 What we have done in the last class We did torque equations Torque equation we have done You are able to see my screen? Yes, sir Great And what else? In the torque equation what we have done? Sir, how long will the link for tomorrow's exam be open sir? Somebody said something Can you speak a bit loud? Rona, please tell everyone what we have done last class Properly tell what all things we have done in 4 hours in the last class Rona, are you able to speak? Yes, you can Skanda, skanda, can you explain everyone what we have done? Skanda left Tripan Sir, you taught us about torque And then You taught us about T is equal to I alpha Not questions based on that equation Okay Alright, so today we are going to discuss about The work energy on the rigid body Alright? Yeah, please write down Sir, can you solve some questions Yes, I am going to do that But right now let us start with this concept We have done couple of questions on torque equation last class as well So let us complete the theory part of this chapter Then we will take up a lot of other questions as well Write down work energy on the rigid body Covering the book You can keep it up on this also, more differences are made But your elbow has to be Elbow? What do you need it again? No, it's okay Check if your elbow is in Especially when you are writing at the bottom of the book When you are writing at the bottom of the book, you have to move it up like this and then keep it up Yeah, your elbow Please don't talk in the background, okay? If you want to talk, please put yourself on mute Fine, so we are going to discuss how can we use work energy concept on a rigid body Alright? We have learned in the work power energy chapter This concept that work done is equal to U2 plus K2 Minus U1 plus K1 Isn't it? Now, this concept Although we have learned, but we have been using this concept For Point masses Or for the masses that are only doing translation Okay? So In this topic, we are going to learn how to use this work energy theorem Or this particular equation itself On a rigid body Alright? For example, fan Can you tell me, is fan having any kinetic energy? Does fan have kinetic energy? Does fan have kinetic energy? Quickly tell me Does the fan has kinetic energy? Yes, Ronald said yes He can't speak See, the thing is that See, even though the fan is not moving anywhere Alright? Fan is fixed Alright? It has a fixed axis of rotation Then also Then also it has kinetic energy Okay? Because all the particles on the fan Let's say this is fan That is rotating like this Alright? All the particles on the fan They are moving with certain velocities Right? So if you add up all the individual kinetic energies You will get the total kinetic energy of the fan The center of the fan is at rest But the particles which are Which are away from the center They have some finite velocity And hence some finite kinetic energy So total kinetic energy Will be the summation of All the kinetic energies of the individual particles Fine? So the problem with this rigid body Is that you cannot write kinetic energy as Half m v square simply The reason being that the entire fan Is not moving with the same velocity Which velocity will you take? Okay? Center of mass is also at rest You can't say half m v c m square Then the kinetic energy will become zero But we have just learned That it has some finite kinetic energy Every particle has some kinetic energy And kinetic energy cannot be negative So when you add up all the kinetic energy It will come out to be a positive quantity Alright? So we are going to learn How to write kinetic energy For a rigid body Fine? So let us try to do that So for a rigid body There are three cases Case number one Only translation What is only translation? Only translation means what? The entire rigid body Suppose this is the rigid body It is moving as it is Without rotating Okay? With velocity Let's say v Alright? So the orientation of a rigid body Doesn't change All the point have the same velocity So when you write down kinetic energy Is summation of all the point Masses kinetic energy Like for example m1 v1 square m2 v2 square and so on Since all the points have same velocity The velocity square by 2 Comes out of summation So it will be like this Okay? And summation of mi is total mass So it will be simply half mv square Alright? So if body doesn't rotate You can treat as if it is like a point mass And the kinetic energy will be equal to Half m Half m into v square Where v is the velocity of the entire body Which is same throughout Fine? All of you understood this? Case number one? Yes sir Alright? Case number two Write down case number two Is rotation about the fixed axis Alright? So the rigid body is not going anywhere But it is just rotating about a fixed axis Suppose this is the rigid body Alright? And this is rotating about let us say That axis This is the axis about which the entire rigid body rotates And the angular velocity of rotation is given as omega Fine? Sorry Angular velocity is omega Fine? So we are trying to find out the kinetic energy Due to rotation of the entire rigid body So I will assume that entire rigid body Is made up of small small point masses Let's say this is a point mass m1 At a perpendicular distance r1 away from the axis This is point mass m2 At a perpendicular distance r2 And let's say this is at a distance r3 A point mass m3 So similarly there can be n different point masses On the entire rigid body So rigid body is made up of all the point masses Fine? Can anyone tell the velocity of point mass 1? How much is the velocity of point mass 1? Aditya Narayan Tell everyone what is the velocity of point number 1 Is angular velocity is omega And perpendicular distance is r1 Aditya Narayan Let me unmute you Sir is it r1 omega? Correct It is r1 omega So all the point masses are moving in a circular path With center on the axis Fine? So if this circular path The velocity of point mass m1 Will be equal to r1 into omega Fine? So kinetic energy of mass m1 Is what simply half m1 into v1 square Right? Where v1 is r1 omega So this will become half m1 r1 square omega Right? Similarly you can write down kinetic energy of The second particle as Half m2 r2 square omega So like that You can write for all the point masses And when you add up all the kinetic energies You'll get the total kinetic energy of a rigid body Which is equal to summation of Half Sir shouldn't it be m1 r1 square omega square? Yes It will be whole square so that's why So mi ri square omega square Right? Now which of the quantities in this expression Are constant at a particular time? Mass and radius Mass of point masses can be different Half is constant and omega is also constant For entire rigid body at a particular moment Omega is constant so half omega square comes out And this will be summation of mi ri square What do you think summation of mi ri square is Moment of inertia Moment of inertia about which axis? About any axis Now look at here you have Ri is a perpendicular distance from this fixed axis So this is the moment of inertia about the fixed axis Magnetic energy is equal to Half moment of inertia about the fixed axis Into omega square this is the kinetic energy So we have just derived the way to write the kinetic energy For a rigid body Which is rotating about the fixed axis Are you getting it? Any doubts? Are there any doubts? No sir Fine Alright so let's take a small numerical on this All of you try attempting this Suppose there is a rod of mass m And length l It is rotating about this axis Which at a distance of l by 4 from one of its ends About this axis which comes out of your screen It is a fixed axis The entire rod is rotating And the angle of loss is given as omega You need to tell me what is the kinetic energy of this rigid body You got the answer? Are you telling the answer? You can type in if you don't want to speak You can type it out Sir is the mass uniformly distributed over the rod? Yes If nothing is said you assume that mass is uniformly distributed always Okay please tell your answers Trippan Did you get? Sir is it 7 mls per omega square per mls Okay Shomik did you do this one? No There is too much See this formula you have to use i about fixed axis into omega square by 2 So basically you need to find moment of inertia about that axis Fine How can I find moment of inertia about that axis? I can use what? Parallel axis term About the center of mass axis I know the moment of inertia is mls square by 12 And from center of mass this distance is equal to l by 4 So this plus m into l by 4 whole square This is the moment of inertia about the fixed axis So substitute that instead of this if And that is the answer So like this you can find the kinetic energy of any object We will take one more numerical here Suppose this is a disk It's a disk mass m and radius r This disk can be rotated about that axis The axis lies on the plane of the disk And this distance is r by 2 Let's say the angular velocity is omega Tell me what is the kinetic energy for this disk Anyone got this? Vibhav is telling one answer Shristi also told Major Shomic It's okay to be wrong If you are getting wrong answer don't Don't think Vibhav is answering So again I have to use the same formula Half moment of inertia about fixed axis Into omega square Now this is the fixed axis The white horizontal line So I have to use the parallel axis theorem To shift this axis from center of mass To here by a distance of r by 2 So this is my ICM Which is parallel to the fixed axis But what I know is moment of inertia About the axis that comes out of the plane So that moment of inertia is m r square by 2 So now I'll use a Pandit axis theorem And I'll say that moment of inertia Which is m r square by 2 Is the sum of Ix and Iy Ix plus Iy And since Ix and Iy are symmetrically placed This will be equal to 2 times Ix So Ix is equal to m r square by 4 This is my ICM So moment of inertia about fixed axis Will become ICM Which is m r square by 4 Plus now I have to shift it parallel to itself m r by 2 whole square So this will be equal to m r square by 2 This is my moment of inertia about the fixed axis So kinetic energy is Half moment of inertia about fixed axis Into omega square So you are getting m r square omega square by 4 R M1 got it correct Aniruddha as well I'm assuming all of you