 We'll take a question on the the previous class concept and then we'll start the class the new concept okay let us see whether you are well versed in this, give it back there is this rod stop talking there is this rod okay the rod mass is capital M and the length is L okay there is another mass a point mass that is fixed at the end of the rod mass small m but it you are giving somehow this rod acquires some angular velocity alright by the way the rod is free to rotate about this axis okay fine so it can swing like this in a vertical plane fine so you're giving it some angular velocity so that one end of the rod acquires a velocity V0 okay you have to find out what is the value of V0 so that the rod just becomes horizontal but it masses are different small m capital M we have done the concept of energy right okay the hint is you have to write down this work with the theorem w is equal to u2 plus k2 minus u1 plus k1 so the work done is 0 stop talking 2 plus k2 will be equal to u1 plus k1 as work done is 0 is work done 0 or not which force will do work see mg is doing work no doubt about that but for mg you're considering potential energy if you're considering potential energy for a force you don't need to find the work done for that force okay and there are forces from the hinge but the point of application doesn't move so the work done by the hinge forces is 0 between this position and that position apply this theorem it is the rod plus small m can be treated as a single rigid body assume this line to be 0 potential energy it will be better than this line is what u2 u2 is 0 let us say okay fine u2 is 0 what about k2 k2 is also 0 it just goes to that position fine so both u2 and k2 are 0 let's try to write down the kinetic energy of this rod along with this mass fine so you can write down the kinetic energy in different ways all right given you can write it as half moment of inertia of the rod about the fixed axis into mega square plus half m into v0 square so you have written kinetic energy of the rod plus kinetic energy of the mass you're writing it separately okay and moment of inertia but the fixed axis is how much mx square by 3 so half l square by 3 omega is what v0 by l okay this is the kinetic energy another way to write the kinetic energy is to find the moment of inertia about this fixed axis for the entire rigid body for the entire rigid body moment of inertia is what mx square by 3 plus small m into l square so kinetic energy will be of inertia which is ml square by 3 plus ml square into omega square this will be the total kinetic energy this is k1 what about u1 u potential is it is better to split otherwise you have to find out the center of mass location for the combined system if you can find out no problem the potential g will be m plus capital M into g into height of the center of mass or you can split it into two parts potential g of the rod plus potential g for the small mass this will be equal to what negative of capital Mg l by 2 it is down the center of mass of the rod is down okay and negative of mg l any doubt this is u1 this is k1 u2 and k2 substitute there you get the answer okay any doubts okay I did not know that we have can you show the notes what we have done at the end of the last class we just started the energy right and we have written kind of energy about the fixed axis as well as rotation plus translation okay right so we have just introduced what is the kind of energy and potential g of everybody so we can continue solving problems right now all right so we have learned that kinetic energy of a rigid body is half moment of inertia about the fixed axis into omega square fine which omega about which axis about the center of mass or the fixed one which one omega omega is same for entire rigid body that is why it is called a rigid body fine if omega one point is different from other point it will no longer remain a rigid body they start okay so this is the kind of energy if you are not able to find the fixed axis no problem you can write it like this half m into vcm square plus half moment of inertia about center of mass into omega square two ways you can write the kinetic energy both ways you get the same answer for the kinetic energy all right potential energy in g into height of the center of mass okay saying w is equal to u2 plus k2 minus u1 plus k1 and remember it like this also that u1 plus k1 plus w is equal to u2 plus k2 so you can do some amount of work to increase the initial mechanical energy but if you're doing the negative work your initial mechanical energy will reduce to u2 plus k2 fine so both ways you can remember right so we are going to take few questions now which will test you on these fundamentals I think there should not be any confusion we have done lot of question in work for energy chapter already the concept for a rigid the methodology is exactly the same the only difference is the way and potential energy that is the only difference so keep it simple in your head it is very easy to complicate all right so we'll take up a simple sim we'll take simple scenarios and we'll build on it you have to find out the angle velocity when it has rotated by an angle of theta it is rotating in a vertical where the center of mass is initially and where it is finally moving like this it reaches here it travels vertically a distance of how much this is l by 2 so this will be l by 2 sin theta in this position and that position fine so step number one write down w is equal to u2 plus k2 minus u1 plus k1 and then substitute work done is zero final potential is what depends what you're taking your initial potential g where you should take initial potential g let's take this potential g so my u2 u2 becomes sorry u1 becomes 0 k1 is also 0 it starts from rest so u2 will be minus of ngl by 2 sin theta plus kinetic energy will be half I know there's a fixed axis so I can directly write half moment of inertia about a fixed axis ml square by 3 into omega square this is equal to 0 so from here I'll get the value of omega this is what you'll get omega after it rotates by an angle theta okay now suppose I time t how much it will rotate then what you will do I'm asking you how you'll solve for angle theta after time t omega is and then bring here one by sin theta one by root sin theta d theta you have to integrate now and that side will be dt so when you if you're able to integrate one by root sin theta you'll get the answer okay you are just similar kind of questions we have done when we learn the torque problem right same problem we have done we got the angle velocity then also calculated angle check whether it is the same thing using torque we have calculated we got alpha and then integrated alpha to get omega we got alpha is 3g cos theta by 2l we got alpha as 3g cos theta alpha is equal to omega d omega by d theta theta so theta you go the theta goes that side you are by 2 and 3g by 2l sin theta 2 and 2 will be going and then you get omega is this so you get the same answer two different ways are applying work with a theorem also you can get the velocity same way here as well fine find out what will be the reaction force of the hinge did we do that okay so quickly we I tell you let's say this is these are the two reaction forces let's say this is f y and this is fx so fx will be what fx will be equal to m omega square l by 2 and f y will be equal to m alpha into l by 2 because alpha into l by 2 is the tangential acceleration and omega square l by 2 is radial acceleration why not central module to track central mass is at a distance of l by 2 entire rod behaves as if it is a point mass located at the center of mass got it so what's the acceleration machine the central mass you have alpha you have alpha right so alpha into l by 2 will be the tangential acceleration it moves in a circle right with angular acceleration alpha so alpha into l by 2 is tangential acceleration we just differentiate or you directly have this alpha now you have to see net force is equal to mass and acceleration of center of mass yes but how did we get that torque equation we have done this torque equal to i alpha where the yes right the mg is a force so mg sin theta is perpendicular component every sin theta into l by 2 is i alpha yes you can tell some okay any doubt you have done all that then we learn the torque equation