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Unformatted text preview: Physics 7A, Fall 2005 1 Summary of Newtonian mechnics Kinematics of Point Object: Giancoli’s chapter 1, 2, 3. Trajectory: path of the point object in space. Vectors: quantities with both magnitude and direction. Position (vector) : ~ r . Velocity (vector) : ~v ≡ d ~ r d t . Velocity is always in the tangential direction of trajectory. Acceleration (vector) : ~a ≡ d ~v d t = d 2 ~ r d t 2 . Scalars: Speed : magnitude of velocity, v ≡ | ~v | . Tangential acceleration : a tan is one component of ~a which is parallel to ~v (in tangential direction of trajectory). a tan = d v d t , (derivative of speed) Normal acceleration : a normal is the component of ~a which is perpendicular to ~v (in normal direction of trajectory). a normal = v 2 R where R is the radius of curvature of the trajectory curve (generally speaking, R at different points on a curve will be different). For circular motion (trajectory is a circle or part of circle), R is the radius of the circle, a normal is usually called centripetal acceleration . Reference frames, relative motion: we won’t consider rotating reference frame. From ~ r A , relative to C = ~ r A , relative to B + ~ r B , relative to C , taking time derivative, ~v A , relative to C = ~v A , relative to B + ~v B , relative to C ~a A , relative to C = ~a A , relative to B + ~a B , relative to C NOTE: In rotating reference frames, the last equation about accelerations is not true. See Giancoli’s section 11-10. Newton’s laws: Giancoli’s chapter 4, 5 First law: can be viewed as the definition of inertial reference frame . In (and only in) an inertial reference frame, velocity of any point mass with zero net force on it, will be a constant. Second law: in an inertial reference frame (this is important), for a point mass m , acceleration is proportional to total force : X ~ F = m~a = m d 2 ~ r d t 2 NOTE: if we have expression of forces in terms of ~ r and ~v = d ~ r d t , Newton’s second law becomes a sec- ond order ordinary differential equation. In principle, we can determine the motion by solving this equation with initial conditions. Physics 7A, Fall 2005 2 Third law: Weak version: ~ F A → B =- ~ F B → A . Force given by object A to object B is opposite to force given by object B to object A (same magnitude, opposite directions). Strong version: ~ F A → B =- ~ F B → A , and the two vectors are in the same line (in the line connecting A and B). Following picture is an example of force pair consistent with weak version but not strong version: this will give you trouble in conservation of angular momentum. ‡ ‡ 1 F 1 to 2 ‡ ‡) F 2 to 1 Is this possible?? Figure 1: Example for third law Momentum and energy : Giancoli’s chapter 7, 8, 9 Impulse-momentum theorem: multiply both side of second law by d t , then integrate, ~ J = ~ p final- ~ p initial Impulse : ~ J ≡ R ~ F d t ....
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This note was uploaded on 04/06/2009 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at Berkeley.
- Spring '08