understood If you have made any error here Even Sharmik got it correct Shristi, Mayur, have you understood Where you went wrong, Anurag? Yes, sir Okay Alright, so let us talk about How to write the kinetic energy Of a rigid body Which doesn't have a fixed axis For example Let's say I have a rigid body like this Alright This is moving forward This is let's say VCM is given to me And also it is rotating Okay While it is moving forward It is rotating with angular velocity omega Alright I have to find out The kinetic energy for this case number 3 Alright So we will try to prove that How much it will be Let's divide this entire rigid body Into multiple point masses Let's say this mass Is from center of mass Its distance is let's say r1 You can say r1 vector Is the position of this mass M1 Okay So I know that When I have to observe a point In a rigid body First I observe with respect to center of mass Then I will add the center of masses Velocity to find the total velocity So V1 velocity will be equal to Velocity of center of mass Plus r1 cross omega Alright This is the velocity Of point mass M1 So kinetic energy of point mass M1 Is half M1 into VCM plus r1 Cross omega Square Okay Square means putting a dot product with itself So this will be equal to Half M1 VCM square Plus r1 R1 sin theta Whole square into omega square See this is a cross product right So it will be r1 into omega Into sin of angle between r1 and omega Fine Plus 2 times VCM Dot with r1 cross omega Now r1 cross omega Here again I will simplify it as r1 sin theta Into omega Can anyone tell me what is the angle between r1 and omega Omega is Coming out It is along the axis Fine r1 is on this plane So the angle is 90 degrees Fine So since angle is 90 degrees I can write it as half M1 VCM square plus r1 square Omega square Plus 2 VCM Okay Dot with Let's say it is r1 omega This is the kinetic energy of point mass M1 Total kinetic energy is summation of that So summation of all the kinetic energies Will be equal to half Summation of MI VCM square Plus half Summation of MI RI square Into omega square Plus 2 VCM Omega Dot with Summation of MI RI All of you understood this Till now Now tell me VCM is constant So it comes out of summation So the first term will become half VCM into summation of all the masses It will be total mass Half M into VCM square Plus What is this term inside the bracket This is what Moment of finish Moment of finish about which axis The center of mass Not the center of mass Because r1 is a perpendicular distance From the center of mass So this will be half ICM Into omega square Now Summation of MI RI Divide by summation of MI This is what Position vector of the center of mass Isn't it Now Position vector from which point I am measuring position vector Of the center of mass When I do that RI is what RI is a Distance from the center of mass So this will Tell me the location of center of mass From the center of mass itself Which is 0 Summation of MI RI Is equal to 0 And hence kinetic energy is this The third term becomes 0 Because this is Equal to 0 This derivation is not in your syllabus But I thought I should tell you everything So Here it is All of you understood how to write kinetic energy Half MVCM square Plus Half moment of finish about center of mass Into omega square Is it clear Okay Now I am going to tell you an easy way to remember the kinetic energy So First of all There are two cases Fixed axis and axis which is moving All right You might be wondering that can I use the Second case to find the kinetic energy And will I get the same answer Okay Let's check that Take for example a rod Which is rotating about one of its ends Let's take mass m and length l This is the fixed axis One of its end is a fixed axis It is rotating with angular velocity omega You need to find out its kinetic energy Which is moment of inertia about the fixed axis Into omega square And this should also be equal to half MVCM square Plus half ICM into omega square Just check both of the equations Are getting the same answer or not Apply both the equations over here And tell me whether you are getting the same thing Getting the same thing Yes sir See this will give you half About one of its end moment of inertia is ML square by three This into omega square Fine Whereas if you apply that formula You'll get half m into VCM square Plus half ICM is ML square by 12 into omega square Right And you can write down this VCM is At a distance of l by 2 Right This distance is l by 2 So VCM Is omega into l by 2 as well Fine And when you substitute it over there You'll get half m into omega square l square by 4 Plus half ML square by 12 Into omega square So you can see that If you take let's say half m l square by 4 Into omega square common You'll get 1 plus 1 by 3 And you're going to get the same answer Half ML square by 3 Into omega square Right So it does not matter which formula you can Use You'll get the same answer for the kind energy Okay You'll get the torque Torque about one axis will be different From torque about another axis But when you find the kind energy of a rigid body It doesn't make sense to say that Kind energy about one axis or the other You should get the same answer Irrespective of whatever axis you take But the formula which you have Is either about the fixed axis Or about the center of mass axis Right So if you have a fixed axis For a rigid body It is advisable to use the fixed axis formula Which is half i f into omega square But if you're not able to locate the fixed axis It is safer to use this formula Okay So this formula will be like You can remember it like that This could be treated like kinetic energy With respect to center of mass Okay loosely you can say that This bracket term is kinetic energy With respect to center of mass Because with respect to center of mass Entire rigid body You can feel as if it is Rotating with annularcy omega So half ic omega square is kinetic energy With respect to center of mass Plus kinetic energy of the center of mass Which is half mvcm square So like that You know it will help you to remember Sir Yeah tell me Sir if it is If there's some body that's rotating Or some axis other than the center of the mass Speak a bit loud If there's some body rotating about an axis Which is not the center of mass axis But some other fixed axis Not fixed axis It's a axis and then it's moving Sir for that the kinetic energy In that also we have to use The second formula And substitute icm itself Yes This formula is generic formula The second one Whereas the first formula can be utilized Only when there is a fixed axis Okay Alright so let us Try to see Scenarios where we can use the second formula For example Let's say it is a sphere Of mass m And radius r Okay this sphere Is Rolling without slipping With angular velocity omega And it is rolling without slipping On a Horizontal surface let us say Okay on this surface It is rolling without slipping You need to tell me What is its kinetic energy How much is the kinetic energy of this Are you seeing any fixed axis here Abhishek are you seeing any fixed axis Abhishek You're on mute Let me unmute you Yeah Abhishek tell us Abhishek Are you seeing any fixed axis Abhishek Yes sir Where What do you think Are you seeing any fixed axis It is okay If you are not seeing any fixed axis You can anyway use the formula Half m v cm square Plus half icm omega square Use this formula to get the answer Simply have to substitute the values Try doing it Shomit you're getting the wrong dimensions How can you add 2 by 5 with omega square Is it 7 by 10 m r square omega square Okay others What is the relation between omega and vcm For condition for no slipping On a fixed surface vcm is equal to omega r For condition for no slipping On a fixed surface So half m into vcm Which is omega square r square Plus half into icm is 2 by 5 m r square it is a solid sphere Okay So you're getting 7 by 5 r square omega square As your kinetic energy Anybody got 7 by 5 7 by 10 sir Sorry 7 by 10 Okay so I can see that Krishna got it And anybody else Okay Vibhav considered it as a hollow sphere That's fine Have you understood all of you Fine let us move to the next topic Just give me one second Yes so now that we know how to write the kinetic energy We need to also learn how to write the potential energy of a rigid body Because if it is a point mass Then we just write mgh as potential energy But if it is a rigid body Okay rigid body can be located from one height to the other height As in it is not a point mass So you can say that this is my rigid body Okay and let's say this is your ground So starting from height h1 You have the rigid body All the way up to height h2 Alright so we should also learn how to write the Potential energy So next is how to write the potential energy of a rigid body Okay so again we will consider the rigid body is made up of point masses So let's say this is mass m1 Which is at a height of h1 This is mass let's say m2 Which is at a height of h2 Okay so potential energy of mass m1 You can say small u1 Potential energy of mass m1 is m1gh1 u2 is m2gh2 Alright so total potential energy will be sum of all these potential energies So u will be equal to sum of mighi Alright assuming g is constant throughout the rigid body It will be equal to summation of mihi Okay and summation of mihi divided by summation of mi This I can say this is height of center of mass Yes or no? This is the formula right? I can say that this axis is my y axis along the height So the h is my y coordinates Anurag can't see anything Others are able to see my screen Summation of mihi Anurag you can rejoin This will be equal to summation of mi into h center of mass Summation of mi is total mass this into hcm So potential energy can be written as mihi is m into hi So mihi is m into hcm So mhcm into g So if I write it like this mg into height of center of mass So this comes out very nicely If I have to write the potential energy of a rigid body All I have to see at what height the center of mass is I don't need to track anything else I just need to see where is the center of mass What is its height And mg into height of center of mass is the potential energy So you can treat as if entire rigid body is a point mass Located at the center of the mass So now we know how to write the kinetic energy of a