Lecture 5: Curve Linear Motion Normal and Tangential Components

Normal and Tangential Components (1 of 2) When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (thus the origin and coordinate system move with the particle). The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Normal and Tangential Components (2 of 2) The positive n and t directions are defined by the unit vectors un and ut , respectively. The center of curvature, O’, always lies on the concave side of the curve. The radius of curvature, , is defined as the perpendicular distance from the curve to the center of curvature at that point. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Velocity in the n-t Coordinate System The velocity vector is always tangent to the path of motion (tdirection). The magnitude is determined by taking the time derivative of the path function, s t  . v  v ut where v  s  ds dt Here v defines the magnitude of the velocity (speed) and defines the direction of the velocity vector. ut Copyright © 2016 Pearson Education, Inc. All Rights Reserved Acceleration in the n-t Coordinate System (1 of 2) Acceleration is the time rate of change of velocity:  t  vut a  dv dt  d  vut  dt  vu v represents the change in the magnitude of velocity and, u t Here represents the rate of change in the direction of ut . After mathematical manipulation, the acceleration vector can be expressed as:  t a  vu v  u 2 n  a t ut  a nun . Copyright © 2016 Pearson Education, Inc. All Rights Reserved Acceleration in the n-t Coordinate System (2 of 2) So, there are two components to the acceleration vector: a  a t ut  a n u n • The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. at  v or at ds  v dv • The normal or centripetal component is always directed toward the 2 center of curvature of the curve. an  v  • The magnitude of the acceleration vector is a 2  an    at  2 Copyright © 2016 Pearson Education, Inc. All Rights Reserved Special Cases of Motion (1 of 2) There are some special cases of motion to consider. 1) The particle moves along a straight line.     an  v 2   0  a  at  v The tangential component represents the time rate of change in the magnitude of the velocity. 2) The particle moves along a curve at constant speed. at  v  0  a  a n  v 2  The normal component represents the time rate of change in the direction of the velocity. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Special Cases of Motion (2 of 2) 4) The particle moves along a path expressed as The radius of curvature, , y  f  x. at any point on the path can be 1   dy dx 2    d 2 y dx 2 32 Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-14 (1 of 3) When the skier reaches point A along the parabolic path in Figure 12-27a, he has a speed of increasing at 2m s 2 6 m s which is Determine the direction of his velocity and the direction and magnitude of his acceleration at this instant. Neglect the size of the skier in the calculation. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-14 (2 of 3) Velocity. By definition, the velocity is always directed tangent 1 2 1 to the path. Since y x , dy / dx  x, Then at 20 10 Hence at A, v makes an angle of x  10 m,dy / dx  1.   tan 1 1  45 with the x axis, Fig.12-27b. Therefore a  6 m / s a A  ut  ( 2 /  ) un The acceleration is determined from However, it is first necessary to determine the radius of curvature of the path at A (10 m, 5 m). Since then [1  ( dy / dx) 2 ]3/ 2   | d 2 y / dx 2 | The acceleration becomes 1 [1  ( x ) 2 ]3/ 2 10   m 1 | | 10 a A  ut   2ut  d 2 y / dx 2  1 , 10 2 u  n (6m / s 2 ) un 28.28 m  {2ut  1.273un }m / s 2 Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-14 (3 of 3) As shown in Fig. 12-27b a  (2m / s 2 ) 2  (1.273m / s2 )2  2.37m / s 2 2  57.5 1.273 Thus, 45° + 90° + 57.5° − 180° = 12.5° so that,   tan1 a  2.37 m / s 2 Ans. NOTE: By using n, t coordinates, we were able to readily solve this problem through the use of Eq. 12-18, since it accounts for the separate changes in the magnitude and direction of v. Copyright © 2016 Pearson Education, Inc. All Rights Reserved