 Fine. So today I plan to complete torque and angular momentum concepts. But before we get into all that, we should be very familiar with the terms which we will be using to understand these two concepts. So I will again revisit few of the concepts that is required to understand the torque and the angular momentum concepts. So let us first analyze how the rigid body actually behaves and how we can analyze the motion of the rigid body. So as I am talking about kinematics of the rigid body. So let's see how we can analyze the rigid body motion first and then we will get into the physics of it. So let us assume that you have a circular object like this. You have a disc suppose. Irreserve is this much? This is better. Fine. So you have a disc which has let us say radius of r. And this disc is not doing anything but just spinning. This disc is spinning with angular velocity omega. Now suppose you take a point over here. You take a point over there which is at a distance of small r. So what this point is doing? This point p, let us say this is point p. What this point is doing? All of you, what this point is doing? This point is doing what? This point is on a disc which is just rotating with angular velocity omega. So what this point is doing? How is the motion of this point? So this point will perform a circular motion. So this point will move in a circle and its angular velocity of this circular motion is omega only. So if you have to analyze what happens to this point after time t, let us say the point, the entire disc has started rotating at t equal to 0. So after time t what happens to this point? This point let us say moves there. This is a new location of point p. Now let us say the angle made by this point, if this is theta, can I say something about angle theta? Theta will be equal to omega into t. See we can get it by, first of all, tell me what this distance is. This distance which is an arc of a circle, let us say that is s. So s will be equal to r into theta. So if I differentiate this, I will get ds by dt is equal to r d theta by dt. Now ds by dt is nothing but speed of the particle. The how fast the particle is moving in a circle, that is the speed of the particle or we can say magnitude of the velocity. This is equal to r times omega. And since omega is equal to rate of change of angle, from here you will get d theta is omega dt. So that is why this relation which we have started with is correct. But if you look at this relation, this one, this is the first one, that is the second one. So in this relation, this is the speed. What kind of speed it is? It is the tangential speed. So right now it is pretty simple. Everything is moving in a circle. What if the entire disk is moving sideways also? Then this particle not only moves in a circle, but also it moves with the disk that side. So this particular velocity doesn't give you total magnitude of the velocity. This just gives you the magnitude of the tangential velocity. So keep that in mind. So tangential velocity is what you are getting as r into omega. And we have theta equals to omega into t also. Now, let us analyze the motion a little bit more. If you have a disk, you have a disk like this and let us say the radius is r. This entire disk is just rotating with angular velocity omega. You need to tell me the directions of the velocity of these points. Now quickly draw it on your notebook. What are the directions of the velocity of these points? It will be tangential? Yes or no? The direction of velocity will be tangential to the circle? Any doubt with respect to that? Correct. So the direction of velocity will be tangential. So I am drawing few of it. It will be tangential. You just have to draw the tangent and this will give you the direction of velocity at that point in time. Now the magnitude of the velocity is fixed. The direction is tangential and the magnitude of the velocity is r times omega, getting it. So this velocity is due to rotation. This is the velocity that is coming because of rotation or you can say spinning rotation or spinning is better. So you can just say rotation or spinning. Now what if this object is not only let us say rotates with angular velocity omega but let us say it also moves that side with let us say vcm. This vcm is the velocity of center of mass. Now can you tell me the velocity of 0.1 then this 0.23 and 0.4? Can you tell me the velocity of all these four points? Initially the object was not going anywhere. Now the object is moving forward and spinning also. Let us first talk about point number one. I will just give you a hint. Velocity of any point will be the vector sum of velocity due to spin plus velocity due to translation. So can you tell me what is the velocity of point number one? The velocity because of the spin is what for point number one radius time omega and in which direction? Velocity of point number one due to spin is what r into omega and this is on the right hand side. Any doubt guys on this? This is because of spin and the velocity and the velocity because of the translation is vcm and that is also in that direction. So basically what you are doing is you are splitting the motion of this disc in two parts. You are saying that this entire motion is made up of these two steps. You can say that you are spinning the object and the object is moving forward without spinning. You can say that when you superimpose these two motions you will get this one. So when you superimpose the velocity gets superimposed because of this motion velocity is r into omega for that point in right hand side direction and because of this motion velocity is again right hand side it is vcm. So total velocity will be some of these two that will be r omega plus vcm on the right hand side. Okay this is total velocity. Now can you tell me velocity of point number two and point number three quickly? Total velocity this is point number one. Total velocity of point number two. This is what? See point number two because of rotation point number two is going in which direction it is going in upward direction. Fine and because of the translation it is going in right hand side direction. Fine so you have two velocities one this way and one that way. So you need to add these two vectors which are perpendicular to each other. Fine the upper side the velocity is r omega and right hand side the velocity is vcm okay. So when you add two vectors that are perpendicular to each other you have to take under root square like this under root a square plus b square fine. So net velocity is in which direction? The net velocity direction you can also find out using vector sum. So you just need to add them vectorally and you will get direction of velocity like this okay. Any doubt guys on these two points? Velocity please type in yes or no? I want everybody to participate okay so let's not have just couple of guys messaging. Anyways total velocity of point number three let us find out. For point number three the velocity because of the spinning is in this direction and because of translation the velocity is in right hand side direction okay. So total velocity in the right hand side is vcm minus r omega okay. So that is total velocity of point number three and similarly you can get total velocity of point number four also. Point number four because of the spinning the velocity is going down and because of the translation it is moving forward. So again 90 degrees so total velocity will be vcm square plus r omega square under root fine and the direction you can find out by doing vector summation. So just construct a parallelogram and this will tell you the direction of the total velocity okay. So like this you can analyze the you can analyze each and every particle also because right now we have just dragged particle on the edge of this disc okay. Similarly there can be a particle which is at a distance small r okay. So everything will remain same because this particle is also moving in a circular way okay just like the particles on the edge okay. So rather than capital r now small r will come in picture but everything else remains the same fine. Now what about acceleration? If we talk about acceleration let us take an example of a disc again. So this entire disc okay let us say it has an angular acceleration alpha fine. Now the way alpha is defined is this alpha is equal to rate of change of angular velocity okay. Now we know that if the disc is just spinning okay. If the disc is just spinning we know that tangential velocity is equal to r times omega fine. So if we differentiate this you will get dVTE by dt is equal to r times d omega by dt fine. So from here you will get tangential acceleration to be equal to r times alpha okay. Now let us try to find out the accelerations