rigid body And how to write the potential energy of a rigid body Now we can use work energy theorem for the rigid body as well Is there any doubt? If there are please tell me Otherwise I will move forward No doubts Are you guys prepared for tomorrow's exam? Some of you haven't Are not attending the class also because of that All of you 100% prepared Suppose this is the rod It can rotate freely about one of its end It can rotate in a vertical circle Like that About this fixed axis passing through its end Total mass is m Total length is l You need to find out what will be its kinetic energy Not kinetic energy You can say what will be its angular velocity When the rod becomes vertical like that It will start rotating and then the rod will become vertical So when it becomes vertical what will be its angular velocity Find out Omega is what? You have to apply work energy theorem between this position and that position Position when the rod is horizontal that is first one And second one is when the rod is vertical Skanda I will talk to you after the class You can put these kind of messages personally You don't need to let everyone know Okay Shamik is giving answer Others Tirpan, Ariman, Ravi, Kiran has also given an answer All of you should attempt Arpita what is your answer? One minute Okay Srishti, Vibhav I remember everybody's name now Karanderu 3GBL Okay most of you are getting that answer only Should I solve? Alright so I have to use work done Whenever I am using work energy theorem I have to use between the two time intervals Okay First The first instant is when it is horizontal Second instant is when it is vertical Okay So I will take this horizontal line As my zero potential energy So whenever I am applying work energy theorem For gravitational potential energy I have to consider an imaginary horizontal line As my zero potential energy So I am saying that this is my zero potential energy UG is equal to zero For that So in the first instant I have to write the potential energy And potential energy of the rigid body is what? MG into the height of centre of mass From the zero potential energy Which at first instant is zero So U1 is zero Connecting energy is how much? Even K1 is zero The entire rod is at rest Okay U2 You can see the centre of mass has moved down By what distance? It has moved down by a distance of L by 2 So what is the potential energy now? Minus MG L by 2 Minus MG L by 2 Like the way we have been doing till now Okay There is no change Minus of MG L by 2 Carnity energy I can clearly see there is a fixed axis So I will directly use the formula That half moment of inertia about fixed axis Into omega square Alright And about the fixed axis Moment of inertia is basically ML square by 3 Okay And how much is W? Are there any forces on the rod that are doing work? Gravity was one of the force But for gravity you have considered the Potential energy So you don't calculate the work done by the gravity Any other force on the rod? Any other force on the rod? And why MG L by 2? Negative See it is negative because it is The height is below the zero potential energy line If the height is above If centre of mass would have been there It would have been positive potential energy If you are below It is negative potential energy That's how we have been using In work by energy chapter also Okay Now there are forces other than gravity The hinge, the axis where it is tied There will be forces Alright But the good thing is that The point on which the force is applied That point is not moving anywhere Okay So displacement of the application of force is zero So even the force is there Displacement is zero And hence work done is zero So when you solve it You will get omega as root of 3g by L Clear Now let's modify this question a bit Assume that Rod has rotated by an angle It has rotated by an angle of theta Let us say Alright Now you have to find the angular velocity Earlier it has rotated by an angle of 90 degree Now it has rotated only by an angle of theta Find out the angular velocity in this case Same methodology you have to use exactly same All of you should get it Just see up to what height the center of mass goes down Okay Mayur got the answer Others Shomik Ariman Fine So center of mass is here right So how much is the height of center of mass From the zero potential energy now This is the height right This distance is how much What do you think This is L by 2 From here to here The distance is L by 2 So this height is L by 2 sin theta Okay So all you have to do in this equation Rather than writing U2 as mg minus of mg L by 2 You should write minus of mg L by 2 sin theta So the factor of sin theta will come in In the expression Rude 3g by L sin theta is the answer for this particular case You can also check your answer whether it is correct By putting theta as 90 degree If you put theta as 90 degree This scenario is same as that earlier scenario You should get the same answer if you put theta as 90 So if you are getting cos theta You can verify that it is not correct Fine so this is how you have to do this particular question Let us proceed further All of you draw this inclined plane There is this inclined plane on which A sphere is kept at the top The height of the sphere Center of masses height from the surface This height is h Friction is sufficient Friction is sufficient for pure rolling Pure rolling as in rolling without slipping Now when this sphere comes down It just comes down When it just comes down You need to find its angular velocity Omega is how much And one more thing that you need to understand here That in pure rolling In pure rolling Friction doesn't do any work The reason is that there is no relative motion At the point of contact Every time instantaneous displacement is zero So that's the reason why The friction doesn't do any work So keep that in mind You need to find out angular velocity When this sphere rolls without slipping And comes to the bottom This is a solid sphere All of you take it as a solid sphere Consider mass as m Apply work music theorem between this instant And when it comes down Then Should I wait? Trippan got something You are shunned Shristi got something Remember when it comes down The center of mass is still at a height Or above the surface Okay, I'll do it now Let's say that this is my zero potential energy line U is equal to zero Right? So I can use work done equals to U2 plus K2 Minus U1 plus K1 Between this point and that point Apart from gravity For which you are considering the potential energy Only friction is there Which can do work But since it is pure rolling Even friction doesn't do work So W is zero K1 is zero These are some obvious quantities U2 is how much? What is U2? So it will be MGR MGR, the height of the center of mass Is R at that point Okay U1 is MGH H is the center of mass K2, with respect to center of mass It will be half ICM omega squared Plus kinetic energy of the center of mass Half M into VCM squared And VCM is equal to R omega Because it is pure rolling on a fixed surface And ICM will be equal to 2 by 5 MR squared So when you substitute all of this You get the answer Okay Clear, right? All of you clear about it? Alright, we'll continue solving questions Suppose there is a rod like this Very similar to One question which you have already solved So this one, most of you should get it right This rod of mass M And length L It can rotate on a vertical plane You need to find out What will be its angular velocity When it has made an angle of theta When you push it a little bit It starts falling down about this fixed axis Alright What will be its angular velocity When it makes angle theta with the vertical Sir, in the previous questions You don't get theta in the answer, sir Yeah, you may not get theta Because directly H is given to you Sir, so angle of inclination doesn't affect the Angular velocity of the object It will affect the height The height is indirectly related to it Done, anybody got the answer? Centromass is here Okay Let's say we are assuming this horizontal line To be zero potential energy Okay So I have to use this formula W is equal to U2 plus K2 Minus U1 plus K1 Fine So W is zero K1 is zero Right U1 is what? Anyone? The height of centromass is L by 2 So U1 is MG L by 2 MG L by 2 Okay, U2 is what? The height of centromass will be now What? This distance should be L by 2 Location of centromass? L by 2 sin theta into MG Why sin theta? Cos theta, right? The height will be cos theta, this distance Isn't it? Yes sir, minus theta Drop a 90 degree, this is theta L by 2 sin theta will be horizontal distance Okay, be careful So MG L by 2 Cos theta is U2 Fine And K2 is Half moment of inertia about the fixed axis Which is ML square by 3 Into omega square Okay So the answer you will get for angular velocity Omega is Are you getting this? Rude 3G 1 minus cos theta By 2L Anybody got this? Sir, I got 5L instead of 2L When you solve this, you get 5 or 2 I am directly writing it This one is correct This one is correct, right? Yes sir Okay So this is the correct answer So when you rearrange the terms, this is what you will get Fine Now One more question On the same scenario is this You need to Sir, the 2 won't be there, the denominator 2 will not be there Okay Rude 3G by L, 1 minus cos theta Okay So let's say that the hinge is Applying force One force it is applying One component of force is a perpendicular rod Another component of a force Is along the rod like this Let's say this is F perpendicular And this is F parallel So you need to find what is the perpendicular component Of the force the hinge is applying And what is the parallel component of the force The hinge is applying Try doing this The hint is you have to use net external force Is equal to mass time expression of center of mass So the parallel component of force along this direction Is there any acceleration Along the rod, tell me No sir No Sir, centripetal Centripetal acceleration Omega square L by 2 Which is V square by R also So the parallel component of force is M into omega square L by 2 Got it? So this is Mg by 3Mg by 2 1 minus cos theta This is the parallel component of force How will you find the perpendicular component of force Mass into The perpendicular component of acceleration This one And how do you get that This component of acceleration