33106 Lecture Notes - Lecture 3: Cross Product, Angular Velocity

69 views2 pages
8 Sep 2015
School
Department
Course

Document Summary

4 september 2015 (cid:126)a (cid:126)b = (cid:126)b (cid:126)a (cid:126)a ( (cid:126)b + (cid:126)c) = (cid:126)a (cid:126)b + (cid:126)a (cid:126)c. Algebraic (cid:126)a (cid:126)b = (ax i + ay j + az. K) (bx i + by j + bz z) = (aybz azby) i + (azbx axbz) j + (axby aybx) k (cid:126)a (cid:126)b = Torque is the force applied times the distance from the axle. The force applied is only the component perpendicular to (cid:126)r. to nd the component force perpendicular to the (cid:126)r is through the use of the cross product. The right hand rule is also used to determine the direction that angular velocity is applied. A circle on this page turning counterclockwise would have a angular velocity vector pointing out of the page. Describing the geometry of motion for ideal particles (points having no size or structure). However, they have mass. x(t) is a complete description of a particle.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related textbook solutions

Related Documents