of different points. Let us say this is point number 1, point number 2, 3 and 4 okay and also at any point in time the angular velocity is given as omega okay. Can you quickly find out the total acceleration of point number 1? Point number 1 is performing circular motion okay. So it will have a tangential acceleration in this direction okay whose magnitude will be alpha into r where r is a radius okay. Yes it is pure spinning Bharat. Is there any other acceleration? Since the particle is moving in a circular motion there is an acceleration towards the center which is centripetal acceleration fine. So that is what v square by r okay and since v is equal to r omega v square by r becomes omega square r also fine. So we have a situation where where the tangential acceleration is alpha r and radial acceleration is v square by r. So since these two vectors are again perpendicular to each other the total acceleration will be what total expression of point number 1 will be equal to under root of alpha r whole square plus omega square r the whole square fine. So like that you can find out the total acceleration of these two points sorry of this point fine. Similarly you can see that everywhere you have that situation like for example point number 2 has alpha r forward and towards the center you have omega square r. So everywhere if it is spinning going on alpha r tangential acceleration will be perpendicular to the radial acceleration fine. So it's a straightforward thing everywhere the magnitude of the acceleration will be same if the radius is same for the particle all right. Now what if what if this entire disk is accelerating forward let's say this disk has acceleration of center of mass as ACM okay. Now I want to find out accelerations of these two points 1 2 this four point sorry 1 2 3 4 okay and I am just interested in finding tangential acceleration only fine. So I want to know the tangential acceleration right now okay tangential accelerations of 1 2 3 and 4. So it is moving forward as well as it is spinning with angular acceleration alpha fine. So you can you know divide this motion into two parts you can say that it is summation of or superposition of two motions where one is pure spinning with angular acceleration alpha okay and here it is just moving forward with ACM like that you can superimpose okay. Now quickly get the answer you can see that tangential direction is this right so because of ACM the acceleration is not tangential it is radial okay. So tangential acceleration is just alpha into r only for point number 1 okay whereas the radial acceleration this is tangential for point 1 radial for point 1 will be omega square r minus ACM okay. So tangential remains just alpha into r what about point number 2 tangential acceleration of point number 2 all of you quickly type in this is point number 2 what is its tangential acceleration guys you do not have to retract your message if you are wrong it is okay nobody gets it correct in the first time as in all the questions okay. So point number 2 has tangential acceleration because of alpha in this direction so alpha will always have a tangential acceleration okay so this is alpha into r tangential and because of ACM it will move in this direction find that is also tangential the total acceleration of point number 2 in right hand side direction is ACM minus alpha r so this is tangential acceleration of point number 2 and radial acceleration of point number 2 will be simply omega square r okay. So point number 3 has similar expression as point number 1 where tangential acceleration of point number 3 is given as alpha r and the radial of point number 3 will be you know omega square r plus ACM because point number 3 has a centripetal acceleration in the direction of ACM okay whereas point number 4 has tangential acceleration of ACM plus alpha r and radial acceleration to be equal to omega square r okay. So right now there is no relation between let us say ACM and alpha okay alpha can be independent as in the object can be spinning independent of how fast it is moving forward the entire object okay but at times there is a relation between how fast the object is moving forward and how fast the object is rotating can you give me an example where that happens see right now how fast it is rotating has nothing to do with how fast it is moving forward like for example alpha and ACM they can be independent of each other so I am asking you an example where alpha and ACM depend on each other no example is coming in your mind nothing ball rolling Ashutosh has given Diwali example Ashutosh that that is nice but then not entirely correct Bharath you just there is a scroll against the video right so just take it and put it on the I mean just scroll it till the very end do you the lag will get off okay no not entirely the spinning football also see how fast the football is spinning how fast the football is spinning has nothing to do with how fast the football will move forward okay so I can understand you want to give example from the football but you can take another guess okay so like for example like one of the student is telling car moving forward okay so how car is moving forward the wheels are rotating okay now how fast the car moves depends on how fast the wheel is spinning isn't it so there is some relation between how fast the wheel is spinning with how fast the car is going forward all right so similarly any object that rolls on the surface for example you take a sphere or you take any disc a ring so how fast the ring or the disc or the sphere spin depends also on how fast it is moving forward fine so there are cases where acceleration ACM and alpha they are related to each other okay but it is not a rule that they should be always related okay so we'll get into all those kind of scenarios little bit later okay so now it is we are good to start the topic talk okay so any doubts on whatever we have done till now please type in quickly yes or no those who are facing lag can just take that there there's a this thing right scroll the button next to the down to the video you just take it and slide it till the very end okay so that the lag is gone any doubt guys please type in quickly okay all right no doubts so we can start this topic torque so like what we have discussed earlier torque is nothing but the cause of rotation okay so this will create an angular acceleration alpha okay and if torque is present there will be angular acceleration if torque is not present angular acceleration will be zero okay angular acceleration is rate of change of angular velocity okay so if initial angular velocity is zero it will remain zero in case torque is absent okay now the way torque is quantified is first of all torque requires I think we have done this last class also so I'll quickly you know revisit this concept how torque is defined so torque depends on force first of all okay what with what amount of force you're trying to rotate a particular object okay let us assume the object to be this stick which is free to rotate about this axis about the axis which is passing through the end and coming out of your screen okay now suppose you're applying force okay you're applying force from here so definitely depends on what is your magnitude of the force that will determine how fast this door is rotating or this stick is rotating isn't it so torque definitely is proportional to what is the magnitude of the force okay and also you may see that when you apply force let us say point number one and same force if you apply at point number two okay to achieve the same angular acceleration you need to apply more force over here and you need to apply lesser force over there to achieve the same angular acceleration right so what is the difference between these two forces if you have same magnitude of force the difference is the distance of the force from the axis of rotation fine so further away the force is from the axis of rotation easier it is for you to rotate fine so the amount of torque that get generated is also proportional to how far away you are from the axis of rotation let us say that this distance is r fine so torque is proportional to how far away you are from the axis of rotation so I am representing it like a vector so that is why I am taking the magnitude of r vector r vector is starting from axis of rotation till the point of application of force okay now consider another case where you have this stick which can rotate about this axis and you are applying let us say force in this direction a large amount of force will this stick ever rotate so you can you know visualize the scenario and you can easily arrive to the conclusion that this stick will never rotate okay so it will not rotate right so not