 Good evening everyone those who are online please type in your names and if you can see a question start solving it. So these are the passage based questions I have taken so these passage based questions they come in J advance they don't appear in names so on a single passage there can be like two or three questions so these three two three questions will be based on one passage only so understanding of passage becomes very important okay so it is okay to spend you know at least two to three minutes on just understanding of passage because it is seen that if you understand the passage and if you answer one question correctly the other two questions related to passage will be solved very very quickly okay so like that good evening you can see a question in front of you there are two more questions based on the same passage so today we are taking up questions on budget body motion and they are passage based questions who else is online type in your names understood the question understood now tell me which concept will be you will you be using here what concept will be used before you even start solving can I use conjunction of angular momentum but there will be friction between the disc right there will be friction between two discs what about that welcome everyone what about that friction between the two discs that will create a torque right then how can you use conjunction of angular momentum there is a torque right so conjunction of angular momentum can be used only when there is no torque yes or no anyone correct so like what Bharat said there will be frictional torque if you consider only one disc fine but if you consider both the discs together these of tors will be equal and opposite so they become internal tors and they'll cancel out each other net external torque is zero if you consider both the disc together so you can conserve the angular momentum by having both discs together as a system okay you cannot conserve angular momentum of a single disc okay now solve this question quickly what is the answer what directions they are rotating that's a very good question all right it is not given and it's not very clear in the diagram also but there is an assumption that both are having direction same direction okay if both are having opposite direction then situation will be different okay so if you conserve angular momentum initial angular momentum about center of mass is i a omega a plus i b omega b fine so if angular velocities are in opposite direction then minus sign will come over here but i'm assuming they are in the same direction okay now if they are moving together they will share the same axle when they start moving together fine like this so their moment of inertia can be added because the axis of rotation remains the same so you can add moment of inertia like this so that's why omega comes out to be this option three any doubt this will be the answer fine yes now what is seen is that in the previous year like this year 2019 j many simple questions came but the problem was that their calculation or intensity you can probably discuss with those who have written j in january okay they will tell you that so that's why the calculation part becomes very important so don't just think that okay fine you know the concept so you can just you know leave it and then you can move it to the new concept so it's not only about concepts fine it is about how good you are in calculation as well so that's what the next question is about find out the passage is same okay there's a disturbance now there is now there is no disturbance right again i get my phone near the mic so disturbance is still there okay solve this you just have to substitute the values given in the passage let's see how quickly you get the answer and how accurately it is poor thing you can scroll the youtube video back you'll be able to see that you can just scroll your video back it is getting recorded so you can scroll it back okay you can scroll it back you'll be able to see it it's a youtube video right i'm sure your experts are watching youtube videos don't use calculators 100 astuto used calculator or astuto is a genius one of these two options don't tell me as it was you got 1 divided by 0.01 that is also 100 okay that's fine all right so 100 is the correct answer you can just check at your end what about this find out the final kinetic energy so here the used calculator 300 see you got omega right you got the final omega never ever used calculator okay that's very interesting i a plus i b into omega square omega you already got 100 fine you know i a and i b so answer is one okay we'll go to the next question we'll go to the next question okay there's one you made a comment and then could you retract it now disturbance is gone still there bar this thing third anyone else one more minute the acceleration of center of mass is net external force divided by total mass of the system that's the hint spring force is it external or internal to the both the blocks spring force is internal force right the total external force horizontally is f only okay and mass is 2 m so acceleration is f divided by 2 m fine and this acceleration is constant isn't it so if accession is constant i can use this formula s is equal to ut plus half at square initial velocity of center of mass is 0 so displacement is simply half into a into t square that is option number three okay fine let us move to the next question now i hope you are clear in case of any doubt quickly type in this one second block okay ashutosh will describe how we're solved what is x yes yes purvik ashutosh keep the book aside focus here okay what is going on okay let us try to solve this question this is y coordinate and this is x coordinate fine now here i am must be more than see wait let let me explain and then you'll