How will you get it It will be alpha into L by 2 Yes And how will you get alpha Torque is equal to iL Torque equation So if this is theta And that is Mg Torque because of Mg is Mg sin theta Into L by 2 This is equal to i alpha Ml square by 3 times alpha 3g sin theta by 2L Which is alpha So M alpha L by 2 is the answer So you will get 3g by 4 sin theta Into M sorry Is it clear to all of you Are you able to understand Any doubts Is it clear So the perpendicular force plus Mg sin theta will be equal to Mg sin theta Correct Sorry about that So the perpendicular force plus Mg sin theta This will be equal to mass time Acceleration perpendicular to the rod So the perpendicular force will be Mg sin theta So that will be equal to Negative of Mg by 4 Negative means the direction will be reversed now So total force will be F parallel square plus F perpendicular square Clear to all of you Any doubts Quick quick Alright so see We will take 2 breaks today Right now we can take Let's take Around 5 to 10 minutes break now We will take another small break in between Let's take 10 minutes break 10 to 10 minutes 2 break will be taking So rather than taking a bigger break One bigger break Alright 10 minutes break We will see you on the other side Okay so all of you are back Yes sir Yes sir Okay So let us Take one more Before moving to the next concept This is a disc It's a disc of mass M and radius R This disc Can rotate about this axis The axis passes through the surface Of the disc Understood So it can flip over So when it flips over What will happen When it flips over this is what will happen It will flip on a vertical plane So you need to find out Its angular velocity when it flips over This distance From the center And the axis distance is R by 2 Same work energy theorem only you have to use Again and again Those who joined just now There is a disc mass M radius R The disc can rotate about this horizontal axis When the disc flips over You have to find its angular velocity I will give you hints You have to use first of all Work energy theorem This is a fixed axis So kinding energy will be Half I about fixed axis Into omega square Work done by all the forces will be 0 Take this horizontal line which is axis Itself as 0 potential energy Now you can write down U1, U2, K1, K2 Let us take out some answer Others Even Xiaomi Should I proceed Alright So we have already calculated Moment of inertia about this axis Horizontal axis which is at a distance Of R by 2 If you remember We have calculated moment of inertia as ICM Which is MR square by 4 Plus M into R by 2 whole square Like this we have calculated And we have found out That about that axis It is MR square by 2 So I can say that K1 is 0 U1 is how much Anyone Mg into R by 2 R by 2 Good This is that K2 I can write it as half Moment of inertia about the fixed axis Which is MR square by 2 Into omega square And U2 Now it is going down U2 will be equal to minus of Mg R by 2 Where W is equal to 0 Now when you substitute this you will get the answer This came in J advanced Couple of years back Where is the answer 200 Ug by R Whatever comes out from here I don't exactly remember But I have given you all the values Right Concept of work energy on a rigid body Now you can take up Different types of questions On work energy How to apply work energy theorem On a rigid body Now let us go to the last Concept of this chapter Which is angular momentum Right down This is the fourth class We are having on this chapter Longest we have spent on Any particular chapter Or angular impulse And angular impulse Do you remember linear momentum What was it Linear momentum is what What does it denote Linear momentum tells you The amount Of linear Motion Quantifies What is the amount of motion And There is a formula also Mass times Velocity This is how we quantify The linear momentum Whereas angular momentum Angular momentum is Amount of Angular motion When I say amount of angular momentum Motion I mean to say that How quickly the Angle is changing So there should be A way to quantify the Angular momentum also But before we get into that Let us take a small example Let's say There are two cliffs There is this Object that is falling Vertically down And someone is standing here And looking at it You will see the object Straight like that Whereas someone standing here When this guy looks at it He has to first look like this Then rotate it Head like that Then rotate the head further Like this So like this you can see that He has to rotate His vision to view the object So what this tells you That this object has No angular motion With respect to this observer Whereas with respect to the second observer The angle keeps on changing Fine So the angular momentum Will be different For different Point of observations Fine and hence Just like torque Even angular momentum Is found About Some Point Or along some Are you getting it So about every point The angular momentum will be different Just like this example So let us now talk about How we can quantify this Fine So clearly This example tells you That if The distance Of the object From the observer becomes zero Angle momentum becomes zero So it is directly proportional To the distance The velocity Vector is making From the point of observation Right and not just any distance The perpendicular distance Alright so angular momentum If you denote by L Should be proportional to the perpendicular Distance Alright and The faster the velocity is Faster will be the rate Angle changes Alright so I can say That angular momentum is proportional To the velocity And the heavier the object More is the amount of motion So it should be proportional to M as well So if I combine everything So if I combine everything I can say L is equal To perpendicular distance Into mass into velocity Okay And in a vector form I can say R is equal to R cross T linear momentum Fine so this is how You write the angular momentum For the point mass Remember this formula is for the Point mass We have not learned about the angular momentum When we Studied the point masses In laws of motion chapter Or work energy chapter We are learning about the angular momentum Of a point mass In the system of particles chapter Fine this is the first time This concept work is introduced So first we are talking about The angular momentum of a point mass Which is Perpendicular distance R Into M into V How to find perpendicular distance R Perpendicular distance R is found out By extending line of velocity And then dropping a perpendicular From the point of observation Okay Let's take a Small numerical on this This is a horizontal surface Fine You have a projectile that is Projected At an angle theta Fine its mass is M Initial loss is U You need to find out Its angular momentum with respect To point of projection Angular momentum with respect to Point of observation When the Particle is at the Highest point Find out Attempted Okay, Kirpan got some answer The projectile will go like this At the top most point Its velocity will become Horizontal What will be its velocity at the top most point Zero U cost theta Horizontal direction velocity remains unchanged Okay So U cost theta Is the velocity at the top most point And I have to find out The perpendicular distance Of the total velocity From the point I am finding the angular momentum Right, I am finding angular momentum About this point So I have to drop perpendicular like this Alright Distance is what perpendicular distance Which is the maximum height also Right Maximum height is my perpendicular distance Which is equal to U square Sine square theta By 2g Right And the formula for the angular momentum Is perpendicular distance Mass into velocity So this will be equal to U square Sine square theta By 2g Into mass into U cost theta So M U cube Sine square theta Cos theta Divide by 2g This is my Angular momentum When the particle is at the maximum height All of you understood this No doubt Now that we know how to Write the angular momentum of a point mass Let us find out How to write the angular momentum For a rigid body Please write down Angular momentum For a rigid body So suppose this is my rigid body So I draw that To write cases Case number 1 The rigid body Is let us say only translation If rigid body do not rotate It is only translating The entire rigid body Can be treated Like a point mass Located at center of mass So just like we Have been writing perpendicular distance Into M into V Same way you write here as well The difference will come From case 2 Case number 2 let us say it is fixed axis And let us say this is the scenario It is a rigid body And let us say entire rigid body Is rotating about This fixed axis About this axis And Angular velocity omega is given Let us say omega is the angular velocity Fine So I can consider the entire rigid body As if it is made up of Point masses So let us say this is R1 This is mass M1 The M2 is at a distance of R2 M3 at a distance of R3 Okay So can you tell me what is the angular momentum Of mass M1 Quickly tell me Pappan, your distance is how much Of this mass M1 from the Point First tell me what will be its velocity V1 is what M1's velocity is what R1 omega R1 omega And what will be its perpendicular distance This is velocity will be like this perpendicular to R1 So perpendicular distances R1 only Okay if perpendicular distance is R1 So angular momentum will be what M into V which is R1 omega Into R1 Right So L1 is R1 square omega Alright So total angular momentum is Adding up everything It is Mi Ri square Into omega If omega comes out because it is constant It will be summation of Mi Ri square Now what is summation of Mi Ri square Anyone This is what Moment of inertia about Which axis Center of mass About the fixed axis See R is a distance from the fixed axis It is not distance from the center of mass Right that is what we have considered So this is If into omega Alright so we have got The angular momentum About the fixed axis To be equal to If into Omega Like we have got the Kinetic energy of a rigid body About a fixed axis Simply as half I f Into omega square Similarly angular momentum also Gives you a simple expression If into omega Okay Now Case number 3 For case number 3 will not do the derivation It will become very lengthy Case number 3 If it is Rotation Plus translation It rotates as well