 Go to the next one see I have to get to you know solve previous year question So I mean some of the unique question that is coming in my mind. I'm doing that first. Okay, so Because you need to also recall some concepts Anyways, so here is one cylinder. Okay, there is this cylinder of mass m and radius r Okay, mass m and radius r. There is this cylinder and there are This there is this string Which is attached to two masses. Let's say this is mass m1 and this is mass m2 Okay, and there is no slipping between the string and this pulley this pulley is actually rotating Let's say m1 is going down Okay, so this pulley is actually rotating like this Okay, and there is no slipping you need to find out You need to find out what is the acceleration of m1 and m2 is Acceleration of m1 same as acceleration of m2 or they are different So they have to be same. You have to be same because it is constrained to move like that There is no so what is the so what is the moment of inertia here? I Said this is a cylinder solid cylinder Okay, see the what I asked is acceleration of m1 and m2 and angular acceleration is related to that So I'm getting m1g minus m2g by m1 plus m2 plus m by 2 Okay Okay, let me slowly start solving this question meanwhile all of you those who have not Got it. Please keep on trying Does the tension need to be same? Tension need not be same. Okay. We have been using this Assumption that it is same string for so same tension Okay, but that assumption is true only when it is frictionless cylinder and there is you know or the You know the string is massless the the cylinder is frictionless the the pulley is massless So all kinds of assumption we are making that is why we You know, we have same string same tension, but when it comes to No problems like these you cannot assume both the tensions to be same. Okay That the reason is this, you know, I'll first draw all the forces and Then let me draw the free body diagram of a cylinder The reason is this there is this force due to you know, this Joint let's say this is t0 force and here you have t. Let's say this is t2 This is t2 and this one is t1 Okay, so talk due to t1 and t2 on the cylinder will be what t1 r Minus t2 r because of perpendicular distance of t1 is r and of the t2 is again r only right So, but then t1 is time to rotate this way, but t2 is time to rotate in opposite way So there has to be subtraction. So this is a net torque for the cylinder. Okay, and if Net torque is zero the angular acceleration should be zero Okay, so if cylinder has to rotate one tension should be more than the other tension Okay, that is the reason why tension is not same if cylinder has a mass and it is rotating As in it has angular acceleration This is equal to i alpha Okay, this is one equation Okay, because the torque due to the t0 is zero t0 is passing through the center of mass So talk the talk due to this force t0 force is zero. All right So, let me write second law of motion equation for m1. So I will get m1 g Minus t1 is equal to m1 a okay, and for M2 we have t2 minus M2 g is equal to M2 into a now how many variables we have t1 t2 alpha Then a so we have four variables, but only three equations Okay, so that is the reason why you need another relation that is actually a constraint relation that is a no-slipping relation Okay, for example this point on the cylinder Let's say there is this point which is on the cylinder it will move with the same acceleration as The string is moving so strings acceleration at this point is same as the cylinders acceleration at that point Okay, so the cylinders acceleration at this point is what alpha into R Okay, so about this point if you see what is the expression of this point with respect to this this point is moving in a circle So it is alpha into R The acceleration of this point and this has to be equal to the acceleration of the string which is a Okay, so you have now four equations and moment of inertia, you know is Capital M R is square by two So all you have to do is just solve these and you'll get the answer Okay Any doubt with respect to this No, sir Okay one more suppose You have a rod lying on a horizontal table Okay, it is a frictionless table this rod is lying on a frictionless table then What else is given is you know the mass is M and the length of this rod is L okay then this Small mass Comes from this side. Let's say the mass is small M. The velocity is V not Okay, so everything is horizontal, you know, so when this mass hits the rod this mass comes to rest Okay, and then rod moves Leaving this small M on the table itself. Okay, so you need to find what is the velocity of Center of mass for this rod After this small M hits the rod VCM is what okay? Tell me one thing. Can I conserve linear momentum over here? How do you want to leave? In this direction, can I conserve linear momentum? Yes, I can conserve the linear momentum see many a times we are very Conscious about the thing that you know since it is a bigger object we cannot conserve lean linear momentum or Something which is valid only for the smaller blocks that we have learned in laws of motion chapter But that is not true. Okay, whatever you have learned before this chapter. Everything is true Okay, so you can use Conservation of linear momentum over here. Okay, but the problem is that The rods velocity at all the points. They are different So which velocity will you take to write the linear moment of the rod mass into velocity? Which you have to write because velocity is different at different points after it After the mass hits the block after the mass hit the rod So the logical choice is we send off mass. Okay, so there's a proof also for that, but we will not go there So, you know initial momentum before this mass Hits the rod is what for the system is m into v0 Okay, and after This hits the rod the momentum of the rod becomes m into vcm and momentum of this small m becomes 0 and before The momentum of the capital M was 0 so you can equate these two and you'll get velocity of center of mass to be equal to a small M by capital M times v0 fine Now you need to also find out the angular velocity of the rod this rod will not just move straight, right? This rod will rotate also. Yes or no to find out. What is the angular velocity of this rod? Velocity be the velocity of the tip of the rod. No, that's not correct when we write a linear momentum of a rigid body We take mass into velocity center of mass So that's not correct That would have