explain you a little situation you have two platforms these two platforms are fixed in their place and this surface is smooth and this surface is also smooth now you are seeing a cylinder in the middle this cylinder is about the horizontal axis which is going inside your screen this cylinder is free to rotate and there is friction between the small m and the cylinder and the other things that are given is that when this small m goes to the other platform it goes from platform 1 to 2 its velocity changes from u and we have to remove the angular velocity of the cylinder so I hope you have understood the question so let's see how we will solve this problem now here you must be seeing that there is some interaction between the small m and the cylinder now if there is interaction then what comes to our mind first our mind comes to the conservation of linear momentum now here we cannot do linear momentum what is the reason for this, friends the reason for this is that the axis of the cylinder will generate an external force when it is interacting with the small m cylinder so that is why we cannot conserve the small m and the capital M's linear momentum now the linear momentum cannot be conserved can the energy be conserved you will find that the energy will not be conserved because the small m and the capital M are sliding so some energy will be converted to heat so neither we can conserve the linear momentum nor can we conserve the energy now what is left you have the angular momentum conservation can we conserve that now you will see that here this axis this axis is generating an external force and that external force will pass through this axis alright and when this small m is interacting with this small m is interacting with the cylinder so there will be a frictional force between these two along with the gravitational force so here you will get an idea that we can think of conserving the angular momentum about the axis but here we will try to solve this question with the angular impulse alright so you will get that when you try to solve the angular impulse so you will be in control of the solution you will be in confidence because when you use a theorem directly like you will use conservation of angular momentum so you will always have this doubt whether my condition is satisfying or not but when you write angular impulse you do not have any restriction that this will happen only when we can use it so when you use angular impulse you will have more confidence so let us keep this in mind first of all let us make a free body diagram of small m this is small m alright this small m when it is interacting with capital M then it will generate a frictional force assume that the value of that frictional force is small f a gravity force will also be there mg and a normal force will be generated alright so vertically something is not changing so normal reaction here will be equal to mg whatever is changing horizontally so I am going to write a linear impulse equation so let us carefully see what is the equation of linear impulse integral f dt is equal to change in the linear momentum final linear momentum minus initial linear momentum ok when you are writing this equation keep in mind that you are talking about vectors here force is also a vector and linear momentum is also a vector so you have to take care of the science like in this situation I have assumed that forward direction is positive so when you are talking about impulse in this problem then f is minus f so minus of integral small f dt this will be your final linear momentum which is small m into v minus initial linear momentum which again is in positive direction minus of m into u this is my first equation ok now this term has come here integral of f dt now I have not given f as a function of time so I will have to see if I have any other equation in which this integral f dt term will come its answer is yes because whenever there is any force its reaction will develop so who is putting this small f is putting a cylinder so when we will make a freeway diagram of the cylinder then this small f will also be there so let's see how we will deal with the cylinder on the cylinder in opposite direction a reaction will develop small f ok this is the mass of the cylinder mass m and radius r and this is its angular velocity now the problem here is that the axis of the cylinder that axis will also develop external force ok because of which when you write the linear impulse equation then the force of the axis is cx direction so this cx will also come the force of the axis in the horizontal direction so that equation will be useless for me because I will write that equation and there will be another variable which is unrelated to the rest of the equation so I will save it so can I write something else here apart from the linear impulse now you will notice here that if I if I try to take out the torque of the force then only the torque of the friction will be there so here we will write the equation of angular impulse what is the equation of angular impulse torque dt is equal to final angular momentum minus initial angular momentum ok here also we will have to take care of the directions and how we take care of the direction of angular variables there are many ways the simplest way is to consider clockwise or anticlockwise as positive and the other direction will become negative automatically ok so I believe that the direction of the torque of F is positive so what is the torque of F? clockwise alright so torque here is perpendicular to F R so R into F into dt this will be my angular impulse left hand side this is equal to final angular momentum I am writing this torque about centre of mass so my impulse equation about centre of mass axis so LF will be moment of inertia which is mR square by 2 cylinder into omega minus initial angular momentum which is 0 now take R out of the integral because friends F dt we have to replace is it not mR square by 2 into omega this is my second equation now you will see from the first equation the value of F dt will be m times u minus v now you just have to do a last step the value of F dt here keep it the second equation of F dt and you will get the value of angular velocity omega so I hope you can understand