understand clearly okay rather than taking my origin here let me take the origin at the center of mass itself okay so let's say my x center of mass is at the origin fine so let's say x equal to 0 represents origin fine so let's say after time t after time t the x coordinate of this first block let's say first block x coordinate is x1 okay and the second block x coordinate is x2 fine so the x coordinate of center of mass will become what x coordinate of center of mass will be mx1 okay plus again both masses are same so mx2 divided by m plus m fine so this is what x1 plus x2 by 2 all right so this is the x coordinate of center of mass isn't it and this has to be equal to the previous thing which you have solved for the previous question which was half at square right so half f divided by 2m into t square okay so this is your first equation okay and what else is given is that extension in the spring is x0 so both the blocks move right let's say this moves this way and that one moves that way fine so x1 the x coordinate of this block m minus x coordinate of that block x2 this i'm saying that this is the extension in the spring okay i'm ignoring the length of the spring initially all right so you have these two equations so when you solve these you'll get option 1 to be correct any doubts yeah of course you can imagine small m will move more than what center of mass will move any doubts any doubts here could be by symmetry kaika symmetry they go use this equation you'll get option 3 for the second block okay don't force yourself to identify you know although you may get the answer at times but you know rely more on equations than symmetry okay next question was in fact that only so yes option three for this one okay fine let's move ahead not this this one option four this one yeah option four is correct for this one it's all this question why we consider x1 minus x2 not x2 minus x1 because the block which is ahead is moving more than the behind one right x coordinate of the ahead one will be more than the behind okay this is a very simple question if you break this problem into multiple parts so good thing is that if you know this problem the first question will take time but the other two if you know if you solve this one other two will be solved quickly because same on the same equations other two questions will also be based upon anyone got the answer none of these niranjan got none of these no no hurry guys just take one by one and you'll be able to solve it draw the free body diagram there's constraint relation i got a very weird answer this itself is weird mass of the plank mass of the plank is 1 kg isn't it given both plank and small m is plank at the top it is written 1 kg even the cylinder is 1 kg you can get the try getting the expression first okay don't worry about the final answer get the expression in terms of capital M small m f and theta and then substitute no one should i solve or you guys are in between 1s what is 1s oh one second one second done plank be lifted from the force force is how much 55 newtons hmm i can understand so okay what you're saying plank so plank can be lifted up but let us say i mean hypothetically solve this in terms of capital M capital M small m and f and theta okay in terms of that solve this okay but yes the plank will get lifted off so this question doesn't make the complete sense but in still can you find the expression in terms of the variables but the answer is you know if i keep on saying you're getting getting wrong so you have only four options right so now even that is wrong none of these i always said it is wrong so only often two remains should i solve a good thing about taking online class another good thing in fact is can draw the diagram very nicely anyways so let us take cylinder and the plank one by one okay so here is the plank now on the plank what are forces are there you have f cos theta that's what barat you have to improve your handwriting so this is f sin theta yellow color then there is mg small mg and on top there will be let's say normal direction n1 find any other force of course there will be frictional forces so this is let us say f1 okay this is f1 okay and then there will be pairs of the forces first pair is normal direction so this is n1 okay and then this is another pair f1 okay this is the center okay and then we have capital mg force like this of cylinders and one normal direction from the floor let's say that is n2 okay and of course there will be a friction from the bottom so let's say that is f2 okay any other force can think of anything you think it is left or this is fine friction is not mentioned and if you if if you say that okay it is given rough on the plane but is not mentioned friction between the plank and the cylinder then the plank can never apply the the initial torque on the cylinder and cylinder will never be able to roll plank will just slide away from the cylinder okay there has to be friction between the plank and the cylinder as well okay fine so now let's mark the accelerations also here so let's say that acceleration of the cylinder is a center of mass okay and angular acceleration of this cylinder is let's say alpha okay similarly planks acceleration let's say over here I'm taking it as a1 fine okay so now let's write the equations for plank okay for plank we have f cos theta minus f1 is equal to m into a1 right and n1 minus mg is equal to 0 this is for the plank okay plank doesn't roll so no no question of writing the torque equation for that okay now for the cylinder for the cylinder we have f1 plus f2 okay wait here it will be there will be f sin theta as well right this is f sin theta is equal to 0 now for the cylinder f1 plus f2 is equal to capital M into acm okay what else what else I can write I can write n1 plus capital mg minus