only the the rotation depends on the magnitude of force and its distance but it also depends on what is the angle between r and f vector okay so if it is 90 degree then it can easily rotate and then if it is for example this angle if it is less than 90 degree angle then you need to apply actually more amount of force to achieve the same angular acceleration right so you need to apply more effort so definitely it has some relation with angle theta okay so if you decrease angle theta which is the angle between let us say this is f and this vector is r so angle between f and r is theta right so if you increase the angle from 0 to pi by 2 okay then your torque should increase fine so it will be easier for you to rotate so there has to be some dependence on angle okay and we can take any function of that angle but that function should be increasing function between 0 to pi by 2 okay so we can think of it as a sine function so we can say that torque is proportional to sine of angle between r and force all right so if I combine all these findings if I combine all these findings we know that torque is 0 when force is 0 no matter what is the angle and distance similarly torque is 0 when r is 0 okay so if I apply keep on applying force along the axis of rotation then there is no effect of that force fine similarly if theta is 0 torque will be 0 okay so any of these quantities 0 should give you torque to be 0 fine so I will take just a very simple expression for the torque which is multiplication of all these three okay now you can argue that why just multiplication of all these three why not r cube into f square divided by or into sine 3 by 2 theta that can also be there okay but everyone has chosen this most simplistic expression which is simply rf into sine of theta okay so this is the magnitude of torque and in case you want to write it in a vector format you can also do that it will be equal to r cross f fine so direction of the torque is given as the cross product direction between r and f all right how you find the direction of cross product you have let us say vector a like this and this is let us say vector b okay then what we do we align your hand in the direction of first vector let us say first vector is a right now and I am trying to find a cross b direction fine so align your hand in the direction of a and then curl your hands towards b your thumb will point towards direction of a cross b okay similarly if you want to find out direction of b cross a then you align your hand in the direction of b and then go towards a so now thumb is pointing inside the screen okay earlier it was coming out of the screen so like that you can easily get the direction of the cross products okay that that is one way you can you know talk about the direction of the cross product the another way to take care of the direction of torque is by finding out whether it is anticlockwise or clockwise torque for example if you take a stick let us say this is f1 and there is another force let us say this is f2 okay now you can easily find out because of f1 the stick will try to rotate in this direction fine so because of f1 the torque you can write it you know direction of torque you can just represent by having this direction you can say that it is like this the direction of torque and because of f2 the stick will try to rotate in that direction so direction of f2 will be like that or sorry direction of torque due to f2 will be like that so clearly these two are having opposite sense of rotation so torque due to these two when you are summing it up you need to subtract because torque due to f1 is in opposite direction of torque due to f2 fine so like that you can have a sense of direction for the torques fine I find this to be simplest of all this is how you can keep track of direction whether it is clockwise or anticlockwise torque okay because you can visualize using this okay any doubt guys anything anything you want to ask please type in okay as you thought it's like saying right hand side is positive okay you can take anything to be positive but yes convention is that usually people take anticlockwise to be positive so let's follow that only okay now let us try to find out the physical interpretation of the torque formula okay now we know that torque's magnitude is magnitude of r magnitude of f into sin theta okay now we will try to find out physically how we can visualize the torque because once you visualize it becomes easier for you to get the value of torque all right this can be written as f into r sin theta okay now let us see in the diagram what is r sin theta so again take the example of the stick only so this is the stick which is able to rotate about this axis which comes out of your screen okay so let us say that you are applying force at some angle okay and this dotted line represents r vector direction and this one is the f okay so this is theta okay this is the magnitude of r this length is r distance between these two vertical lines is r okay now where is r sin theta let us try to do that let us try to find that so if you drop a perpendicular from this point which is axis of rotation onto this line let's say this is 90 degree okay now this angle is theta so this is also theta okay this length is magnitude of r okay so this distance is what this distance this distance will be magnitude of r into sin of theta isn't it so this is right angle triangle this is happiness this is theta okay now let us find out what is this r sin theta okay now r sin theta if you can clearly if you carefully watch it r sin theta is the perpendicular distance of line of force from axis of rotation isn't it this is a line of force so you just extend the line of force and drop the perpendicular from the axis of rotation the length of that perpendicular is r sin theta okay so how you get the perpendicular distance so all you have to do is just extend line of force this is step number one and then drop a perpendicular on line of force from axis okay and that perpendicular length is perpendicular distance fine so if you know that if you know the perpendicular distance torque is simply f into that perpendicular distance all right because r sin theta is perpendicular distance only all right so you can quickly find out the magnitude all right and the direction you can easily track using sense of rotation like this fine so you can track the direction and magnitude separately you can say it is clockwise torque and this much magnitude like that fine so this is one way to find that torque you just find out the perpendicular distance of the line of force from the axis of rotation and multiply with the force you'll get the value of torque okay any doubt guys please type in i'll move to the next interpretation meanwhile line of force that passes through the axis if line of force passes through the axis what is its perpendicular distance from the axis it is zero right so if line of force passes through the axis of rotation the torque will be always zero okay there should not be any confusion on that let's interpret the torque like this torque is r into f sin theta let us see what is f sin theta physically so again we'll take the example of the stick difficult to draw the stick horizontal hmm so about this axis suppose this stick is able to rotate this is r vector okay and you have a force acting at that point let us say this is f vector fine this is the angle between these two vectors theta okay so this is theta now f sin theta is what you can say that you can split this force in two direction one is vertical direction and one is horizontal component fine so you can see that horizontal component of the force is f cos theta and this component is f sin theta okay so it's like you have divided the force you have divided the force into two you're saying that your f is the vector sum of f cos theta and f sin theta okay and clearly f cos theta is passing through the axis of rotation so f cos theta cannot produce any torque whatever torque is produced is because of f sin theta okay and what is f sin theta f sin theta is the force component that is perpendicular to r vector fine so you can write torque as r into component of force that is perpendicular to r okay so like this also you can find out the value of the torque so any of these two ways you're comfortable with you can find the value of torque any doubt guys anything that you want to ask now so let's take a few numericals now so suppose you have a stick okay which is let us say three meters long this stick is three meters long it's a massless stick okay there is no mass you are applying force from here let us say this force is of the magnitude of 10 newtons and then you are applying another force this has the magnitude of 20 newtons