how we have the angular momentum concept here we have used it very easily now we will ask one more question after that your concept will be very solid like a rigid body so this question I will tell you and I will make a diagram so the situation here is something like this that there is a sphere like this so here this sphere is very small in size it is very small ok so this sphere this sphere believe me it is coming down it is about to touch the surface at that time I have given its velocity like this its velocity this velocity is u and friends its angular velocity is given like this believe me it is omega knot ok and after the collision it is something like this believe me it collides at an angle theta ok after the collision it is bouncing like this it is bouncing it reaches here in a parabolic path friends and after reaching here what you get to see is that it retraces its path after the collision ok so you have to remove that if so you have to remove that when this path is retracing this variable your u omega knot theta how it is connected with each other ok so basically here I am trying to know from you friends how you how you will use angular impulse or use linear impulse alright so first of all because of the symmetry because of the symmetry of the problem when this ball will hit here friends so here also its velocity should be u ok and angular velocity omega knot should be ok why it should be like this because clearly here there is no energy loss on this path again and again this mass m spherical mass m is repeating its path so this way will be the situation ok and here you will see that on this location on this location angular velocity is anticlockwise and here you will see that angular velocity has been reversed ok so all this needs for symmetry otherwise this will not be like this ok so let's assume that this is your spherical mass m this is going to collide ok when it collides with this surface its velocity is in this direction u so what will be the component in the vertical and horizontal component if I take the velocity then the vertical direction velocity will be u sin theta just before the collision and horizontal component will be u cos theta ok now I am representing the velocity here with white color and with yellow color I will represent force here one force will be gravity here mg force one here will be normal reaction ok and the other force will be friction here and when the friction will be there let's assume that the coefficient of friction is mu ok I am not with clarity in the beginning I have been able to tell you because my focus here angular momentum impulse and linear impulse these two equations are to be written ok so let's assume when colliding then on this object on this sphere will be static friction or kinetic friction will be there at this point when it will hit friends you will see that there is some definite velocity tangentially and the surface is on the rest then there will be kinetic friction because it is sliding there is some instantaneous sliding so in this direction the friction will be that kinetic friction that is mu into n friends all these forces will be there alright so if we just look at the ball then the vertical impulse is there and the horizontal impulse is there friends if I see the impulse equation just see impulse equation let's mark it here y axis and x axis if I see this equation on y axis then one thing you always keep in mind when collision happens and this kind of impact is happening then normal reaction is much more than gravity so you can ignore the gravity force ok so the sudden change in momentum will be due to normal reaction just before and after collision so we will only talk about this impulse if we are talking about sudden after collision so I have taken the upward direction with positive hence n dt is a positive quantity so this is equal to momentum final what will be the final momentum immediately after collision everything should be reversed only then it will come back after collision from the other side ok so to reverse velocity will be reversed and angular velocity which was anticlockwise now it will be clockwise ok sorry this is my linear impulse friends it has nothing to do with angular velocity it has nothing to do with linear velocity so linear velocity in upward direction u sin theta was coming down now it is the same velocity so this is equal to mass times minus velocity which is u sin theta minus m into negative velocity final momentum minus initial momentum so this is equal to 2 mu sin theta friends ok and if you look horizontally then it will come to you integral of mu n dt it will come to you m into same logic if you put 2 mu cos theta this comes to you so with these two equations you can in fact eliminate n dt ok ok now friends this sphere it is rotating an angular impulse is coming here friends to get an angular impulse here we have to see torque about center of mass so about the center of mass angular impulse equation which is integral tau dt final angular momentum minus initial angular momentum how will we write angular momentum change is equal to torque torque is mu into normal reaction into r radius dt ok this is equal to the moment of finish angular momentum about the center of mass what happens i center of mass into omega this is what happens angular momentum about the center of mass icm solid sphere is done 2 by 5 m r square so that into omega the final omega is minus initial omega ok so i take out mu and r so this is my equation friends 4 m r square by 5 so you will see you have 3 equations in one equation in one equation the value of integral entity in one equation same integral entity is 2 mu cos theta by mu and same entity which is sorry i left omega knot same entity here 4 m r square omega knot by 5 mu r so if we have to retrace this then we will write its condition 2 mu sin theta should be equal to 2 mu cos theta by mu and that should also be equal to 4 m r square omega knot by 5 mu r so these 3 are equal then this situation is possible so i hope you have got enough information about angular impulse i will suggest that you should ask yourself at least 10 to 15 questions so that your concept becomes solid thanks for watching this is diraj from centrum academy signing off bye for now