n2 is equal to 0 and I can write the torque equation also okay torque will do i alpha I can write about center of mass so f1 talked into f1 is in the direction of alpha so f1's torque is positive so r into f1 minus r into f2 should be equal to cylinders moment of inertia about the center of mass so mr square by 2 into alpha okay fine now you'll see that if you just look at the equations you have useful equations are this this and that because normalization anyway you don't need to calculate so using these three equations I need to solve for the expression of cylinder which is acm now count the variables capital f f1 f2 okay 3 and then acm and alpha okay so you have five variables and just three equations all right sign part of just like force in the direction of acceleration is constant is positive similarly usually we assume that the torque in the direction of angular acceleration is positive so torque due to f1 is in the direction of angular acceleration alpha so f1 is trying to rotate the cylinder in this way same way as alpha is rotating okay but f2 is trying to rotate in opposite direction so torque due to f1 is positive but due to f2 it is negative okay so hence negative sign over here all right now I have to write the constraint relations constraint equation of constraint relation let's say this is point one and this is point two okay since it is pure rolling happening so point one and point two relative acceleration has to be zero fine so at point number one the total acceleration is acm minus alpha into r so this should be equal to zero this should be equal to grounds acceleration so that is why this is at point one at one okay now look at point number two point number two is acceleration on the cylinder is acm plus alpha r so this is acm plus alpha into r okay this should be equal to acceleration of the plank which is a1 fine so this is at point number two fine so like this you have to write down these equations and then you'll be able to solve it now please solve it and get the answer quickly all of you understood where you guys were making mistakes quickly solve get the answer that's wrong Bharat anybody else nirangin that's why that's probably because you're attending both the sessions chemistry as well as physics okay so option two is correct all right let's move to the next question now see the next question is to find the frictional force at a so once you solve those equations you get all three questions correct so now if this is frictional force at b right is it sorry Bharat okay so these kind of questions don't think that for just one question you are spending a lot of time because you solve the first question you automatically get the other ones easily this one what about the previous question there was a previous question which one no no solve this question option three what else anybody else it's an application of parallel axis theorem should I solve or should I wait there are n rods okay so I need to basically find out moment of inertia of all the rods about the center over here okay and then I'll add moment of inertia of the ring which is there for the ring it is easy okay because rings moment of inertia is simply mr square so you can see in all the options there is one mr square is added that is of the ring so the problem is tricky because of the polygon that all right and another good thing is that you just have to find out moment of inertia because of the one link of the polygon and then multiply that with number of links because all the links are placed symmetrically about this axis okay so all I have to do is to find out moment of inertia due to this one link and multiply that with n okay now there are n rods of mass and each okay so the total number of links are n so if I if I have to find out this angle this angle will be how much 2 pi by n okay 2 pi is the total angle divided by n sides so this will be that angle okay but if I drop a perpendicular over here why I'm dropping perpendicular because I'll be using perpendicular sorry parallel so I'll be translating this moment of inertia from here to this point so it will help me okay this angle is pi by n okay it is half of 2 pi by n that is pi by n okay now what else is given length of the rod is not given well metal ring of radius r so what is given is this radius is given because when the ring will be welded it will be welded at the corners and center will be this only so this distance from here point number one to point number two this distance is r that is given fine so let's say that distance is small d okay so small d is given as r cos of pi by n okay now moment of inertia I have to find out for one of the links I'll be using i is equal to i center of mass plus small m into d square okay now i center of mass is ml square by 12 okay now mass is small m but l is not given okay but in this right angle triangle this length you can see that this length is r sin of pi by n okay so total length is two times of that okay so icm becomes m into l square which is two r sin pi by n whole square ml square by 12 okay plus m into d square uh that is small m into d square that is r square cos square pi by n okay so this you can see that you can further simplify it and it will become m uh mr square if you take that common you'll get it as sin square pi by n by three plus cos square pi by n okay now there are how many links n links are there so just multiply this total moment of inertia will be n times of that okay so not only just this plus the rings moment of inertia which is mr square the ring is also valid that's why option one is correct any doubt no doubts this one how will i solve this one i whatever i have got in the previous question that was uh icm of the entire system isn't it okay so i need to find out basically moment of inertia about one of the corners