JEE Mains 2026 Physics | Angular Velocity of a Rigid Rod System | 23 Jan Shift 1 скачать в хорошем качестве

JEE Mains 2026 Physics | Angular Velocity of a Rigid Rod System | 23 Jan Shift 1

Comprehensive solution for the System of Particles and Rotation problem from JEE Mains 2026 (23rd January, Shift 1). Question: Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is : Options: 1. 3L / 2md² 2. 2L / 5md² 3. 4L / 3md² 4. 2L / md² --- Detailed Solution: 1. Find the Centre of Mass (CM): The CM divides the distance 'd' in inverse ratio of masses. Distance of mass 'm' from CM (r1) = (2m / (m + 2m)) * d = 2d/3 Distance of mass '2m' from CM (r2) = (m / (m + 2m)) * d = d/3 2. Calculate Moment of Inertia (I): The rod has negligible mass, so only the balls contribute. I = m(r1)² + 2m(r2)² I = m(2d/3)² + 2m(d/3)² I = 4md²/9 + 2md²/9 = 6md²/9 = 2md²/3 3. Calculate Angular Velocity (ω): Angular momentum is L = I * ω. L = (2md²/3) * ω ω = 3L / 2md² Final Answer: Option 1. --- Timestamps: 0:00 - Question Overview 0:35 - Locating the Centre of Mass (CM) 0:57 - Deriving the CM distances (r1 and r2) 1:27 - Calculating total Moment of Inertia (I) 1:58 - Applying the Angular Momentum Formula (L = Iω) 2:10 - Final Result & Option Selection