and this angle is 30 degrees okay you need to find out the torque about point number one due to all the forces acting on it this in that stick is hinged at point one all right the this point of application is the center of the rod okay as in this distance is 1.5 and this is at the edge okay 45 newton meter all of you see because of the 10 newton first of all the torque sense of rotation is in this direction okay and because of the 20 newton the sense of rotation is this direction so both the directions are same so I need to add the torques due to both of them okay so torque due to 10 newton is what I'm assuming that this is 90 degree okay so torque due to 10 newton is 3 into 10 which is 30 the SI unit is of course newton meter force and length they are multiplied okay and torque due to 30 newton how will you find that you know you can find it in two ways you can find the perpendicular distance on the line of force or you can find the perpendicular component of the force along the length so finding perpendicular component of the force in this case is easier so I'll do just that so perpendicular component of the force is f sin 30 okay and r is 1.5 1.5 into f sin 30 which is 1 by 2 okay so this comes out to be 15 newton meter so total torque is some of these two because these two are trying to rotate in the same direction so that is 45 newton meter okay this is torque due to 20 let us take up a few more this is a square of side length two meters this is two meter square okay this square can rotate about this center axis which comes out of your screen so let us say that you are applying forces like this this is let's say 10 newton okay then you have a force of 20 newtons then you have a force of 10 newtons again this angle is 45 degrees okay and then you have a force of 20 newton like this and this angle is 30 degrees okay that axis which is which is the which is passing to the center of the square coming out of your screen okay we have one answer 30 plus 10 root 2 10 plus 2 20 root 2 so we have different answers we are getting any other answer okay now we have three different answers any other answer 20 minus 10 root 3 another answer okay let me do it now all of you please pay attention so first of all torque due to 10 newton the sense of rotation because of 10 newtons is like this okay sense of rotation due to 20 will be like that fine so torque due to 10 will get added on to torque due to 20 and what is other there are two 10 newton forces let me call this torque due to 10 newton which is the next 10 newton this will be equal to what torque due to this force is what what is the sense of rotation for this torque this torque sorry this force is passing through the axis okay so torque due to this 10 newton is 0 because it passes through the axis okay now you can see that 20 newton force is making some angle okay so what I will do I will split this force into two components one like this which you can say that it is 20 sin 30 okay and one like that this will be equal to 20 cos 30 okay so this 20 newton force you can say it is a combination of these two forces 20 sin 30 upwards and 20 cos 30 horizontally okay now you can take them separately you can see that torque due to let us say 20 cos 30 is trying to rotate in this way isn't it 20 cos 30 is trying to rotate the square in that way which is in opposite direction of torque due to 10 and 20 and similarly torque due to 20 sin 30 this also is trying to rotate in that direction okay so total torque will be torque due to 10 plus torque due to 20 minus torque due to 20 cos 30 plus torque due to 20 sin 30 why I have split the force of 20 newton because it will be easy for me to visualize then this is the total torque fine now torque due to 10 newtons is what this is a line of force drop a perpendicular from the axis this perpendicular length is how much this perpendicular length is half the side of the square which is 1 okay so 10 into 1 is torque due to 10 newtons plus torque due to 20 is what this is a line of force drop a perpendicular on line of force okay so that perpendicular distance again is 1 so 20 into 1 so this is torque due to 10 plus torque due to 20 now torque due to 20 cos 30 is what 20 cos 30 again the perpendicular distance is 1 only this is line of force of 20 cos 30 okay so 20 cos 30 which is 20 into root 3 by 2 into 1 plus 20 into sin 30 which is 1 by 2 into 1 okay so I can cancel out 20 like that and this is 10 so we'll get 10 plus 20 minus 10 root 3 minus 10 so you will get 20 minus 10 root 3 okay so we have this as the answer any doubt guys anything anything you want to clarify about this nothing okay fine now we are going to learn about something more important about the torque so if total torque total torque is 0 it means alpha is 0 why is the other 10 0 why is other 10 0 okay so you know when I am speaking you guys should focus on the screen so don't get engrossed in solving the question okay so first listen and then you can anyways so this 10 Newton this 10 Newton force is passing through the axis of rotation you can extend the line it passes through the axis of rotation because this angle is 45 degrees and hence the torque due to 10 Newton this 10 Newton is 0 okay fine so if total torque is 0 about the axis then angular acceleration alpha is 0 this situation is called rotational equilibrium rotational equilibrium okay and similarly if summation of all forces is 0 acceleration is 0 fine this is called translational equilibrium all right so it may be possible that the object has translational equilibrium but no rotational equilibrium all right for example this one can you quickly find out the torque let's say you have a stick like this you are applying a force in this direction this is f and this is f okay the total length of this rod is l okay now find out the torque about any point about any point along the length of the rod what is a torque okay let's assume that there is a point which is at a distance of x this is x okay so this length will be what this length will be l minus x okay so torque due to this force this torque will be x into f okay and torque due to that force will be l minus x into f okay now let's try to see that the sense of rotation about this point so this force is trying to rotate there like this and that force is trying to rotate like that okay so that the sense of rotation is same so the torque will get added up so total torque is x into f plus l minus x into f okay this is the total torque so total torque comes out to be equal to l into f fine so here you can see that the object has net four zero one is down and other is up so net force is zero but net torque is not zero so this object is in translational equilibrium but it doesn't have rotational equilibrium got it similarly if you have two forces if you have one force like this other force like that f and f total torque about about any point on the axis is zero okay this rod will not rotate but it will move forward because net force is not equal to zero getting it so the object that has rotational equilibrium need not have translational equilibrium and vice versa all right but when I say that I say that rigid body if a rigid body is in equilibrium when I say that statement it means that it has rotational plus translational equilibrium fine both are there otherwise I would specify that it has just translational equilibrium or rotational equilibrium okay so if it has rotational equilibrium I can find the torque about any axis okay I can find the torque about any axis or about any point and equate that to zero okay similarly if I have translational equilibrium I can find the net force along any direction and equate that to zero find these these two things are clear that if a rigid body is in equilibrium I can find the torque about any axis and equate that to zero and similarly I can find the force in any direction and equate that to zero so this is the condition for the overall equilibrium of a rigid body if net force is zero it will not move anywhere the center of mass will be stationary and if net torque is zero then it will not rotate also so these are the two conditions for a rigid body to be under equilibrium any doubt guys anything no doubts okay so we can start solving questions right on equilibrium okay the first one so this is a seesaw typical seesaw which you play the total length of this seesaw okay which can be assumed as a rod is l okay now here you are applying a force f all right which is at a distance of l by 3 from the axis of rotation the axis of rotation is horizontal coming out of your screen so this is l by 3 so what force should you apply there for this rod to not to rotate what should be the force so that the rod is under equilibrium any such question is to draw free body