4 Torque about point & Torque about axis | Rotational Motion Class 11 | RBD | JEE Mains & Advanced

Watch Complete YouTube Lectures Distraction-Free Download the Competishun App 👉 https://play.google.com/store/apps/de... 🌐 Login for Courses & Test Series: 👉 https://i.competishun.com/ Popular Courses for IIT-JEE: Pratham | Prakhar | Prakhar Integrated I | Praveen | Pragyaan C3 | Champ | CATS | ABC | PYQ Express UTS | ATS 👉 Explore all: https://competishun.com/courses/ 📞 Support: 8888-0000-21 | 7410-900-901 | 7410-900-908 Explore: praveen.competishun.com 𝑷𝒓𝒂𝒗𝒆𝒆𝒏 (𝑫𝒓𝒐𝒑𝒑𝒆𝒓 & 𝑿𝑰𝑰 𝑺𝒕𝒖𝒅𝒚𝒊𝒏𝒈 𝑩𝒂𝒕𝒄𝒉): Comprehensive one-year online course for JEE Main + Advanced preparation. -------------- Topics covered in this Rotational Motion Class 11 lecture by ABJ sir IITD with timestamp: 00:00 - 10:08 Torque: Torque measures the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force. 10:08 - 18:03 Line of action of force: The line of action (also called the line of application) of a force (F) is a geometric representation of how the force is applied. It is the line through the point at which the force is applied in the same direction as the vector F. ABJ Sir also explains the direction of the torque with the help of the right-hand thumb rule. 18:03 - 20:36 Important Point about torque: Torque is a vector quantity. It is an axial vector. The unit of torque is N-m (Newton-meter). ABJ Sir also explains the dimension of the torque. 20:36 - 28:17 RBD Problem 1: Based on the torque. In this problem, we have two points, O and P, and their coordinates are given. A force is applied on point P, so we have to find the torque on point P wrt point O. 28:17 - 34:04 RBD Problem 2: Based on the torque. In this problem, we have a point P (0,5), a force F=10N is applied on the P at 37 degree angle. We have to find the torque on the point P about origin. 34:04 - 37:51 Important Point: If a force F is applied along the position vector then torque will be zero. We can also say if LOAF passes through point O, then torque about point O will be zero. 37:51- 42:31 Torque about an Axis: The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation. Take a point O on the axis and find torque about point O. The component of this torque along axis is defined as torque about axis. 42:31 - 48:21 Torque about an Axis (Case 1): LOAF passing through the axis: If Line of action of force passes through any point on the axis, then torque about the axis will be zero. 48:21 - 51:32 Torque about an Axis (Case 2): LOAF parallel to the axis: If Line of action of force is parallel through the axis, then the torque about the axis will be zero. 51:32 - 57:39 Torque about an Axis (Case 2): Axis and LOAF are skew lines: The formulae of torque for this case is derived by ABJ Sir, in which torque is equal to the force and perpendicular distance of the point from the axis. 59:03 - 01:09:03 RBD Problem 3: In this problem, we have a smooth hinge and uniform rod of mass m and length l, and we have to find the torque due to hinge force and weight about hinge. With the help of these examples, ABJ Sir explaining the method to write torque correctly. 01:09:03 - 01:14:17 RBD Problem 4: In this problem, we have a ball of mass m and radius R, and a force F is applied on the top most point and we have to find the torque about the axis 1 & axis 2. With the help of these examples, ABJ Sir explaining the method to write torque correctly. 01:14:17 - 01:24:21 RBD Problem 5: In this problem, we have a ball of mass m and radius R on the inclined plane , and a string is connected to the ball as given diagram. Tension in the string is T. We have to find the torque about the axis 1 & axis 2. With the help of these examples, ABJ Sir explaining the method to write torque correctly. 01:24:22 - 01:32:19 Newton's Second Law: The relation between torque and angular acceleration can be established using Newton’s second law, which is force exerted is directly proportional to the acceleration produced, force is the product of mass and acceleration.