as Moves forward For example there is a Rigid body like this This is the center of mass The center of mass has some velocity VCM And it is Also rotating with angular velocity Omega And suppose I have to find out the angular momentum With respect to this point Okay So the position vector Of the center of mass With respect to that point Is RCM From this point Okay So the angular momentum in this case Will be equal to ICM Into omega Plus RCM Cross So we have Got the Angular momentum Of a rigid body That is Rotating as well as moving forward You can remember this also like The angular momentum with respect to Center of mass I can say it is ICM omega Plus angular momentum Of the center of mass So this you can further modify as ICM omega Plus perpendicular Distance of the velocity of center of mass Into M into VCM But remember these two terms Are vector quantities So probably this direction may not be Equal to that direction But anyways I can write it Like this both directions are same Getting it Now can you tell me what is the Angular momentum with respect to center of mass If I have to Find with respect to center of mass only Then what will be RCM RCM will be what RCM will be 0 then If I have to find About let's say O is same as center of mass At this point O At this point O is center of mass only Over here Then RCM has become 0 So LCM will be Equal to ICM omega So The angular momentum With respect to center of mass axis Is always equal to ICM omega It looks similar to Angular momentum about the fixed axis Any doubts till now Anything So we have quantified the Angular momentum For the rigid body In different different cases Sir can angular momentum be conserved Angular momentum is conserved Yeah that is what We are going to discuss next See if there is no physical Law related to angular Momentum then the definition of angular momentum Is useless I can randomly define something like You know let's call Some XYZ as Linear momentum into Angular momentum Then I can just randomly put a formula And if physically It doesn't mean anything Or there is no Law like conservation of angular momentum Like that Then the meaning of that is not there So we can't use it Right now we have just quantified The angular momentum of particle And rigid body For three different cases Is it clear to all of you Yes sir Okay Now we are coming to Law pertaining to Angular momentum And this law is called conservation of angular momentum Please write down Fine so let's talk about this So I know that For the Point mass angular momentum Is equal to R cross M into V Fine so if I differentiate it DL by dt This will give me R cross M into dV by dt Assuming R is constant Fine so this will give me R cross M into A This is R cross Net force Right So what is R cross F Torque R cross F is perpendicular distance Of the force into F R sin theta into F Right so Torque is actually equal To rate at which the Angular momentum Changes Okay Just like the case Where force Was equal to rate at which Linear momentum changes Similarly Torque is equal to rate at which Angular momentum changes and we have Observed it in case of particle Right now For a rigid body About the fixed axis For rigid body by the way we use Capital L for angular momentum And for a particle we use small L For angular momentum So this is equal to moment of inertia About fixed axis Into omega Okay so if I differentiate it DL by dt If constant with respect to time I get IF into D omega by dt What is D omega by dt Alpha So IF Into alpha And IF into alpha Is torque about the fixed axis Right so when I differentiate The angular momentum for a rigid body Then also I am getting the Torque itself And I also know that the angular momentum With respect to center of mass Is equal to ICM into omega So if I differentiate This also I am going to get ICM Alpha Which is torque with respect to center of Mass So The way the angular momentum is defined I can say that Torque is equal to rate At which angular momentum Changes Fine And just like The force This force will be equal to M into dv by dt Plus v into dm by dt Even mass can change Right Every time we ignore dm by dt term We say that force is equal to mass M exploration only Similarly torque Can be observed when Moment of inertia also changes So torque will be actually Equal to When you differentiate We are keeping is constant So torque can be equal to Omega di by dt Plus Id omega by dt Fine So don't be surprised when you see these Kind of things But they are Rarely You know This is a question but you should know That torque Not only depends on Rate at which omega changes But it also depends on rate at which The moment of inertia changes So The first physical meaning of angular momentum Is If you differentiate it You will get torque Is it clear Clear to everyone Okay So The second physical meaning So for a system For a system consisting of multiple Objects I know that the total angular momentum Will be equal to some of the angular momentum L1, L2 Suppose it has two point masses And two rigid bodies Your system has four objects Two point masses Let's say two point masses like that And let's say two rigid bodies Fine This is your system Four objects Okay The point masses are moving here and there And the rigid bodies Are rotating As well as Moving like this Fine So this is a total angular momentum At any point in time Some of the individual sums Of the angular momentum If I differentiate it The total angular momentum I will get dL1 by dt Plus dL2 by dt I will get it like this Right The first derivative This one Is Torque On the first object This is torque on the second object This is torque on the rigid body Let's call it as three And this is torque on the second rigid body Tau 4 This is my total Angular momentum And we have already discussed that internal Torques when you add Will become zero So when you add all the torques on Individual objects This will give you total external torque Because the sum of the internal Torques Will become zero External torque The sum of external torque is what? Rate at which total angular momentum Of the system changes Fine So if External torque is zero If external torque is zero This will tell you That the dL by dt Of the system is zero It means that Angular momentum of a system Is constant So the implication of this Is that when they collide Their velocities and angular velocities Could change But they will change in such a manner That whatever was the sum of angular momentum Before Remain the sum of angular momentum After the collision as well Their angular momentum will be conserved The only condition For angular momentum to be conserved About a particular axis Is that The net external torque About that axis Should be zero Getting it all of you clear about it So if net external torque About an axis is zero You can conserve the angular momentum About that axis Is it very clear Any doubts? So you have any doubts? Abhishek said something In online class You are like very very quiet But in In a classroom setup Makes so much noise So this is the conservation Of angular momentum Let us take couple of Questions to understand It in a much greater depth Let's say that You have This is a rod Of mass m and length l Okay There is a small mass m That is moving with velocity u0 Hits the bottom most part And sticks to it It hits the bottom most part And sticks to the rod You need to find The angular velocity of the rod After collision First you need to tell me Is there an axis For which Net external torque is zero Is there such axis in this case Is there an axis For which net external torque is zero Simple The axis of the hinge Sudha is saying Okay Are there forces from the hinge Can hinge exert any force Sir it exerts force on the rod It exerts force But The torque about the hinge For that force will be zero or not No sir It will be zero because whatever force the hinge exerts It passes through the hinge So its perpendicular distance from the hinge axis Is zero And what about the torque because of mg Is that also zero There will be mg force also right Is the torque about the mg force About the axis also zero No sir No The perpendicular distance of mg from the axis Just before the collision Sir before its zero but later it won't be zero So I am talking about Just before and after collision So just before and after collision The line of action of gravity The line of action of gravity Is passing through the hinge Isn't it So the net external torque Just before and after Is zero About the hinge axis So I can conserve I can conserve the angular momentum Of the system Angular momentum By the way when this mass M collides the rod Is this mass exerting a force on the rod Yes sir So there is a torque on the rod External torque is there right The mass is exerting force on the rod Can I conserve angular momentum about the axis Sir but the rod will exerting Equal force back So if you consider your system Yes If you consider your system as Rod and mass together Then the force Between small m and capital M becomes internal force There is no external force But if you consider only rod Then there is an external torque Okay But if you consider rod and point mass together There is no external torque Even the force between them Become internal force So now it is very clear That I can conserve angular momentum About this axis But can I conserve kinetic energy About this axis Can I do that Just before and after collision Sir it depends whether it is elastic What kind of collision it is right now Elastic Sir it is elastic because they stick together Sir It is inelastic because Sorry So most of the collisions Will be inelastic Or partially inelastic Elastic collision is a very special case Which usually do not happen Okay so whenever such kind of collision Happens some kind of Energy will be lost In terms of some Mechanical energy will be lost In terms of heat energy Okay and hence Conserving energy is not a good idea But angular momentum You can conserve because The only criteria for conservation of angular momentum is Net external torque About that axis should be zero And which is fulfilled over here Okay so please conserve the angular momentum Before and after collision And tell me What will be the angular velocity After the collision Okay Trippan got it But both are getting different answers Can I conserve linear momentum