diagram okay even if it is like simplest of all so the free body diagram of the rod will draw this is the rod so this is force f okay and this is the unknown force let me call it as f1 we don't know what is that force there will be a force from the hinge okay the hinge force is let us say n is the hinge force okay now all of you have missed one thing or all of you have assumed one one thing is that the rod is massless you've assumed it randomly okay just because it is not given or you might be in a hurry of getting the answer you have ignored the mass of the rod so if there is a mass of rod your answers are not correct all right so right now the assumption is the mass of the rod is zero now there is first of all rotational as well as translational equilibrium so for translational equilibrium I'll have net force along the y-axis to be zero so f plus f1 minus n this is equal to zero fine so this is equation number one all right and I can have sorry rotational equilibrium as well okay now rotational equilibrium I can write the torque equation about any point about any axis all right so it is up to me about which axis I can find the torque the easiest one of them to consider is about this point okay what is so special about this point to find the torque why we are trying to find the torque about that point and equating to zero why not other points can you message so if you find the torque about that point where the normal reaction is applied then the torque due to normal reaction will become zero so when you write torque equation normal reaction the n will not come in the equation itself okay so that is the only reason if you write torque about any other point also you'll get the same answer so torque about point p is zero so f1 is trying to rotate that way and f is trying to rotate in this way so both sense of rotations are in opposite directions so when you write the total torque you need to subtract it okay so torque due to the left hand side force is l by 3 into f minus this length from here till there it is 2l by 3 now so minus of 2l by 3 into f1 is equal to zero so from here you look at the value of f1 to be equal to f by 2 okay so just by using force equation you'll not be able to find f1 because there is another variable n is coming up okay now when you use f1's value on the equation 1 you can even get the value of n also n is equal to f plus f1 this will give you 2f sorry 3f by 2 okay so like this you can solve this particular question okay in case of any doubt please type in yes or no see this is the first time probably many of you are using torque equation to solve for the forces okay till now we have been using net force equal to zero or force equal to mass emigration and that was sufficient for us to solve the entire question okay but now since the object has a rotational equilibrium also we need to equate the net torque also to be equal to zero okay otherwise we will not be able to solve the equation because there are more number of variables okay now there is a small concept on center of gravity write it down so that we can solve different questions okay so center of gravity is a point which acts as if entire weight of the object concentrated at that point or you can say that the value mg that mg force because of the gravity is applied through that point okay this is different from center of mass because center of mass is a point where entire mass is concentrated and here in center of gravity this is a point where the entire weight is concentrated so these are the two different things okay so let us try to find out what is the location of the center of gravity so you just randomly take a big rigid object like this okay it is made up of multiple point marks point masses let us say m1 m2 m3 and so on are the point masses on this big rigid object okay and let us assume that center of gravity somewhere over here this is you can say center of gravity okay now if entire weight of the object is concentrated on this center of gravity now can you tell me what is the net torque because of gravity about the center of gravity what should be the net torque about the center of gravity because of the gravity it should be zero fine so if this is the point where entire mg is acting then the torque due to the gravity force which is mg should be zero about that point okay so let us try to find out the location of such point by tracking the torque due to all the points with respect to this one okay so let us say that this is point m1 okay so I can take this as origin all right and if that is the origin this is my r vector okay and this is my mg okay so one thing is pretty clear here that mg will be always vertically down okay no matter what point you take all the time the forces will be vertically down okay so whenever you have to find the torque the perpendicular length will come out to be the x coordinate of that particular mass getting it so m1's x coordinate is let us say x1 so perpendicular length this is x1 if you take another point here let us say you take m2 on the left hand side this is x2 and this is let us say m2g okay now you can say that torque due to m2g is in opposite direction of torque due to m1g so they should subtract okay that is automatically taken care by sign of x coordinate so x coordinate of m2 is negative it is on the left hand side if this is origin then x1 is positive x2 will be negative fine so torque due to m1g will be equal to x1 into m1g all right torque due to m2g it will be equal to x2 into m2g all right and so on and so forth so these these are the torque due to all the mg is acting on different point masses so total torque about this point is sum of all the torques torque due to m1g torque due to m2g and so on so this will come out to be m1x1g plus m2x2g like that okay this should give you zero okay now if I rearrange the term I will get m1x1 plus m2x2 and so on multiplied by g to be equal to zero what is the assumption here g is a constant okay your rigid body can be large enough so that g doesn't change so if g is so that g changes so if g is a constant then only you can take it outside hence if I if I further rearrange this term I will get m1x1 plus m2x2 so on divided by m1 plus m2 like that this also will become zero from here okay so the center of gravity should be such point whose weighted average with mass and its x coordinate should become equal to zero okay and what is that point for which this is equal to zero that point is center of mass only guys okay so if g is constant if g is a constant throughout the rigid body then center of mass coincides with center of gravity okay this is the case with many of the objects like you know if they do not specify whether g is constant or not you should assume it to be a constant and then you can consider that the center of gravity coincides with center of mass okay any doubt on this particular concept please type in yes all right so let's take questions okay so now suppose you have a rod like this whose mass is m mass of rod is m you have a hinge over here this rod can rotate about this point you are applying a force from this end as f okay this end the force is let us say f1 to balance the rod fine so the rod is able to rotate about this axis which comes out of your screen this length is l by 3 where l is the length of the rod you need to find what is the value of f1 okay so we have two different answers Bharat has retracted his answer why you guys have to delete whatever you write nobody's going to judge you okay Pratik has messaged 2f same thing as what Sukrit has written Sukrit is live now announcing it to everyone 2f okay almost everybody getting the same answer so I'll just quickly solve it here so first step is what draw the free by diagram okay so this is f this is f1 okay and here you have normal reaction n all right and then you will have weight of the rod that is applied from the center of mass which is mg okay so this length is l by 3 see when we have done laws of motion chapter we never cared about the point of application of force how far it is from the axis and things like that but in this chapter we need to not only keep track of the direction of force but also exactly where the force is applied so that the torque is found easily okay now first we'll write the force equations so f plus f1 plus mg minus n is equal to 0 this is a long vertical direction horizontally there is nothing to write so no point writing net force along 100 direction to be 0 okay now I want to find the value of f1 so I'll write the torque equation about this point about point p because if I write about point p n will not feature in the equation since torque due to n is 0 about that point okay however you can write the torque equation about any other point you look at the