Lecture 3: General Curvelinear Motion

Applications (2 of 2) A roller coaster car travels down a fixed, helical path at a constant speed. How can we determine its position or acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car? Copyright © 2016 Pearson Education, Inc. All Rights Reserved Section 12.4 General Curvilinear Motion General Curvilinear Motion A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are usually used to describe the motion. A particle moves along a curve defined by the path function, s. The position of the particle at any instant is designated by the vector r  r (t) Both the magnitude and direction of r may vary with time. If the particle moves a distance along the curve during time interval t, the displacement is determined by vector subtraction: r  r ’ - r Copyright © 2016 Pearson Education, Inc. All Rights Reserved Velocity Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment t is vavg  r / t . The instantaneous velocity is the time-derivative of position v  dr dt The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length approaches the magnitude of differentiating the path function r as t  0 s the speed can be obtained by ( v  ds dt). Note that this is not a vector! Copyright © 2016 Pearson Education, Inc. All Rights Reserved Acceleration Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’ over a time increment that increment is: t, the average acceleration during a avg  v / t  ( v  v’) / t The instantaneous acceleration is the time-derivative of velocity: 2 2 a  dv / dt  d r / dt A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Section 12.5 Curvilinear Motion: Rectangular Components Curvilinear Motion: Rectangular Components It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference. The position of the particle can be defined at any instant by the position vector r  xi  yj  zk . The x, y, z-components may all be functions of time, i.e., x  x  t  , y  y  t  , and z  z  t  . 2 2 2 0.5 The magnitude of the position vector is: r  (x  y  z ) The direction of r is defined by the unit vector: ur  1 / r  r Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rectangular Components: Velocity The velocity vector is the time derivative of the position vector: v  dr / dt  d ( xi ) / dt  d ( yj ) / dt  d ( zk ) / dt Since the unit vectors i, j, k are constant in magnitude and direction, this equation reduces to where   v  vxi  vy j  vz k  v x  x  dx / dt, v y  y  dy / dt, v z  z  dz / dt The magnitude of the velocity vector is v  [( vx ) 2  ( vy) 2  ( vz ) 2 ]0.5 The direction of v is tangent to the path of motion. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rectangular Components: Acceleration The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector). a  dv / dt  d 2 r / dt 2  a x i  a y j  a z k where     a x  v x  x  dv x / dt, a y  v y  y  dv y / dt,   a z  v z  z  dv z / dt The magnitude of the acceleration vector is a  (a x ) 2  (a y ) 2  (a y ) 2 The direction of a is usually not tangent to the path of the particle. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (1 of 3) Given: The box slides down the slope described by the equation y  (0.05x 2 ) m, where x is in meters. v x   3 m / s, a x   1.5 m / s 2 at x  5 m. Find: The y components of the velocity and the acceleration of the box at x = 5 m. Plan: Note that the particle’s velocity can be found by taking the first time derivative of the path’s equation. And the acceleration can be found by taking the second time derivative of the path’s equation. Take a derivative of the position to find the component of the velocity and the acceleration. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (2 of 3) Solution: Find the y-component of velocity by taking a time derivative of the position y  (0.05x 2 )     y  2  0.05  x x  0.1 x x Find the acceleration component by taking a time derivative  of the velocity y       y  0.1 x x  0.1 x x Substituting the x-component of the acceleration, velocity at x=5 into   y and y. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (3 of 3)  Since  x  v x   3 m / s, x  a x   1.5 m / s 2at x  5 m       y  0.1 x x  0.1  5   3  1.5 m / s   y  0.1 x x  0.1 x x  0.1  3  0.1  5  1.5  2 At  0.9 – 0.75  0.15 m / s 2 x  5m v y  –1.5m / s  1.5m / s  a y  0.15 m / s 2  Copyright © 2016 Pearson Education, Inc. All Rights Reserved Practice Problems Examples: 12-9; 12:10; 12:11; 12:12: 12:13 Fundamental problems: F 12-19 ; F 12-26 Exercise Problems: 12-72; 12-99 Copyright © 2016 Pearson Education, Inc. All Rights Reserved