so simply i'll be using i is equal to i center of mass plus total mass of the system into d square so now i'm applying this for the entire system okay icm we have already found out plus total mass of the system is capital m plus n times small m into distance is actually r square isn't it the corners distance from the center is radius so option two is correct fine let us move to the next question now this one option four nirajan hanugar now it is should i solve this one or should i wait okay let us solve this one now so i'll take the components this is mg sin theta mg cos theta this is friction okay this is acm okay and it is rolling so i need to show alpha as well so this is alpha fine so net force equal to mass m expression so i can write mg sin theta minus friction this is equal to m into acm okay of course there will be a normal reaction also so fine and n minus mg cos theta will be equal to zero fine so these are the force equations and then there will be a torque equation only friction is creating torque about center of mass so r into fr is equal to i into alpha okay and then there's pure rolling so at this point alpha r minus acm this is a net acceleration this should be equal to zero okay fine so just need to solve this one and for limiting case once you get the friction okay friction should be you know less than or equal to mu times normal reaction okay now to find the minimum value of coefficient of friction you just substitute the value of friction to be equal to mu times n okay so once you get the friction from these equations substitute the friction value as mu times n and you get the value of mu where n is mg cos theta okay which option you're getting now option 4 this is what you're getting solve this so Kirti is answering for discussion of the previous one if nothing is said then you have to assume sphere is solid remember this if it is not mentioned you have to assume it is solid sphere anybody else see don't get worried by looking at the size of the question or some weird diagram it's ultimately a simple question any question that asks you to find the moment of inertia you I mean it can't be very difficult okay the ring is mass less so only moment of inertia is because of the sphere only okay so all the spheres have mass m and radius r so they are symmetrically plays about the center of the ring so just have to find moment of inertia you know because of one of the sphere and then just translate it okay so numerator should be a multiple of four okay and the denominator should be multiple of five and and sadly all four options satisfy that we need to solve it further so moment of inertia of the sphere is 2 by 5 m r square okay but we don't know this distance okay this distance is what we need to translate the center of the you know sphere to the center of the ring so this distance is 2r isn't it this one is 2r okay this distance is equal to that distance all right so using Pythagoras theorem let's say this is d so this also has to be d so you have 2r whole square is equal to d square plus d square is equal to 2d square fine so d will come out to be equal to root 2 times r okay so for one of the sphere it will be 2 by 5 m r square plus how much I have to translate a distance of d so m into d square so that is 2r square total moment of inertia will be this multiplied by four because there are four such spheres okay so this will be how much 10 12 48 so 48 by 15 sorry 48 by 5 48 by 5 m r square so that's our option four clear right we will move to the next question now this one all of you attempt option two this thing is in equilibrium look at this diagram let's say this is t okay so this thing is in equilibrium so this is also t you have capital MG so t minus MG is equal to 0 okay and okay and this this thing this structure is not rolling as well because everything is in equilibrium so the torque due to the force okay so f into the radius not the radius of the sphere this distance I am talking about this one okay let's take it as d only then it will be better this is d so f into d all right minus t into 2d this distance is 2d this is equal to 0 so I have equated torque to be 0 about this point because this is not rolling so I can use torque is equal to i alpha about any point so alpha is 0 so net torque about any of the points has to be 0 so this is a convenient point for me because friction will not come in the equation unnecessarily friction will come out otherwise now substitute the value of t over here so you'll get f into d minus MG into 2d is equal to 0 okay so the value of force will come out to be two times of MG so like this you have to solve this question so option 2 is correct over here this one this one option 4 is what Purvik is saying others yes yes Bharat omega is as indicated in the diagram option 4 is not correct see look at the direction of omega and velocity anybody else right now tell me what is the velocity of the bottom most point at this moment when omega is the angular velocity and v is the linear velocity of the center of mass what is the velocity of point of contact is it is it 0 how much it is it is vcm minus omega r or vcm plus omega r how much it is they look at the sense of rotation with respect to this point this point is moving in a circle and going like this isn't it this point with respect to center of mass is going with the velocity of r into omega this way okay so total velocity will be r omega plus velocity center of mass both are in the same direction okay you can imagine also suppose it is rolling without sliding then the sense of rotation has to be like this but it is in opposite direction of the rotation okay so the point of contact is definitely not at rest and it is not a pure rolling okay because a point of contact has the velocity of zero so hence what will happen the process