same answer but then you have to then use utilize this equation as well to solve that torque equation because n will come then anyways so torque about p is 0 so due to f the torque is in this direction due to mg the torque is in that and due to f1 it is like this so due to mg and f1 the torque should get added up and because of capital F the torque is in opposite direction so it gets subtracted so torque due to mg plus torque due to f1 minus torque due to f magnitude wise okay I'm not adding them like vectors I'm adding their magnitudes this should be equal to 0 about point p now torque due to mg is what mg into this distance from here to here this distance is l by 2 so that distance is l by 6 so mg into l by 6 plus due to f1 the distance is 2l by 3 so f1 into 2l by 3 minus torque due to f that is f into l by 3 this should be equal to 0 okay so like this you can solve and then this l is also gone so f will come out to be f1 plus mg by 4 oh I need to find f1 not f so f1 will be equal to f by 2 minus mg by 4 okay so like this you can do this particular question now let us try out another one any doubt guys please type in quickly I'm moving to the next one okay here is the next question you have a ladder this red color is the ladder red color line this ladder has mass of 10 kg this ladder has mass of 10 kgs it leans against a smooth vertical wall so this is a wall that is smooth so mu is 0 okay it it makes 53 degrees with it so this is 53 degrees okay the other end is rested on the rough horizontal floor so this is rough so it has friction you need to find out the normal force and the friction force that the floor exerts on the ladder so what is a normal force here and what is the force of friction over there you need to find out assume that length of the the ladder is uniformly distributed mass and total length you don't need it actually but you can assume it to be l okay what else is given is sine of 37 even though you should remember it but I'm giving it right now sine of 37 is 3 by 5 okay Sriram is asking explain the third equation see I have what I've done I've added talk to mg and talk to f1 why because mg is trying to rotate about this point the rod in this direction and f1 is also trying to rotate in the same direction so talk to mg and talk to f1 will get added up okay and then talk due to f will get subtracted because f is trying to rotate is in opposite direction okay so this is anti clockwise and this is clockwise about point p so this is what this equation is about talk to mg talk to f1 minus talk to f is equal to 0 yes so that the ladder is stationary you need to find the value of normal force and friction so that the ladder is stationary first step draw the free by diagram that is friction you are getting yes that's correct others who is busy plays busy plays what does it mean no friction is not 60 okay should I do it gaurav is busy busy oh busy is like wizard wizard plays what plays plays as in drama so drama of a wizard okay let me do it now so first step draw the free by diagram this is the ladder okay so there will be normal direction from the vertical wall in which direction perpendicular to it that that's why it is called normal reaction isn't it so this is let us say n1 okay and you will have from the horizontal surface normal reaction as well as friction now this is a static friction okay even if coefficient of friction is given you can't write it as mu times n so just assume it to be something like you know fr it's a variable gaming username that's nice okay so now you will have mg force okay is there any other force there's no other force right so now let me assume my x and y axis horizontal and vertical okay so this is my x this is my y now net force along horizontal direction should be equal to zero okay net force is what n1 minus fr in horizontal direction okay so n1 minus fr this has to be equal to zero so this is my first equation okay along vertical direction I'll have n2 minus mg is equal to zero this is my second equation okay now let's count number of variables n1 n2 and fr there are three variables but I have only two equations from where the third equation will come third equation will come from the torque equation it is under rotational equilibrium fine so I'll find the torque equations I'll write the torque equation now if I write torque equation about this point about the bottom point then both n2 and fr torque due to both of them will become zero okay so my torque equation will be simpler so I'll write torque equation about point p you can write the torque equation about any point of your like okay so torque about p we need to write now mg is trying to rotate the rod in this way whereas normal reaction is trying to rotate that way if you if you see with respect to p all right so torque due to mg get cancelled out from torque due to n1 all right now torque due to mg magnitude of that will be what just extend this line of mg and drop a perpendicular from the axis of rotation okay so this distance is a perpendicular distance which is what if this is 53 this has to be 37 degrees okay this length is l by 2 okay so the perpendicular distance is l by 2 cos of 37 okay this perpendicular distance multiplied by the mg this is a torque due to mg okay minus normal reaction n1 multiplied by its perpendicular distance now how you get perpendicular distance of n1 just extend this line of force okay I am extending the line of force like this and then drop a perpendicular from the axis so this is a perpendicular distance this one this distance is a perpendicular distance for n1 okay this distance is what this distance is l into sin 37 okay so n1 into l sin 37 this is equal to 0 so l will get cancelled out and you will have cos 37 as 4 by 5 because sin 37 is 3 by 5 okay so 4 by 5 into 1 by 2 into mg this is equal to n1 into sin 37 is what 3 by 5 okay so I have 5 getting cancelled out and this is 2 so from here I will get the value of n1 to be equal to 2 mg by 3 fine so from equation 1 friction force should be equal to n1 so that will be equal to 2 by 3 mg okay and n2 we have found out initially itself which is equal to mg okay so like this you have to solve this particular question any doubt guys with respect to this please type in quickly which you place any doubt okay let us quickly take one more question then we will move to next concept this is a rod of mass m of mass m and length l okay this rod is hung using the string so string should be thin so you have hanged the string like that all right so this angle is 30 degrees this angle is 60 degrees okay you need to find the values of tensions what is the value of tension in these two strings and also you need to find out condition you need to find the condition so that it remains stationary first find the value of tension the tensions are different okay there are two different strings mg root mg mg root 3 see this this rod is stationary right so the net force acting on it will be 0 the what are the forces acting on it you can split the force saying that this is mg let's say t1 and this is t2 okay so horizontally you can say that t1 cos 30 minus t2 cos 60 should be equal to 0 okay so t1 root 3 minus t2 should be equal to 0 is the first equation and vertically t1 sin 30 which is vertical component of tension plus t2 sin 60 should give you the value mg okay so t1 plus root 3 t2 will be equal to 2 mg this is your second equation okay so just solve these two questions you can multiply the second equation by root 3 so it become root 3 t1 plus 3 times t2 is equal to 2 root 3 mg okay and then subtract this equation from that you will get 4 times of t2 to be equal to 2 root 3 mg okay or t2 will come out to be equal to root 3 mg by 2 Bharath don't be in a hurry to solve the question you will all the time get negative marks so this is t2 so t1 is t2 by root 3 so t1 is t2 by root 3 so that will be mg by 2 fine so this is how you solve this particular question to get the value of t1 and t2 all right now this rod is under rotational equilibrium also okay so the total torque about any point should be equal to 0 so you can say that torque about this point p let us say that is 0 okay so like that you can get some condition to be satisfied okay any doubt guys no doubts so the next concept is torque equation so torque is what torque is cause of rotation okay the torque is the cause of rotation so if cause is absent rotation will not happen but suppose the cause is present okay suppose the torque is there so we don't know right now how much torque will be applied to cause how much angular acceleration away it is almost 11 now so let's continue anyways this will be getting recorded so in case you feel a little exhausted you can take a break and you can scroll it back so that you are