here Anyone Is it possible to conserve linear momentum Trippan what do you think Can I conserve linear momentum Reason Sir because there will be normal forces Which are external force Because there is an external force From the hinge Yes So I cannot conserve the Linear momentum I can conserve the angular momentum Because even though Net external force is there The torque is zero So initial angular momentum Will be equal to the final Angular momentum So initial angular momentum of the system Is what only this mass is moving Everything is at rest So it will be equal to M into U0 Into its perpendicular distance Which is L Plus angular momentum of rod This is equal to the final angular momentum So when it becomes a part of a rod The entire thing is just one rigid body Moment of inertia about the fixed axis Into omega Moment of inertia about the fixed axis Is L squared by 3 For the rod Plus I have to add The particles moment of inertia also Which is this So when I substitute it I will get M U0 L This is equal to M L squared by 3 Plus M L squared Times omega 1 L is gone So omega will be equal to 3 M U0 Divided by M plus 3 M Times L Anybody got this answer? Yes sir So Kirthina Kirthina got it Others did not get Durban you got your mistake Shomik Yes sir What was the mistake you were making? Consider the I added the masses Then it is M L squared by 3 No that Mass is not distributed along the length It distributes itself Equally along the length Then it becomes part of the rod Shomik what mistake you made? Okay he is on mute Alright so let us take one more question On this There is a Rod of mass M There is a rod of mass M and length L Okay A bullet comes from here A bullet of mass M Comes with velocity v0 It hits the rod At a distance Of L by 4 So this distance is L by 4 And the bullet Comes out The bullet comes out with half the velocity Okay it Pierces the entire rod and comes out Fine you need to find out By what angle The rod will swing This angle theta is what? The bullet pierces And comes out immediately Find out the Angle the rod will swing Mass of the bullet same After passing through the rod Yes mass is same Mass of the bullet remains same So first you need to Find out the angular velocity of the rod And then conserve the Mechanical energy But you can't conserve mechanical energy Before the collision and after the collision After collision happened From that point onwards you can conserve mechanical energy Should I solve Should I solve or wait Wait Okay I am waiting for Another Few moments Before and after collision Just before and after collision You have to conserve Angular momentum Then After collision To the final point You have to conserve mechanical energy Anybody close to the answer Or got the answer Shamak you got any answer Anjali you got anything Arpita Not able to solve this The answer is very long Okay no problem I will just write down the equations Alright so I have to first conserve The angular momentum So initial angular momentum Of the mass plus rod Should be equal to the angular momentum Of the mass plus rod final So initial angular momentum Is M v0 Into 3L by 4 Its perpendicular distance Plus rods is 0 Is equal to the final angular momentum Is M v0 by 2 Into 3L by 4 Its perpendicular distance is 3L by 4 From the axis about which you are conserving The angular momentum Okay Plus I about fixed axis Which is M L square By 3 into omega Okay So this will give you omega Once you know With what angular velocity it starts You can find out theta By using work energy theorem Which is W equals to U2 plus K2 Minus U1 plus K1 Okay So W is 0 K2 is 0 Finally it comes to rest K1 is Half I omega square There is a fixed axis so I can use Kinding energy About defects axis formula Half I omega square And let's say that this Is my 0 potential energy The horizontal line passing through the center Of mass So my initial potential energy becomes 0 Whereas This point Will move slightly up because it is going In a circle So this distance Will be equal to what I know that This one is L by 2 Cos theta Total is L by 2 So L by 2 Minus L by 2 cos theta So U2 is Mg L by 2 1 minus Cos theta Right? So from here you get the value of Theta This is what you did Terpan Yes it is Others Is it clear to everyone Please go through it once Sir you sent the bullet like Rips through the rod or whatever Sir On the moment of inertia change On the rod Yeah we are ignoring the little Mass that reduces For the rod You are saying that some mass of the Rod will Fall on the ground Yes sir The assumption is that This part L by 4 Remains intact You might have assumed this L by 4 Falls on the ground Yes sir No problem I am assuming it is intact Any other doubt? Alright one more question then I can see still there is Little bit of discomfort With this concept We will keep on solving Questions All of you here is the next scenario There is A rod that is kept on the Horizontal table Okay this is the rod The rod is kept on the Horizontal table Then there is A ball That is coming towards A rod With velocity u0 Okay ball of mass m coming Towards a rod with velocity u0 After hitting the rod After hitting the rod The velocity of the Ball Immediately becomes Downwards With velocity u0 by 2 Okay The rod has mass m and its Dimension is L The length is L Okay It is kept on a horizontal Frictionless surface You need to find the velocity Of center of mass of the rod VCM is what And what is the angular Velocity of the rod Where does the Sir where does the Board It is at a distance L by 4 All hits a rod At a distance L by 4 From one of its ends Is it done Can I conserve the linear momentum In this case Can I conserve The linear momentum Take the pole Can I conserve the linear momentum Once you take the pole The screen will be visible to you Immediately after taking the pole Take the pole and submit Okay So this is what People are saying Now tell me is there an external force Horizontal direction On the system System is your rod and the mass m Yes there is normal force Normal force is perpendicular to the rod Normal force The rod is kept horizontally And it is smooth No friction Okay sir So horizontally there is no external force On the system they are exerting force On each other while they are colliding But that is internal force So I can conserve The linear momentum Along This direction In fact I can conserve linear momentum Along any direction because not external force Horizontally Any direction it is 0 So initial linear momentum Of the mass is m into u0 This will be equal to The final angular momentum m into vcm Sorry linear momentum m into vcm for the rod Plus Now it is negative of m into u0 by 2 Because the final angular momentum of mass m changes the direction Okay so m into vcm Becomes equal to 3m u0 by 2 Alright So vcm is 3m u0 By 2m Okay Now tell me Can I conserve the angular momentum I can conserve Along the center of mass axis In fact any horizontal axis I can conserve the angular momentum But If I write about center of mass Angular momentum formula is pretty simple ICM into omega About center of mass And there is no fixed axis So better to write about center of mass So I will conserve about center of mass The angular momentum Initial angular momentum of the particle Is Let's say m into u0 L by 4 Its perpendicular distance from the center Is that Final angular momentum of a particle Is negative of m into u0 into L by 4 u becomes u0 by 2 Got it? The initial angular momentum of a rod Is 0 Final you can say Is ml square by 12 1 plus capital L1 is equal to Small L2 plus capital L2 That is the conservation of angular momentum Have you understood this? Both the things? Okay now tell me How these particles will move The velocity over here At a distance d Let's say this is at a distance d What will be the total velocity of this point How will you calculate It is going to rotate like this Omega will be in this way And vcm is forward So vcm like this This will be vcm And Omega into d like that Fine so total velocity Is vcm minus omega t At a distance d Okay so I can find A distance Of a point which will be at rest Immediately after collision If it is at rest Immediately after collision Then vcm minus omega d should be equal to 0 Total velocity should be 0 Okay like that you can analyze After the collision as well Let's move to Now the final concept In this chapter It will take just 5-10 minutes After that the chapter's theory Is over That is angular impulse We have learned In work energy chapter That the linear impulse Is equal to the Change of linear momentum Okay so The I know that F is equal to dp by dt So integral f dt Is integral of dp Which is actually Change in the linear momentum Integral f dt is called J which is linear momentum This is equal to Delta p change in linear momentum Similarly Torque is equal to dl by dt For a rigid body Torque dt integral Is equal to the integral Of the angular momentum Which is change of the angular momentum Right This term Is referred as angular impulse Okay so linear impulse Is change in momentum And angular impulses change in angular momentum Okay let's take a small numerical On this to understand The implication of this How we can deal So suppose You have a rod Same rod that we have taken earlier Like horizontal Rod like this And it is kept on a smooth Table and I am giving it An impulse From here J J is the impulse given to this Rod of mass m And length l You need to tell me The velocity of center of mass And the angular velocity After the impulse is given Try doing this quickly Alright so I know that The linear impulse J will be equal to change in the linear momentum Let's say that The center of mass is vcm So J will be equal to m into vcm Minus initial Linear momentum is 0 So vcm is equal to J by m This will give you v Now Angular impulse how to find J is basically Integral of F dt And the angular impulse is Integral of tau dt So torque with respect to Center of mass I have to find Which is equal to l by 2 F dt Integral So this can be referred as L by 2 integral of F dt So you can see that The linear impulse into its Perpendicular distance from the center of mass Is the angular impulse What you need to understand The linear impulse Multiplied by the perpendicular distance From the center of mass Or about any axis you are finding The angular impulse Perpendicular distance