been correct if one end of the rod is fixed and the other end is moving with velocity v Then angular velocity can say is v by L, but here that is not correct Okay, kushal is giving an answer m v0 L by 2 others Some getting a whole root of 12 Into v0 square by L square into 1 minus m cube small m cube by m square big m square Oh Acha first tell me Which concept should you use here? You're trying to find angular velocity So in which concept the angular velocity comes in the expression or which equation that angular velocity term comes in Work energy that work energy theorem is one now Are you assuming here that energy is conserved before and after Energy may not be conserved due to the collision. Some energy might have got lost Right, so you cannot conserve energy before the Before the collision and after the collision because during the collision energy is lost Okay, so here the more appropriate Expression other than work energy theorem The only expression left with you to use where angular velocity comes in the equation is conservation of angular Okay, so that is what we will be using here Okay, now about which axis should I use conservation of angular momentum? About an axis about which net external torque is zero Okay, now external to what external to the rod external to the small m or external to both small m plus capital m Excellent to both Okay, so whatever is the force between small m and capital m that is internal force Because of that there is no external torque that is getting produced because I'm assuming my system to be small m and capital m together Okay, so I can choose any axis to write moment of sorry. I can use any axis to write The angular momentum of the rod Okay, but if I use any other axis other than center of mass axis to write angular momentum of the rod Then the expression is bigger, you know the expression is r cross m into vcm Plus icm into omega This r is a distance of the axis from the center of mass Okay, so you have to take perpendicular distance of the axis from the center of mass So unnecessarily why to go beyond the center of mass? So we'll try to find angular momentum about the center of mass axis itself so that this r is zero Okay, hence We can say that angular momentum of small m initially plus angular momentum of capital m initially This is initial should be equal to angular momentum of small m Plus angular momentum of capital m finally Okay, now what is the angular momentum of small m about this axis before the collision It is a particle, right? It's a particle Whose perpendicular distance is what this is a perpendicular distance of its velocity from the axis Right, which is what? l by 2 Okay, so it's angular momentum of this particle before the collision is l by 2 into m into v0 All of you following me Yes Plus angular momentum of the capital m is zero before the collision the rod is not doing anything Plus angular momentum of small m after the collision is zero It comes to rest after collision Plus angular momentum of capital m after the collision After the collision angular momentum of the capital m is simply icm into omega Getting it now icm is what m into l square by 12 for the rod into omega fine, so You will get this comes to six and this is Like that, so you'll get omega to be equal to six m v0 divided by Capital m into l Okay, so this is the angular velocity Anyone got this no one Anyways, no problem Just it is good that you are getting to learn these things But keep in mind when you do it later on similar question So just like you are free to use conservation of linear momentum Similarly, you should be you know very much familiar with conservation of angular momentum also Any doubts till now any one of you any doubts? I use a concept, but I took it as m l square by 3 poor wick Don't do that Okay, I assume no doubts. Oh, I can move ahead. Okay. So the same question. Okay, same question You need to find out a point on the rod which is At rest immediately after the collision the point on the rod which is at rest where it is located All the points on the rod They are having different velocities So which is the point on the rod which is at rest immediately after the collision So is it a point at a distance of l by 6 from the center mouse? Where left or right anybody else getting See the points on the right if you look at it with respect to center suppose you are taking a point at a distance of r Fine. So let's say this is the point Which is at a distance r. Okay. So its velocity with respect to center of mass If this is r Is omega into r Okay with respect to center of mass its velocity is omega into r Then vcm is also in that direction. So total velocity is this plus vcm Okay, so will this ever go to zero? No, it will not go to zero. But if you look at the left hand side At the same distance, let us say at a distance of r omega into r is down Okay, because it is rotating like that Fine. So with respect to center of mass its velocity will be down omega r And also this is going forward with vcm Okay, so total velocity on the left hand side at a distance r is omega r minus vcm Okay, so We have to find out where this becomes zero All right, so R is nothing but vcm divided by omega So you have the value of omega you have the value of vcm Even just substitute the values you'll get it as l by six from The center towards the left Okay Any doubts any doubts with respect to this question now we can take up, you know past year questions If you have any doubts on this, please quickly raise it Okay Suppose to conserve angular momentum from the point where vcm Is equal to omega r see shweta you can conserve The you know angular momentum about any point, you know, you can Because there is no external torque on the system About any horizontal axis Fine Why we are taking center of the rod you're taking center of the rod so that You can write angular momentum of a rod Simply as icm into omega if you are writing The angular momentum about any other axis for the rod Then it will not be equal to i into omega about that axis Okay So it will be m in r cross m vcm R is a perpendicular distance of that axis from the center of mass Into m into vcm plus icm omega so unnecessary an extra term is coming Okay, so that is the reason why we are conserving angular momentum about the center of mass axis Yes moment of inertia will change that is why to compensate that there isn't Another term r cross m v So you will end of the day when you add up these two You get the same thing