you know you can catch up later and also so it is only getting recorded so torque is a cause of rotation fine we are going to find out now how much cause will create how much effect this is cause and the effect is angular acceleration alpha fine so this is torque which is tau so an equation between cause and effect as in the equation between torque and alpha will be the torque equation just like the equation between force and acceleration is force equation equation between torque and alpha is torque equation so this is very important this is what we are going to learn now okay and since we are talking about relation between torque and acceleration we can't say torque is zero and then try to find the relation right so we'll just first assume that there's a torque and because of torque the object will rotate fine and if you are considering a rotating rigid body if you have to consider a rotating rigid body then you can take the first example as the spinning itself see like i told in case you have to eat you can eat it you know you can scroll back the class and then watch it from there okay so suppose it is spinning fine or let me take a simplistic example i'll take a regular shape object let me take a disc a disc that is spinning so let's say this disc is spinning about this axis that is coming out of the screen fine now since this is spinning let us assume it has angular acceleration alpha all right now what i'll do i'll try to find out the total torque acting on this spinning object about the axis fine let me add up all the torques for the point masses and then see what expression we will get all right so first of all let us take this small mass over here this is small mass let us say mass is m1 okay so the torque on m1 is what r cross force on m1 all right so this is r and the force on m1 is let us say in this direction fine you can assume any direction of force from where this force is coming this force could be an external force to the disc or nearby particles are exerting this force could be anything now this r is in radial direction isn't it this r is from the center to this point okay so the direction of r is radial so keep that in mind that cross product with radial force plus tangential force i have divided the force into two components radial and tangential fine so once you take a cross product of r with the radial force how much that will be r cross f r will be what all of you how much will be this okay so i think all of you taken your breakfast should we start so value of this is zero because r is also in radial direction and force f is also in radial direction this component of force okay you guys are able to hear me right others are you guys there i can see 19 people are online only one is applying okay so torque due to this force radial component of force will be zero so r cross f r is zero okay and what about r cross f tangential you can see that tangential force is perpendicular to r so i can just write it as r into f tangential because cos of 90 will come which is one anyways all right and i also know that tangential force is mass into tangential acceleration using newton's second law okay and tangential acceleration can also be written as alpha into r okay why i have written tangential acceleration to be alpha into r is because the angular variables including alpha will be a constant for entire rigid body fine so entire rigid body will have same alpha so it is better to better to write down in terms of alpha okay so torque will be equal to r square m into alpha right now i am talking about particle one let us say that it's the distance is r1 so torque due to the force acting on m1 will be equal to m1 r1 square into alpha fine similarly torque due to the force acting on m2 will be equal to m2 r2 square into alpha so like that you can just keep on writing fine so when you sum it up all the torques you'll get total torque okay total torque to be equal to m1 r1 square alpha plus m2 r2 square alpha and so on okay so you can take alpha common it will become m1 r1 square plus m2 r2 square and so on into alpha now what is that term inside the bracket this term inside the bracket is summation of mass into the perpendicular distance from the axis square okay so this is the moment of inertia about the axis let us say this axis is axis p so i can say that axis p moment of inertia into alpha all right so this is equal to the torque this is total torque now the total torque will have two torques torque that is due to the force that is external to the system and this one will be added to the torque which is internal to the system what do i mean here is that you know the force acting on m1 could be coming from outside from someone from outside might be applying force or the particles near to m1 may also apply force on m1 okay so when you add up all the torques some internet torques will also get added up okay now let us see what is that sum of all internet torques in order to understand that we'll just take example of two masses two point masses they are applying forces on each other suppose this is force due to the second on to the first then according to newton's third law equal and opposite force will be applied on to the second due to the first okay now suppose i want to find the torque about this point okay so let's say this is position vector r1 and this is vector r2 so if this is f this automatically becomes minus f all right so right now i'm trying to find what is a total torque on the system due to just these two forces on these two masses okay so torque on m1 is r1 cross f okay and torque on m2 is r2 cross minus f so like this it will come okay so total torque will be sum of these two so it will be r1 minus r2 cross with f okay now r1 minus r2 is what this is r1 this is minus r2 so this vector that this vector is the line joining two and one okay so this is r1 minus r2 and r1 minus r2 is parallel to f only hence the cross product becomes zero fine and that is the reason why summation of all internal torques will become equal to zero okay the summation of all torques is simply summation of all external torques only and that is equal to ip into alpha so if you find out the total external torque about us are you able to hear me guys okay so if i find out the total external torque about the axis then sum of all external torque will be equal to moment of inertia times alpha okay now we have made some assumptions while deriving this what is the assumption the axis is fixed the object is just spinning about the axis okay so this equation summation of all external torque is equal to i about alpha you cannot use about any random axis this is valid write down valid only for two cases case number one about the fixed axis see in case you are facing a lag issue please drag that what is that cursor below the video to the extreme right i think those who are facing lag they are minimizing and maximizing the video the video the window of the video again and again so just be online all the time live as in so it is valid only for two cases about the fixed axis or about the axis that passes from center of mass in case it is still not getting resolved you can close the window and restart the video okay so all of you clear about it this equation is valid about the fixed axis or about the axis that passes through the center of mass all right so the torque which is r cross f will change depending on which axis you are taking as well as moment of inertia will change depending on about which axis you are finding the or about which axis you are applying this particular equation any doubt guys please type in yes or no quickly but at the for the second condition where external torque is equal to icm into alpha there is another derivation for it okay so we are not getting into derivation of it right now but then right now just take it as it is all right and if you are interested in deriving it you can write that acceleration of any point p will be equal to a cm vector plus you know r cross alpha so from here you can derive okay but then we are not getting to derivation of all that it's not necessary it will take some time and then you will arrive to this equation also okay so this equation you can apply about the fixed axis or about the axis that passes through the center of mass okay fine Bharat I'll share the derivation for the second equation also okay don't worry but I don't want to spend 15 minutes on that right now anyways so let us try to use this equation torque is equal to i alpha for different conditions okay so this equation is very similar to that force is equal to mass times acceleration so I hope you guys still remember how much trouble this equation has given to you okay now double that this is what the trouble this equation will give