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Physics I Class 14 Cross Product, Torque, and Angular Momentum Rev. 09-Oct-04 GB 14-1 Review of Angular and Linear Acceleration MUST express angles in radians. s r v r a tangential r a centripetal v 2 2 r 2 2 r r r The radial direction is defined to be + outward from the center. a radial a centripetal 14-2 Review of Torque For linear motion, we have “F = m a”. For rotation, we have I The symbol “” is torque. We will define it more precisely today. When the rotation is speeding up, and are in the same direction. When the rotation is slowing down, and are in opposite directions. Torque and angular acceleration are always in the same direction in Physics 1. 14-3 Motivation for the Concept of Torque We know that when we try to throw an object by applying a force off center, that we end up imparting an angular velocity to the object as well as linear velocity. For a given force, the greater the distance it is applied off center, the greater the resulting angular acceleration. We need a mathematical technique to calculate the “off-center” effect of a force and relate it to the direction and magnitude of the resulting angular acceleration. 14-4 The Vector Cross Product We learned how to “multiply” two vectors to get a scalar. That was the “dot” product: d a b | a | | b | cos( ) Now we will “multiply” two vectors to get another vector: c a b ; | c | | a | | b | sin( ) The direction comes from the right-hand rule. It is at a right angle to the plane formed by a and b . In other words, the cross product is at right angles to both a and b . (3D thinking required!) 14-5 The Vector Cross Product 14-6 Torque as a Cross Product rF | | | r | | F | sin( ) r is the vector from the axis of rotation to where the force is applied. The torque can be zero in three different ways: 1. No force is applied (| F | 0 ). | 2. The force is applied at the axis of rotation ( r | 0 ). 3. F and r in the same or opposite directions (sin( ) 0 ). 14-7 Example of Torque - A Rod Suspended by a String Off Center | | | r | | F | sin( ) | | r m g Gravity force on “extended” bodies can be viewed as acting on a point mass (with the same total mass) at the object’s center of mass (balancing point). (out of page) r F=W What would be the direction of the rod’s angular acceleration? 14-8 Angular Momentum of a Particle center of rotation (defined) r p mv Angular momentum of a particle once a center is defined: l r p (What is the direction of angular momentum here?) Once we define a center (or axis) of rotation, any object with a linear momentum that does not move directly through that point has an angular momentum defined relative to the chosen center. 14-9 Class #14 Take-Away Concepts I 1. : Speeding up, slowing down. . 2. Definition of vector cross product: c a b ; | c | | a | | b | sin( ) 3. Torque as a cross product: r F . 4. Angular momentum of a particle: l r p . 14-10 Class #14 Problems of the Day __1. One of your classmates in Physics 1 is sitting on a stool that rotates on a vertical axle. She is facing north and puts out her right arm toward where you are standing to the east of her. You push her arm forward (north) and she begins to spin counter-clockwise as seen from above. The direction of the torque vector from your push is A) B) C) D) E) F) G) H) North. South. East. West. Up. Down. Clockwise. Counter-clockwise. 14-11 Class #14 Problems of the Day 2. A force of 5 N is applied to the rim of a wheel at an angle of 30º with respect to the outward radial direction. (The force is in the plane of the wheel.) The radius of the wheel is 0.3 m. What is the magnitude of the torque on the wheel? 30º F 14-12 Activity #14 Torque and Angular Momentum Objective of the Activity: 1. 2. Think some more about rotation concepts. I Determine the direction of torque from . 3. Determine the direction of torque from r F . 4. Calculate momentum of a particle from l r p . (No data measurements today, only paper and pencil.) 14-13