is like this the process is that the angular velocity will decrease okay and linear velocity will increase fine what will happen is that omega will go from omega not to zero and then it will increase in opposite direction from zero to whatever is a requisite value all right now tell me during this process what kind of friction will be applied is it static friction kinetic friction it's a kinetic friction so all the while it was sliding fine and hence the friction will be constant and its value would be mu time mu times normal reaction okay so now let's see the other things as in the equations of motion so which direction the friction will be applied right now see this point is sliding that way so friction will be of course applied this way in opposite direction this is a force of friction which is mu time normal reaction with normalization happens to be mg you can see the force balance mg is down and normal reaction is up okay and there is no other force acting on this object fine so mu mg will be equal to mass time acceleration of center of mass so acceleration of center of mass will be equal to mu times g this is your first equation and the torque mu mg times r this is the torque this will be equal to mr square by 2 which is moment of inertia into alpha okay so alpha will come out to be 2 mu g by r any doubts so this process omega not to zero how much time it will take i'll be using omega is equal to omega not minus alpha into t and put omega to be zero this is equal to omega not minus 2 mu g by r into t okay so let's say this is t 1 t 1 so t 1 comes out to be omega not r divided by 2 mu g okay this is a time taken for angular velocity to become zero okay now it could happen in other way round also like velocity could become zero and then velocity increases in opposite direction fine so let's see whether velocity becomes zero first or angular velocity becomes zero first so if i use v is equal to v not minus at okay so v is zero and initial velocity is omega not r by 4 minus a is what a is mu g okay this into t dash so t dash comes out to be omega not r divided by 4 mu g okay so rather than omega becoming zero and then rising from zero to the required value of omega for which pure rolling starts what will happen is the velocity will become zero and then increase in the opposite direction okay because velocity is becoming zero first getting my point no doubts right which part see understand first thing that angular velocity is in this direction and velocity is in forward direction okay so one one thing has to change its direction okay either the velocity has to change its direction or angular velocity has to change its direction then only pure rolling can start okay now the whichever becomes zero first will be able to change its direction first okay now you can see that i mean the change in direction will only happen when it becomes zero first right it will become zero and then reverses direction so you can see that angular velocity becomes zero at later time than velocity becoming zero okay so velocity will become zero and then it will change its direction first so angular velocity will not change its direction but at least one of angular velocity or linear velocity has to change the direction for pure rolling to start this you understand okay now let's see that okay let us say that the velocity is changing its direction now so velocity will increase from zero getting it so now velocity will be equal to v is equal to u plus at accelerate the frictional will friction will increase the linear velocity initial velocity is zero it is changing its direction and now it is in the direction of friction so plus acceleration which is mu g into t2 let us say okay okay and what about angular velocity can you find the angular velocity omega will be equal to omega 1 minus alpha into t2 t2 will be same for both of these but omega 1 is not zero omega 1 is omega after this time okay after when the velocity becomes zero then you are applying these equations okay finds omega 1 how will you find omega 1 omega 1 is omega not minus alpha is 2 mu g by r into t t dash which is omega not r divided by 4 mu g okay so this comes out to be omega not by 2 omega 1 is omega not by 2 okay so this is omega not by 2 minus alpha is 2 mu g by r into t2 fine this is omega okay and this is v so if pure rolling starts after t2 then omega into r must be equal to v fine so just use this you will get the value of t2 and total time will be t1 plus t2 any doubts quickly let me know option 2 is correct no doubts right let's move to the next question this one how did you get this super its option 3 now will the angular momentum be conserved will the angular momentum be conserved about any axis you take that passage through this line will it be conserved yes or no okay so initial angular momentum will remain equal to the final angular momentum in that case okay so can you find out the initial angular momentum how much it is you remember the formula right m into vcm into perpendicular distance plus icm into omega this is the angular momentum okay but then you need to take care of the sign probably this is of opposite sign of this then it there could be plus or minus just check whether you are getting like this also the same answer 3 actually in this case m vcm into d about this point the angular momentum is like this okay and icm omega is in opposite direction so there will be minus sign in this case okay no no this is the formula okay the actual formula is this r cross m vcm plus icm into omega this is the formula this is the formula for angular momentum about a point from which the center of mass is position vector is r okay so you can see that you'll get the same answer m into vcm is v0 that is omega 0 r by 4 into d d is what r minus icm it's a disc so mr square by 2 into omega