into the linear impulse Is your angular impulse So this Should be equal to i omega Because Change in angular momentum Is i into final angular velocity Initially it was at rest i is ml Square by 12 That into omega Okay So from here You will get omega as 6j By ml Alright guys so this is The Completion of the chapter We have done in great detail Typically people don't do So much for a single chapter But since it is The biggest chapter of The entire physics 11th and 12th combined So I thought we could spend About a month on this chapter So we have spent 4 classes That is 4 weeks on this So you have got 4 weeks And This is the last class of this chapter So you have another week If you have not done Practice for this chapter I urge you that First Do practice from this chapter Make sure at least basic level Or up to mains level You are comfortable with the problem solving Of this chapter You are not going to get time later on You may get time For school level preparation For this chapter You will never get time for the j level Preparation which you have to do for this chapter Okay and there are multiple Topics in this chapter It is not very easy to comprehend In a very short span of time That is another reason we have spread it Into 4 weeks So you have another week by the time Next class happens for the physics You must do the basic problem practice On your own At least solve 50 to 60 questions Your own Otherwise whatever you have learned Everything will evaporate And then you have to learn from scratch From 0 So we will take a break now And after the break We will have last 1 hour Of problem solving of the advanced level Is it fine? 10 minutes break We will meet Right now it is 342 We will meet at 352 Is it fine to all of you? Are you guys there? I should unmute Right so we will meet Yes sir Those who are interested in solving Advanced level problems Stay back Others can leave Others can leave Okay We will be there We will be there Are you guys there? Are you guys there? Are you guys there? Okay Stay back if your focus is Or if you want to learn Something more On this chapter Okay all of you there? Are you able to hear me? Yes sir Okay So let us start Suppose you have a horizontal surface On that There is a sphere Which can roll And you have wrapped You have wrapped a string On it And then pulling the string with force F Alright It is a solid sphere Mass M and radius R Okay The friction is sufficient For pure rolling You need to find out Extension of center of mass Alpha And you have to find out How much is the friction on this Just draw a free-boy diagram And write down the equations The solid sphere sir What? Solid sphere sir Whenever I don't mention It is a solid sphere So the force is Mg normal direction Friction F And write down the force equation And torque equation And pure rolling equation Straight forward Anybody close to the answer or should I do it? Sir once I will just draw the free-boy diagram Meanwhile See you might be wondering which direction the friction will be applied Whether forward or backward You don't need to break your head on that You assume any direction If your assumed direction is wrong The value of friction will come out to be negative So don't worry about Direction of static friction All of you drawn this free-boy diagram only Yes sir Okay Net force in the forward Direction is F-FR This is equal to M into ACM This is the Translational equation Newton's second law equation Vertical direction N-Mg Is equal to 0 See the second equation is useless Because friction is not related to normal reaction Because we don't know what is the value of friction It is a static friction It can be anything between 0 and mu n It just introduce a fresh variable Which is not connected to any equation So you may not even write The vertical equation So these are the two Translational This thing And the sphere is rotating So net torque is equal to i alpha About the center of mass Because apparently there is no fixed axis So I will play safe And write down torque equal to i alpha About the center of mass What is the net torque about center of mass Quickly R friction plus R into F It is clockwise Plus R into FR That is also clockwise This is equal to i alpha i is what 2 by 5 MR square So F plus FR Becomes equal to 2 by 5 MR alpha This is first equation This is second equation Do I have any other equation It is rolling without slipping On a fixed surface So there is another equation A is equal to alpha R Solve these three equations quickly And tell me what is the value of friction FR is equal to what I got minus 3 by 10 MR alpha A and alpha are unknown You have to get in terms of F MR like that Friction is how much Minus 3 by 7 F 3 by 7 F All of you are waiting I am waiting for others to answer Not sure what others are doing Sir I got A is equal to 10 F by 7 F Not A Friction FR is equal to what If you get A just substitute In one of these equations Get the value of FR 3 by 5 F Yes sir even I got 3 by 7 F So assuming this is correct I know that force of friction Minitude of that Should be less than or equal to Mu Mg So For pure rolling to happen 3 by 7 F Should be less than or equal to Mu times normal reaction Normal reaction happens to be Mg over here So Mu Mg Should be less than or equal to 3 by 7 F So it is not that For every force F There will be a pure rolling This inequality must be Satisfied If you increase your force F to be A very high value Probably it will become larger than Mu Mg Then it will not do pure rolling Alright Next one There is a sphere of radius R Doesn't look like sphere There is a sphere of radius R Mass M radius R It is a solid sphere Friction is sufficient For it to pure roll Okay The This distance Height of the center of mass You need to find out how much it should be So that It should be able to Complete Circular path When it comes down there is a circular path also It should be able to complete the full Circular path When it is coming down it will Not only Have VCM it will also have omega So you need to find out what should be This height so that It completes the full circle over here Got it Got the question Okay do this Imagine that path Comes back to the ground You don't need to worry about coming back to the ground Imagine this is like a full circle When it goes in Suppose there is a path to go in But as soon as it goes in the loop completes You need to find out What should be this height For it to Complete this full circle What is the condition that It will complete the full circle Anyone Where is the highest chance that it will Leave the circle We are at the opposite Sir exactly at the top Exactly at the top This is where it could leave The circle If it doesn't leave the circular path over here It is not going to leave anywhere Okay So you have to write down Condition for it to just leave The circle at the top And for that condition Normalization should be zero From the top normalization will come like this If this normalization becomes zero It means that it has left that path Okay so you will get Some relation among the velocities And everything And remember center of mass Is moving in a circle of R minus capital R radius Okay so imagine The top most point the velocity Is V Have you written Force balance equation over there That N plus Mg net force Is equal to mass time acceleration Mv square by R minus R Have you written this equation Anyone So at the top most point When it just reaches let's say Normalization is zero So Mg Is equal to M into V square By R minus R So velocity at top most point Should be such that V square is equal to G times R minus R Right So I have got velocity At this point And velocity initially also is zero Both ways I have got it So I will apply conservation Of mechanical energy Between point one And point two Okay so let's do that Since No work is done I can say U1 plus K1 is equal to U2 plus K2 Because W is zero this can be modified like this U1 is Mg Into H K1 is zero What is U2 anyone U2 is what Mg into Mg into 2R minus capital R Okay this height you will take Center of mass is where Total this height is Two times smaller That minus radius of the Sphere Plus kinetic energy Which is half ICM omega square Which is 2 by 5 MR square Okay ICM into omega square Plus half M into VCM Square V square And since it is pure rolling Omega square Pure rolling so omega Into R is equal to V So you can use it over there And along With equation three You can This is equation three You can solve the equation number four Is it clear to everyone Any doubts on this Kepan you had a few doubts right Initially you were saying something You can ask now Sir can you just solve Some random sums from that DPP I didn't get any of them You didn't get any Sir some one or two I got not much Okay let me open that Hmm DPP 2.3 right Yes sir This is what Yes sir this one Alright let's take the fifth one A cubical block of side length L and mass M is placed on a Horental floor Fourth F is applied At the end of the plane Sheet PQ which is The block What should be the minimum value Of PQ So that the block may be tipped About an edge See whenever such kind of questions are asked You need to first identify What is the condition for it to tip Over the edge What do you think the condition will be What should happen Okay No it will then translate Right it will not I am saying When it is about TORQ is there NET TORQ is there There has to be NET TORQ And what else What is the other condition If TORQ is not there it will not rotate NET TORQ has to be there And suppose it is just about to rotate Then also NET TORQ is zero Isn't it Just when it is about to rotate At that time also NET TORQ is zero When force is very less Then also TORQ is zero So what is the difference between these two scenarios When you draw the diagram No idea See when it is about to rotate The normal reaction Will shift From O because it is about to rotate From point O So normal reaction will be applied from here This portion Is about to lift Or it has just lifted a bit When it is about this point So normal reaction will shift that side Getting it TURPAN Yes How do we use that How do we use that You can now Find out the TORQ about this point And equate that to zero So when you do that TORQ because of normal reaction Will come out to be zero Earlier you might have used TORQ from here Sorry in