 All right, so here is a question on rigid body motion. Let's see how we can go about this horizontal circular platform of radius 0.5 meter. So radius of the platform is given 0.5 meters. The mass of the platform, which is disc, is given as 0.45 kgs. Two massless spring toygons each carrying a steel ball of mass 0.05 kgs are attached to the platform at a distance of 0.25 meters from the center of the either sides of the diameter. So this distance is 0.25. So the toygons are at a distance of 0.25 meters and they have inside it a steel ball of 0.05 kgs. Each gun simultaneously fires a ball horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the ball have horizontal speed of 9 meter per second with respect to ground. So basically, the velocity of the steel ball is given, which is 9 meter per second. You need to find rotational speed of the platform in radians per second, where the platform is free to rotate about this axis that passes through the center of the disc. Now, again, you will, I think, appreciate that this is a clear cut case of conservation of angular momentum about that axis. So if you have a system where the system's parts are disc and the steel balls that are coming out of gun, then whatever are the forces that is between them only, and that is internal force, and that will not create any external torque. So if I consider my system to be disc plus steel balls, I can conserve the angular momentum. So when the guns were placed here and they haven't fired the steel ball yet, then the angular momentum of entire system is what? The angular momentum of entire system was 0. This is equal to angular momentum of the disc plus angular momentum of the steel balls. Let's say steel ball 1 plus steel ball 2. Now, angular momentum of the disc can be written in a straightforward manner. That is, I about center of mass into omega. Now, the steel ball is a point mass. So point mass's angular momentum is written like this perpendicular distance into the linear momentum, which is mass into velocity. Now you can see that the formula for angular momentum is actually r cross mv in a vector form. So r cross mv for this steel ball will have the direction downside. Similarly, r cross mv for this particular steel ball also has the same direction. So angular momentum of these two steel balls will get added up. So r perpendicular into mv into 2. Now ICM is what? ICM for the disc is mr square by 2. So ICM is mass is what? 0.45 kgs mr square, which is 0.5 square by 2. So this into omega. Now you can see that when the steel ball were placed on the disc, then the moment of inertia of the disc plus steel ball were different. You need to add the moment of inertia of the steel ball also. But then during that time, the angular momentum was 0. So that doesn't feature on the left hand side. Now the steel balls have left. The mass of the disc is just 0.45 kg only. So like that only you have to write here. So this is ICM into omega. Now r perpendicular into mass into velocity multiplied by 2. This how you can write perpendicular distance is 0.25. Mass of the steel ball is 0.05 into velocity is 9. So this is rmv into 2. So this will come out to be this is 0.5 into 0.05 into 9. This is 0.025 into 9. So this will give you 0.225. This is the angular momentum of the two steel balls. Now this you can simplify and get it as 0.05625 into omega. So 0 should be equal to 0.05625 omega plus 0.225. So omega will come out to be minus of 0.225 divided by 0.05625. Now negative sign is coming because the angular momentum of the disc will be in opposite direction of the angular momentum of the steel balls. So this comes out to be 4 radians per second. Now you might be slightly overwhelmed with the kind of calculation that is involved here. But then you don't need to right away get the value of this term to be equal to something and rmv into 2 to be equal to this. You can just write down in this equation. And then you can simplify also. Like let me show you. You can write down 0 is equal to 0.45 into 0.5 square by 2 times omega plus 0.025 into 9. You can write like this. And then 0.5 square is what? 0.5 square is 0.25. So this is 10. And then 9 fives are 45. So you can write 0.5 there. And then you can see that it comes out to be 2.5. And then you will get omega to be equal to minus 4 radians per second. OK. So like this, you can simplify the whole thing and get the answer. Don't get overwhelmed by the calculation of the question. Thank you.