fine so basically you'll get you're getting minus of mr square omega 0 by 4 now plus or minus is just for our reference okay we have assumed something to be plus or something to be minus so if you just talk about the magnitude often 3 will be the correct one why you need break Bharat okay we'll be having a break at 630 don't worry this one okay what is the answer over here the answer is 3 how will you solve this one is what the rate of work done so dw by dt is power fine so after time t total work done will be p into t and this should be equal to change in kinetic energy fine the change in kinetic energy is proportional to velocity square fine so p into t is proportional to v square or you can say v square is proportional to t power is constant so velocity is proportional to under root t fine let's go to the next one okay let's take a break you know let's take a small break we are right now the time is 625 okay we will meet at 640 okay are you guys there able to hear me solve this question which direction will be the normal reaction between the rod and the floor it will be vertical isn't it and it is smooth so there is no question of friction okay if friction is there then you can't say vertically upward but right now friction is not there so the only force force of impact between the rod and the horizontal this thing is along the normal okay so and that normal is along the common normal so it will be vertically upwards like this okay that's our option number two okay let's move this one okay others why you keep on retracting your message you don't need to be so much self-conscious you have to use the fact that it is an elastic collision okay so coefficient of restitution is one but the problem is it's not that two point objects are colliding one object is extended fine and it will have some angular velocity so it will not only I mean at every point it will have different velocity isn't it so when you try to write coefficient of restitution equals to one which points velocity will you consider on the rod tell me when you write coefficient of restitution is equal to one that is velocity of separation is equal to velocity of approach then which points velocity will you consider because each and every point has different velocities center of mass is that the correct answer we need to consider the velocity of point that underwent collision for coefficient of restitution formula okay point of impact so a is velocity but another issue is with a is velocity is that okay is that you can find the component of a is velocity in various directions so which direction I should consider which direction of a is velocity which direction I should find the component of a is velocity then I write velocity of separation and velocity of approach Sushant that's not the correct question I should be asking why center of mass okay because here collision when it happens it's a contact okay in a way so when you write velocity of approach who is approaching the point of contact is approaching and that is getting anyways now tell me along which direction along which direction along line of collision Bharati is saying along line of collision is that correct I mean what does this mean what does this even mean line of collision because if I tell you there is this situation suppose one disc is there and suppose there is bigger disc fine so now this disc is coming like this the smaller disc will hit the bigger disc and bigger disc was initially going like that so along which direction you'll consider coefficient of restitution formula when these two disc will hit which direction you might have played the Karen board right when you hit the striker you know not head on then after the collide probably it they'll change their velocity it'll go like this and this will probably go in this direction after hitting so which direction I should write velocity of separation and velocity of approach I should be writing at the common normal fine along the common normal you write velocity of separation and velocity of approach okay so here the common normal luckily happens to be vertical only see point A has no direction normal a point okay points tangent direction and normal direction is undefined so you look at the other surface of contact that is horizontal so for that the normal is defined so when it comes to common normal it has to be normal to both so that's why the vertical direction I'll be considering okay under should write no doubt okay before collision speed is v0 this is the speed of a before collision this should be equal to speed of a going in this direction immediately after collision okay now after the collision the rod also gains angular velocity so let's say it has angular velocity omega okay and velocity of center of mass is v dash okay now tell me which direction the velocity of center of mass will be can I tell something about direction of center of masses velocity v dash do I know this is it vertical does it need to be vertical can it be like us you know inclined from the vertical can it be like this velocity center of mass can it be like this why what is the reason exactly no force in horizontal direction normal reaction is acting on the rod which is vertically upward so it can only change the velocity in the vertical direction if initial velocity is also in vertical direction it will remain vertical okay it can change the direction of velocity to upward but more or less it but it will be remaining in the vertical direction itself okay but if friction would have been there then you can't say that after collision its velocity will be vertical fine so v dash is vertical so let us say after collision also its velocity is downward only and it is v dash fine now because of omega what is the direction of what is the magnitude of velocity of a the magnitude of