it to you okay so you can expect a lot of varieties of questions which uses this particular equation okay so we'll start with the simple straight forward ones okay so suppose you have a rod this is the rod okay mass m and length l okay now this rod is free to rotate about this end the axis that is horizontal coming out of your screen Bharat it's not that difficult I just exaggerated it okay don't take it like that so now find out the value of angular acceleration for this rod once you release it from the horizontal position step number one is what identify if there is any fixed axis so that you can use torque is equal to i alpha okay you can have this as a one of those fixed axis so about this axis we are going to use torque is equal to i alpha right now when you see the forces you'll have forces which are like this you have a force from the center of the rod which is mg okay then there will be hinge forces let's call this hinge force as f y and let's call this force as f x okay now you can apply torque into i alpha to find the value of alpha because alpha is what is asked alpha will not appear in the force equation so you have to use torque equation so about this point p if I write torque this is 90 degree right so torque due to mg is mg into l by 2 this is torque which is left hand side this should be equal to i alpha i about p into alpha you have to equate now i about p which is passing through one of the ends is ml square by 3 into alpha okay so alpha will come out to be 3g by 2l okay what is the unit radians per second square fine now I want you to find out the values of fx and fy these are the hinge forces how much the rod starts from the rest others what is the answer let's first see fx now along x direction is there any acceleration if this is the rod if omega is 0 along the x axis there is no acceleration had there been omega then along x axis readily the acceleration would have been omega square r okay then if this is the force let's say this is fx then I can say fx is equal to minus of m omega square r because acceleration is a positive force so minus will come in so direction of fx has to be in this way but omega is 0 right since omega is 0 fx is equal to 0 okay now let's see fy if we take this rod like this I am trying to use net force along vertical direction is equal to mass into acceleration of center of mass this is mg this is fy okay and that is what we have taken as fx okay now net force in vertical direction is what mg minus fy okay this should be equal to mass into acceleration of center of mass what is the acceleration of center of mass over here acm acm is what acm is alpha into l by 2 isn't it it has angular acceleration this point moves in a circle so acceleration of center of mass is alpha into l by 2 just because velocity is 0 doesn't mean that acceleration is also 0 for example the projectile at that I mean I mean if you throw a body up vertically the velocity will be 0 at highest point but acceleration is still g similarly here omega is 0 but alpha is there so alpha into l by 2 is the acceleration of the center of mass alpha is 3g by 2 l so it will be equal to m into 3g by 4 so this is mg minus fy okay so from here you will get the value of fy to be equal to mg by 4 okay so like this you have to solve this particular question always always look at what center of mass is doing for the rigid body fine so let us move to next question any doubt guys please type in let's take up another routine one let us assume let us assume you have a cylinder of mass m and radius r okay it's a cylinder freedom that is net force is equal to mass acceleration so just find out net force write it down as mass into acceleration of center of mass acceleration of center of mass is alpha into l by 2 okay suppose you have two masses that are hung like this this is m1 this is m2 fine friction is sufficient so that this string does not slide okay the string does not slide on the cylinder the cylinder is hinged at the center and cylinder can rotate so assume that you know it is moving like that find out the acceleration of m1 and m2 quickly now here even though even though the the string is same but tension is not same throughout the string you can just take a pick off it take a pick off your notebook and send me or whatsapp okay anyways so let us solve this particular question so we have okay i don't see many of the guys who requested the session to be online live okay so next time onwards please don't expect me to oblige to any such request coming from the guys who have not attended the live session today okay so this is let us say tension T1 and this is m2g and let's say this is T2 okay tensions are not same this is T1 this is T2 okay so let us assume that m2 is going down with acceleration of A because of the constraint m1 will also go up with the same acceleration A isn't it that the constraint has nothing to do with the physics here it is just a systemic constraint that they both have to move together with the same acceleration okay so if I write the force equation for m2 I'll have to write it as m2g minus T2 is m2A okay and for m1 we have T1 minus m1g is equal to m1 into A okay these are the two equations now if you count the number of variables T1 T2 and A they are three variables fine so you can't solve just these two and get the answer so and then you can see that the cylinder is there which is rotating so can I use the torque equation and get the value of alpha and then relate alpha and A okay because this acceleration depends on how fast the cylinder is rotating isn't it so there has to be some relation between how fast it is rotating with what is the acceleration of m1 and m2 okay hence we need to understand what this cylinder is doing so let's take this as pulley okay there will be a force from the hinge let's say that force is T okay and then here we have T2 tangentially applied and from there we'll have T1 that is tangentially applied okay and since we have assumed that the acceleration is like this automatically the sense of rotation has to be like that okay so T2 is in the direction of angular acceleration the torque due to T2 is in the direction of angular acceleration so torque due to T2 I'll take positive just like you take force in the direction of acceleration to be positive similarly torque due to the force which is in the direction of angular acceleration should be taken as positive fine so T2 into R or let's say R into T2 because T2 is tangential perpendicular distance is radius only similarly T1 is tangential perpendicular distance is R only so RT2 minus R of T1 because T1 is trying to rotate in opposite direction so it has to get subtracted this should be equal to I times alpha fine now moment of inertia of the cylinder we all know is mr square by 2 so R into T2 minus R into T1 is mr square by 2 times alpha okay so I will get T2 minus T1 as mr alpha by 2 now even though I have got the third equation I have also introduced a fourth variable which is alpha okay so hence I need another equation and preferably an equation which make use of the variables which are already there in the equation and not introducing another variable fine so if you see this point let us let me put it some other color this point moves with the cylinder as well as with the string and there is no slipping fine so acceleration of this point if you look at the string is what if you look at the string acceleration is a okay and if you look at the cylinder acceleration of this point is what alpha into R fine and if there is no slipping a should be equal to alpha times R acceleration of the cylinder at that point should be equal to acceleration of the string at that point so a should be equal to alpha R so that is why T2 minus T1 should be equal to capital m a by 2 so now this fourth equation can be solved along with first and second to get the answer getting it so you can quickly do that in fact now that there is just couple of minutes left so if I subtract if I just you know add them up I will get T1 minus T2 plus m2 minus m1g is equal to m1 plus m2 into a okay T1 minus T2 is minus of capital m a by 2 plus m2 minus m1g is equal to m1 plus m2 times a okay so a will come out to be equal to m2 minus m1 times g divided by m1 plus m2 plus capital m by 2 okay so like this you have to solve this particular question so today we could just talk about briefly about the torque equation next class I think again we have to spend at least one hour of problem solving on torque equation and then we will talk about angular momentum for one hour and then then next couple of hours afterwards we will start talking about energy consideration from this chapter okay so I guess in in one one and a half class this chapter will get over okay so that's it for today