 S.H.M. After this we go to S.H.M. Focus. I will share with you what is S.H.M. Focus. Continuous distribution of mass, moment of inertia. Continuous distribution of mass. Think with discrete masses. What is the discrete mass continuously? Look at it. It is not a point mass. It is not a point mass. So how to find moment of inertia of this continuous mass distribution that is what we are going to do now. You remember center of mass of continuous mass distribution? Remember that I understood that by dividing the entire mass into small small masses of mass dm each. Similarly distributed mass can be divided into point masses of dm each. And let us say its distance from the axis is r square, dm r square. Summation I have to do. So this summation becomes integral. For a continuous mass, if we have discrete masses then simply I do m1 r1 square plus f2 r2 square. But if it is continuous mass, if infinite such masses exist, small small dm. So I cannot do summation. I have to take help of integral now. Understood? So now we are doing objects and we will apply this formula of integral r square dm. Right? What are you talking about? What is that? m and radius r. Correct. What is the correct question to r? I cannot just say find the moment of inertia of ring of mass m and r. We have to ask about which axis. Okay? So we are determining about the axis which is passing through the center and perpendicular to the plane like this. What is the moment of inertia of this ring? It is very easy this one. Into small small masses. Right? So this is the small mass. Let us say this is the small mass dm. What is the moment of inertia is integral of perpendicular distance which is r square into dm. Is r 1? Yes. The perpendicular distance of all the dms are same only. So r square will come out of integral. This will become r square integral of dm. Integral of dm is what? Some of all the masses, total masses. Okay? So we have got moment of inertia of the ring. About that is passing through the center and perpendicular to the ring. And so let's say we got it. How much it is? It is mass. Mass. If the entire mass is located distance r. m r square. Simple. If the perpendicular distance of all the masses is same. Now tell me for hollow syrinx stack it is m r square. It is the moment of inertia of ring and hollow syrinx are same as you mean masses are equal. Right? Yes. This end will not be equal to that end. Okay? So we have got the moment of hollow syrinx till now. Okay? Now please write down moment of inertia of the disk. The space of mass m and radius r. About the axis. Shh. About the axis passing through the center and perpendicular. How do you analyze this? How your dm will be? Yeah. dm will be ring. This is about the size. So I will just give you a hint. You need to assume shh a ring of radius r and width r. Small r and your small r will change from 0 to capital r. This is the moment of inertia of the dm in terms of r. The mass thing we have done for the semi disk. Similar way you have to do here as well. How you have half disk. The way we will pi small r square. Pi small r square is area inside that ring. Inside that ring. You can take this strip and make a ring in terms of dm. This is mass per unit area into area of the strip. Yes or no? Mass per unit area into area of the strip. What is the idea? This is my dm. And the same thing we have done. Center mass also for the semi disk. What is the idea? This is my dm. And the same thing we have done. Center mass also for the semi disk. And mass per unit r square. You cannot escape this. If you think that you will ignore and actually exam. You will not learn. If you have any doubt ask. About an axis passing through the center and perpendicular to the disk. Exactly same thing we have done. When we calculated the center of mass of the semi disk. So, what are you facing? Difficulty. Difficulty. When we calculated the center of mass of the semi disk. So, what are you facing? Difficulty. And this difficulty will continue. If you do not do anything of school work. This will not going to help you anyway. If you are not doing this. This will not improve. Isn't it very straightforward? And in class 12 same mass could become charge. So, you have to find charge per unit area into the area of the strip. It will be again total charge q divided by 1. Then it will come and hit you hard in the exam. And you will not get zero. You will get negative. Minimum part is not zero. What? Minimum part is not zero. Can you go further? Minimum part is not zero.