velocity of a is omega into this distance yes or no this distance is l so omega into l in which direction omega into l along this direction which is perpendicular to the rod okay so this is omega into l all of you able to understand and this angle is theta fine so I can say that v naught is equal to this is velocity of approach now velocity separation is omega l cos theta fine minus v dash I have taken component of omega l in vertical direction okay so this is what I'm getting all right but none of the option satisfies that so I'll relook at my assumption my assumption was that after collision v dash is downwards okay if I flip the assumption and I say that after collision v dash is upwards then the same thing will become plus and then you'll realize option four is correct any doubts cos theta because omega into l the direction of that is perpendicular to rod okay that's that's a relative velocity perpendicular to rod so vertical direction makes angle theta with the perpendicular to rod direction so you have to take component of omega l okay this question the angular loss of the rod before striking is zero fine so about this you know to find the angular momentum so you remember angular momentum formula omega is zero and this thing translates into you know m into v cm multiplied by the perpendicular distance of the point from the line of velocity of center of mass okay so this is the line of velocity of center of mass and this is the point so this is the perpendicular distance this thing is l cos theta that's why it is l cos theta option two the cos theta fine let us move to the next few questions what is this solve this one others okay should I solve now or should I wait solve okay according to the book all your answers are not correct answer is three okay let us see whether it's all right okay so um fine so this is a statement which everybody knows the point of contact plays a crucial okay and that's also a informed disc m and r v not omega not as shown okay so omega not is two v not by r fine and velocity is v not so we need to find the graph of angular velocity of the center of mass sorry angular velocity angular speed of the disc about its center I mean this is not required about its center angular speed will be same about of any point on the rigid body it need not be about a point anyways so you know we need to first understand the graphs okay what is happening in the graph is this here omega has some value it goes to zero and then it suddenly picks up and becomes constant can this ever happen suddenly the velocity picks up suddenly the angular velocity can it pick up until unless there is an angular impulse this can never happen fine and here you can see that and here you can see that there is no angular impulse fine so one is not possible at all all right now uh you know even two you can see that it is you know after becoming angular velocity becoming zero angular velocity is gaining some uh and it is suddenly gaining all right so I would be keeping this also in check all right I would be confused between third and fourth all right but then uh you know this is just a guess right so let us see how we can actually solve this question okay now why this what is the difference in this and this graph tell me anyone when this will happen and when that will happen difference between this and that graph is what yes four picks up reversal of the motion so basically what is happening is that if angular velocity and the linear velocity both become zero together fine then reversal will not happen if both both v and omega goes to zero together okay then reversal will not happen then it will just stop fine so you can see that uh the acceleration due to friction is mu times g over here fine so uh the it will become zero at what time so v is equal to v not minus mu g into t which is zero so t is equal to v not by mu g in this time the linear velocity becomes zero and what about angular velocity first we need to find the torque okay torque is mu mg which is torque sorry which is friction force into r this is what this is uh disk this is a disk so this is equal to mr square by 2 into alpha right so alpha is mu g or 2 mu g by r okay now i can find out the time in which omega becomes zero this is equal to zero and omega not is 2 v not by r minus 2 mu g by r into t this is equal to zero so here also i'm getting t as v not by mu g okay so both linear velocity and angular velocity they are becoming zero at the same time okay so what will happen it will once it comes to rest it will not uh you know i mean it will not it can't start its motion on its own right because of inertia that's a option 3 is correct both angular velocity and linear velocity become zero and hence reversal will not happen are you able to understand what is going on any doubts here that's one same now reversal will happen try solving this one your doubts will be cleared will linear velocity and angular velocity become zero together in this case same angular velocity how much was the angular velocity omega was equal to 2 v not by r so other torque was mu mg into r this is torque and this is equal to i alpha so 2 by 5 m r square into alpha fine so you'll see that m goes away r r so alpha is 5 mu g by 2 r fine so this is alpha okay and a remains equals to mu g so time in which the velocity becomes zero is this v not minus mu g into t so t comes out to be v not divided by mu g okay and time in which angular velocity becomes zero will be like this 2 v not by r minus alpha into t that is 5 mu g by 2 r let's say t1 okay so r r is gone so t1 is 4 v not by 5 mu g okay so you can see that the angular velocity becomes zero first fine so angular velocity will become zero and because the linear velocity is still there the this thing the disc will be you know sliding and because of that slowly and