upward direction This is the normalization Earlier you might have done like that But when it is about to rotate This point is no longer in contact Only edge contact is there So from there normalization will be applied So many such kind of problems are there Everything else remains the same You have to balance the force And equate NET TORQ to zero But the condition for it to just topple From point O Is the normalization has shifted Over here When you solve it I think you have to assume MG The TORQ because of the weight Will become zero Then only you get the answers No no not like that Let's try to solve it MG is like this This is MG Then Force F is applied Like that So You can split the force into two parts It will be easy then This is F Cos 30 One the F Cos 30 Keep it attached to the ground And not let it drop Because it is acting downwards I have split the force into two components What is the problem in that Yes sir But the F Cos 30 is acting downwards So it will keep the block Attached to the ground But It will topple No it will have a torque also Suppose that length A is very large So there is a NET TORQ About O or not It will try to rotate about O But you are correct that It will make sure the block is touching Some portion of the ground But as well as It will make sure that It is about to rotate So Let's Balance the torque about O The torque because of F cos 30 Is F cos 30 Into A The torque because of F sin 30 Is in opposite direction of F cos 30 Or in the same direction Opposite Minus of F sin 30 Into perpendicular distances L Alright The torque because of Mg is in which direction Same direction as F cos 30 Or F sin 30 About this point Minus of Mg Into L by 2 Is equal to 0 Alright So from here you get the value of Minimum value of A PQ which is A So F cos 30 Is F Quickly solve this Minus Mg L by 2 Is equal to 0 So root 3 F Is equal to Mg L plus F L So A will come out to be Mg L Plus F L Divided by root 3 F This is the answer But answer is given in terms of L So You need to do the force balance also The force balance when you do You will have F cos 30 Plus Mg Oh wait wait You have to find Minimum value of A You don't have to find the minimum value of force Okay Sorry about that Let me erase this right now So basically What is happening is that This line of force Is going like that Okay So if this The value of A becomes large Let's say A becomes this much Then the line of force Could pass through O Okay If line of force passes through O If line of force passes through O Then the torque Because of this force will be what About that point Zero Yes Then will it be able to rotate Ever about O Because of the force No right It will never be able to rotate The normal reaction Will not shift here actually It will be somewhere here Understand If there is a rotation because of this force It has to be Like this But torque because of Mg is in opposite direction Mg resist the rotation But if this force Direction passes through only Then the torque because of force will become zero So you need to make sure That this line of force is Little bit away from this line Okay The same condition is such That this line Just passes through O That will be the That will give you the minimum value of A So if you draw a line like this And this line The red horizontal line Is the minimum value of A This one If it goes like this Then this force Will have a torque Which tries to rotate This body like that Even Mg is trying to rotate like that So this will not rotate So at least It has to be This line of force has to be away from this line All of you understood this concept Sir even if the block rotates It will rotate about the other Corner It will not rotate about O It will put force on that side When, right now Yeah Yeah, right now It will rotate about this point Only when this line of force Goes like this There has to be a perpendicular distance From this point In that direction Then only there will be a torque About this point like that Right now about this point Because of Mg And if this line of force Is this way Because of that also the torque is in this way About this point Okay So both Mg and the force Torque in the one direction So this force is not helping it to rotate You are getting it Yes sir understood So this distance is the minimum distance And This angle Is 30 degrees Sir you have to use trigonometry to find the Yeah, trigonometry This is 30 This is 30 30 This is L So that distance will be equal to L 10 30 L by root 3 Fine What is our question Sriphan Sir can you do the next question Next one Thin uniform rod of mass M and length L is moving Translationally with Acceleration A due to Anti-filer force Minimum value of acceleration for which The rod may move Translationally Minimum value of acceleration Okay So basically if it is moving only Translationally what does it mean Alpha is Zero Understood So if alpha is zero Net torque about center should be equal to Zero So f1 into 2A0 Minus f into This will be how much Total length is L So f into This distance will be L by 2 minus A0 This should be equal to zero And I know that f plus F1 Is equal to M into Acceleration is given A F1 F1 minus f is equal to M What F1 minus Oh sorry about that Correct F1 F1 minus f is equal to M into A So you have to find out A so A and F are the two Variables two equations you can find it What else It will start like if we are finding The minimum value of A Then A0 has to be Zero or some it won't come Properly so can you do it Okay So this will be f1 into 2A0 Minus f I have to eliminate F from here F is equal to Or you have to eliminate f1 So f1 is equal to MA plus F So just substitute it here The value of f1 Okay When you substitute here you will get The value of A0 You will get the value of A0 As a function of Probably L And Maximum value of Acceleration for which The rod may move Translationary So then Then You have to maximize that A Alright you can change the value of L Okay So just get A as a function of L And change the value of L and find out for what L A will be maximum Don't go by the solution Do it like this And then see whether you are getting it or not What else This question looks Interesting 10th Okay try doing the 10th All of you I will put it on the On this It will be easier for discussion I will draw it here Is chapter is easy or The Fluids was easier Before you did this Chapter Fluid was the toughest one This is Small m This is a sphere of mass m Solid sphere Everywhere pure rolling is happening And let's Assume that small m is going down This angle is theta You need to find The value of alpha ACM The value of expression for the mass m This ACM is for This solid sphere And the value of friction That is there between The wedge and the sphere All of you do this Sir there won't be any Talk due to m v sin theta Talk about what About the center of mass Yeah Entire mg And all its component Acts through the center of mass So no talk due to that So no talk due to that Because it passes Through the center of mass Simply write down the equations I got the equations What you are not getting equations But I got it Why solving it now I will also write the equations Alright This is t This is t Assuming this pulley is Friction less If this pulley has friction Then tensions on both sides Will not be equal This is friction less Okay mg sin theta Let's say friction is backwards So this is Fr It has Let's say angular equation alpha And equation ACM Center of mass And this is equation A This is mg So for the small m I can write down as mg minus t is equal to m Into a any doubts here No For the sphere I can say t minus The force of friction minus mg sin theta Is equal to m into ACM How many of you got this equation Did you get these two equations Yes sir Okay Now I mean the Along the normal reaction the equation is redundant But you can write that down as well n minus mg cos theta Is equal to 0 And now the rotation equation Torque equal to i alpha Torque because of tension is t into r Torque because of friction is r into Fr This is the total torque This is equal to i alpha i is 2 by 5 m r Square times alpha These are the four equations But the problem is The number of variables How many are they A is there, ACM is there Alpha is there Friction, tension Normal reaction There are six variables only four equations What can we do now then What is the equation can we write Torque is equal to r ACM not a ACM is equal to alpha r This is my fifth equation What will be my sixth one This point Whatever attrition of this point Should be the attrition of this mass or not Both these points will move together The top most point of the sphere And this point on this mass They are both connected with the string They both move with the same Displacement, same attrition Have you understood this What is the attrition of this top most point Tell me ACM Plus alpha r Alpha r is the attrition of top most point This should be equal to A Now look at this These equations and let me know If you have any doubts Just go through it once Properly Is it correct I mean you have any doubts Arpita, Aditya Skanda So The chapter is over And the basic problem practice is also done And you might have Experienced That This chapter is not very Straight forward You cannot wake up on a particular day And just do a little bit of practice And will not be able to Master it It will be a continuous process So even if you are I mean even if you have done Let's say 60 to 70 questions On this chapter It is fine You can move ahead for the next chapter You can never master this chapter completely Okay But at the same time You cannot just Come for classes And hope that One day before the exam I will sit and understand everything You have to work Very hard And make sure that this chapter Is manageable for you You cannot leave this chapter Big weightage is there On this chapter for whichever exam You are going to take Whichever exam you take Big weightage For this So that's all from my side And the homework Is J mains level exercises From the center module And if you are aiming for advance You should also Solve the advance exercises from the center module Okay And the chapter name You know right Whatever we have done The chapters are center of mass And rotational dynamics for both the chapters There are not many questions Fine So that's it We'll meet next week now Bye