 All right friends, so this is Dheeraj and live again. So this is a doubt which was asked by one of our NPJ Sankur students. So this is again J Main's January 9th paper. So you can read the question, we have two masses of M and M by 2 that are connected at the two ends of the massless rigid rod of length L. So you can see the figure and understand what is going on. The rod is suspended by a thin wire of torsional constant K. So this wire which you see here has a torsional constant K. Now if you don't know torsional constant is just like a spring constant where the normal spring constant gives you force if the spring get extended by let's say a length of x or get compressed by length of x then there will be a force that get developed of magnitude K into x and if there is a wire which has a torsional constant K then if it gets rotated by an angle theta then a restoring torque get created whose magnitude will be K into theta. So but then still I mean if you have done good number of question from this chapter you will be aware of whatever I have just spoke about. So this wire is connected at the center of the mass of the rod system. The center of mass is not the center of the system because one mass is M by 2 and other is M. So the center of mass is near this mass. Now because of this torsional constant the restoring torque K into theta for angular displacement theta gets created. So if rod is rotated by angle theta 0 and released you need to find out the tension in it when it passes through the mean position. Now first of all you need to understand there is an angular SHM that is happening over here. So torque for any angle theta will be K into theta. This torque should be equal to moment of inertia of the system into alpha. So I can get alpha to be equal to K by I into theta. Now if you have to write down in a vector form since it is a restoring torque there will be minus sign over here because if you rotate the system anticlockwise a clockwise torque will be created. So they are opposite in direction hence negative sign. Now if you compare it with SHM equation which is this you will get omega to be equal to under root of K by I. This is the angular frequency. So here I also need to find out what is the moment of inertia. So let us say this is X1 and this distance is X2. So moment of inertia will be equal to M by 2 into X1 square plus M into X2 square. But I need to know what is X1 and X2. So for that we know that X1 plus X2 should be equal to L and if this point is center of mass of the system then M by 2 into X1 should be equal to M into X2. So from here you will get X2 to be equal to X1 by 2. So we have equation 1 and equation 2 now. So we can substitute it the value of X2 on equation 1 we will get 3 times X1 by 2 to be equal to L. So X1 will come out to be 2L by 3 and X2 will come out to be L by 3. So now I can substitute these values over here to get the moment of inertia. So now I am aware of what is the value of omega angular frequency. Now you might be knowing already that the relation for velocity in a simple harmonic motion is what omega times under root of A square minus X square. Now this is for simple harmonic motion. Now here if it is like angular simple harmonic motion then instead of velocity you will have angular velocity fine remember this is not angular velocity this is angular frequency just that okay. So angular velocity let me write it as let's say omega naught just differentiate from this angular frequency will be equal to angular frequency into instead of amplitude A I have theta naught which is angular amplitude minus theta square fine and at the mean position at the mean position the value of theta is 0 okay. So omega naught which is the angular velocity at the mean position will be omega into theta naught alright. So we have the knowledge of now the angular velocity. So remember this is angular frequency omega which you need to substitute over here moment of inertia you will get from this equation okay just substitute value of X1 and X2 from here okay. So I am assuming that we have angular velocity known now okay. Now if you draw a simple let us say free body diagram of mass m okay if you do that you will see that it will experience a centrifugal force of m omega naught square times X2 this way okay and this way there will be tension t alright. Now similarly you could have drawn free body diagram for mass m by 2 as well for m by 2 this way it will be m by 2 omega naught square X1 and this way you will have tension okay. Now I know that m by 2 into X1 is equal to m into X2 so both the cases you get the same value of tension okay. So you can use any of these two so I will use the above one and I will write tension is equal to m omega naught square into X2 okay. So the value of omega naught will come from here and value of X2 is l by 3 alright. So like this you can find the value of tension okay alright so like this you have to solve this question I hope you have learned something new today in case you have any doubts feel free to get in touch with me okay thank you.