slowly it will reverse its angular velocity in opposite direction okay fine and what will happen is that after some time it will start pure rolling when omega is equal to v by r or v is equal to omega r all right so this is what will happen it will go like this then it will come back to this position where pure rolling starts where angular velocity increases in opposite direction and after that point onwards pure rolling starts so omega is equal to v by r okay fine so that's a option four and why not option three anyone why not option three no assume that is the same omega okay omega final is equal to omega initial then is option three possible if this is happening whatever was the initial angular velocity is the same final as well then it can this happen then also this is not possible okay it will break the conservation of energy if you measure the initial kinetic energy okay so initial kinetic energy if you measure half mvcm square plus half icm omega square omega was 2 v not by r okay fine and finally finally omega should be same according to the graph so omega should be 2 v not by r only and at the same time it is pure rolling as well so omega should also be equal to v not by r so this contradicts okay so omega should be actually half of the earlier omega at you know then only it makes sense fine so like this you don't need to be mathematical all the time you can visualize the scenario and then answer it fine is this the end this one so we have just 10 more minutes we'll solve three or four more questions for today okay symmetry option two right due to symmetry if you take the components horizontal and vertical all the while the horizontal component between p and q this is p by the way this this is p and this one is q so all the while between p and q yeah by net displacement also very good or you can visualize like this that all the while the component of velocity along the horizontal is always this way okay but when you take the vertical component from here to this point it is upward and from there to q it is downward so vertically net displacement is also zero and you can see the velocity also changes direction so net net average velocity along y direction is zero and now x direction is this okay let's go to the next one is it option four only acceleration is centripetal acceleration fine so anywhere it goes it will have acceleration towards the center so you can make out the horizontally their component will get cancelled away but vertically their components will get added that's our option four all right okay so one of those routine easy questions towards the end try solving it accurately and fast four option four punch this is one of those routine questions acceleration of center of the sphere is asked okay now if you are comfortable you can you know don't even involve friction here it is saying that it rolls without sliding okay so if you draw the freeway diagram you will see that this point is at rest so you can use torque equal to i alpha about this point because it is at rest okay so we have mg sin theta over here and we have normal reaction and mg cos theta fine and then of course we have the friction as well fine you can see that all the force passes through this point except mg sin theta fine so you can directly get the torque without even writing for friction so mg sin theta sorry you can directly get the angular acceleration without you know bringing friction in picture if you write torque about this axis all right the mg sin theta into r is it's a sphere right is equal to i about that point into alpha where i about that point is 2 by 5 m r square plus m r square parallel axis theorem okay so like this you can see that this is 7 by 5 m r square this is i so alpha is 5 g by 7 r into sin of theta okay so this is alpha and we know that in a pure rolling a center of masses alpha into r so 5 g by 7 into sin theta is 3 by 5 so 3 g by 7 come here option 4 fine all right let us move to the next question no doubts right if there is any doubt quickly type in option 3 yes it will be 0 all the time it's a pure rolling right it is pure rolling so point of contact will be at rest because the surface on which it is placed that is also at rest relative velocity has to be 0 okay all right so final question for today one more question is there then we are done this one last question happened to be the easiest of all and this is usually the case in exam also so we have this habit of you know solving question paper from question number one and then sequentially going so that's not how you should attempt the objective question paper how much did you get the acceleration of center of mass 3 g by 7 yes or no that's what that was the first question right a center of mass we have already found a center of mass was 3 g by 7 okay and this is h so this distance is how much this distance is h by sine of 37 okay so 5 h by 3 is a displacement so s equals to half 80 square so 3 g by 7 into t square initial velocity is 0 okay so you will get 75 to 35 to 70 by 9 we're getting factor of 70 by 9 so option 2 is correct so I won't even bother to complete the answer because you have to just pick the correct option so this is the answer for this question fine all right guys so that's it for today so we had a session on the comprehension type questions so you might have realized that there is a different way you have to approach these kind of questions and always the first question of the comprehension will take slightly more time than the other slightly get it takes much more time the others so don't get panicked if the first question of reading comprehension takes time because once the first question of comprehension is solved the other questions can be you know done very very quickly because everything is based on single scenario okay all right bye bye