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Curated by: Mohit Tyagi Lectures organized (87 videos)
🎓 COMPLETE YOUR JEE PREPARATION WITH COMPETISHUN Watch complete JEE lectures FREE with PAARAS Batch and access Competishun Courses, Test Series, PYQs & free Study Material — all in one place. 🚀 📱 COMPETISHUN APP 👉 https://play.google.com/store/apps/details?id=xyz.penpencil.competishun 💻 STUDY ON WEB 👉 https://i.competishun.com 🌐 EXPLORE COMPETISHUN 🎯 All Courses 👉 https://competishun.com/courses/ 📝 JEE & Other Test Series 👉 https://competishun.com/test-series/ 📚 Competishun Books & Study Material 👉 https://competishun.com/buy-books/ 🎁 Free Study Resources 👉 https://competishun.com/free-resources/ 🌐 Official Website 👉 https://competishun.com ━━━━━━━━━━━━━━━━━━ 📞 ADMISSION & SUPPORT 8888-0000-21 7410-900-901 7410-900-906 7410-900-907 7410-900-908 🔔 Subscribe to the channel for complete JEE lectures, preparation strategies, exam updates and expert guidance from the Competishun faculty. --------------- Topics covered in this Rotational Motion Class 11 lecture by ABJ sir IITD with timestamp: 00:00 – 04:55 RBD Problem 1: Based on the angular momentum. In this problem, we have a solid sphere of mass m and radius R, having a velocity u on a rough surface. So we have to find the angular velocity so the ball finally stops. To solve this problem, the conservation of angular momentum is used, and with the help of the equation, the value of angular velocity is found. 05:08 – 09:26 RBD Problem 2: Based on the angular momentum. In this problem, we have a solid sphere of mass m and radius R, having a velocity u on a rough surface and angular velocity as shown. So we have to find the final velocity of the center of mass and the angular velocity of the ball. To solve this problem, the conservation of the angular momentum is used. 09:26 – 19:22 Toppling: When a torque is applied on a body resisting change of motion (usually due to friction), the instantaneous axis of rotation shifts (usually due to the shift of normal). The body topples if the instantaneous axis of rotation falls outside the body. To clarify the concept ABJ Sir took an example, in which we have to find the torque of normal force about the center of mass, according to the given condition. Using this example, ABJ Sir explains the concept of Toppling. 19:22 - 21:40 Toppling Case 1: In the first case of toppling there is a block of mass m and height h and length l at rest. In this case, applied normal force will be passed from the center of the block, and normal force on the body is balanced by the weight of the body. So there is no toppling on the block. 21:40 – 24:15 Toppling Case 2: In the Second case of toppling, there is a block of mass m and height h and length l on the smooth surface. A force F is applied at the center of the block. In this case, applied normal force will be passed from the center of the block, and the weight of the body balances the normal force on the body. So there is no toppling on the block. 24:15 – 28:09 Toppling Case 3: In the Third case of toppling, there is a block of mass m and height h and length l on the smooth surface. A force F is applied at the block’s topmost point according to the diagram of the block. In this case, the applied normal force will be shifted from the center of the block, and the minimum force for which toppling will occur, can find out. 28:09 – 32:20 Toppling Case 4: In the Fourth case of toppling, there is a block of mass m and height h and length l on the Rough surface. A force F is applied at the block’s topmost point according to the diagram of the block. In this case, there will be a friction force and the applied normal force will be shifted from the center of the block, and the minimum force for which toppling will occur can find out. 32:20 – 35:55 RBD Problem 3: Based on the toppling. In this problem, we have an equilateral prism of side l and mass m on a rough surface according to given diagram, and a force F is applied on the top of the prism. So we have to determine the minimum force value for which toppling will occur.