Dynamics 14th edition by r c hibbeler section 12 8

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS Today’s Objectives: Students will be able to: Determine velocity and acceleration components using cylindrical coordinates In-Class Activities: • Check Homework • Reading Quiz • Applications • Velocity Components • Acceleration Components • Concept Quiz • Group Problem Solving • Attention Quiz Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved READING QUIZ In a polar coordinate system, the velocity vector can be written as v = vrur + vθuθ = rur + rθuθ The term θ is called A) transverse velocity C) angular velocity B) radial velocity D) angular acceleration The speed of a particle in a cylindrical coordinate system is B) rθ A) r C) Dynamics, Fourteenth Edition R.C Hibbeler 2 (rθ) + (r) D) 2 (rθ) + (r) + (z) Copyright ©2016 by Pearson Education, Inc All rights reserved APPLICATIONS A cylindrical coordinate system is used in cases where the particle moves along a 3-D curve In the figure shown, the box slides down the helical ramp How would you find the box’s velocity components to check to see if the package will fly off the ramp? Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved APPLICATIONS (continued) The cylindrical coordinate system can be used to describe the motion of the girl on the slide Here the radial coordinate is constant, the transverse coordinate increases with time as the girl rotates about the vertical axis, and her altitude, z, decreases with time How can you find her acceleration components? Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CYLINDRICAL COMPONENTS (Section 12.8) We can express the location of P in polar coordinates as r = r ur Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, θ, is measured counter-clockwise (CCW) from the horizontal Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved VELOCITY in POLAR COORDINATES) The instantaneous velocity is defined as: v = dr/dt = d(rur)/dt v = rur + r dur dt Using the chain rule: dur/dt = (dur/dθ)(dθ/dt) We can prove that dur/dθ = uθ so dur/dt = θuθ Therefore: v = rur + rθuθ Thus, the velocity vector has two components: r, called the radial component, and rθ called the transverse component The speed of the particle at any given instant is the sum of the squares of both components or v = (r θ ) + ( r ) Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved ACCELERATION (POLAR COORDINATES) The instantaneous acceleration is defined as: a = dv/dt = (d/dt)(rur + rθuθ) After manipulation, the acceleration can be expressed as a = (r – rθ )ur + (rθ + 2rθ )uθ The term (r – rθ ) is the radial acceleration or ar The term (rθ + 2rθ ) is the transverse acceleration or aθ 22 The magnitude of acceleration is a = (r – rθ ) + (rθ + 2rθ ) Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CYLINDRICAL COORDINATES If the particle P moves along a space curve, its position can be written as rP = rur + zuz Taking time derivatives and using the chain rule: Velocity: vP = rur + rθuθ + zuz Acceleration: Dynamics, Fourteenth Edition R.C Hibbeler aP = (r – rθ )ur + (rθ + 2rθ )uθ + zuz Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE Given: The platform is rotating such that, at any instant, its angular position is θ = (4t 3/2 ) rad, where t is in seconds A ball rolls outward so that its position is r = (0.1t ) m Find: The magnitude of velocity and acceleration of the ball when t = 1.5 s Plan: Use a polar coordinate system and related kinematic equations Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE (continued)   Solution: , , 4t 3/2 , 6, At t=1.5 s, r 0.3375 m, 0.675 m/s, 7.348 rad, 7.348 rad/s, 0.9 m/s 2.449 rad/s Substitute into the equation for velocity v = r ur + rθ uθ = 0.675 ur + 0.3375 (7.348) uθ = 0.675 ur + 2.480 uθ v = Dynamics, Fourteenth Edition R.C Hibbeler 2 (0.675) + (2.480) = 2.57 m/s Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE (continued) Substitute in the equation for acceleration: a = (r – rθ )ur + (rθ + 2rθ)uθ a = [0.9 – 0.3375(7.348) ] ur + [0.3375(2.449) + 2(0.675)(7.348)] uθ a = – 17.33 ur + 10.75 uθ m/s a = 2 2 (– 17.33) + (10.75) = 20.4 m/s Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CONCEPT QUIZ If r is zero for a particle, the particle is A) not moving B) moving in a circular path C) moving on a straight line D) moving with constant velocity If a particle moves in a circular path with constant velocity, its radial acceleration is A) zero B) r C) − rθ Dynamics, Fourteenth Edition R.C Hibbeler D) 2rθ Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING  Given: The arm of the robot is extending at a constant rate = 1.5 ft/s when r = ft, z = (4t ) ft, and θ = (0.5 t) rad, where t is in seconds Find: The velocity and acceleration of the grip A when t = s Plan: Use cylindrical coordinates Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING (continued)  Solution: When t = s, r = ft and the arm is extending at a constant rate = 1.5 ft/s Thus ft/s 1.5 t 4.5 rad, z t 36 ft, 1.5 rad/s, rad/s t 24 ft/s, ft/s 2 Substitute in the equation for velocity v = r ur + rθ uθ + z ur = 1.5 ur + (1.5) uθ + 24 uz = 1.5 ur + 4.5 uθ + 24 uz Magnitude v = Dynamics, Fourteenth Edition R.C Hibbeler 2 (1.5) + (4.5) + (24) = 24.5 ft/s Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING (continued) Acceleration equation in cylindrical coordinates a = (r – rθ )ur + (rθ + 2rθ)uθ + zuz = {0 – (1.5) }ur +{3 (0) + (1.5) 1.5 } uθ + uz a = [6.75 ur + 4.5 uθ + uz] ft/s a = 2 2 (6.75) + (4.5) + (8) = 11.4 ft/s Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved ATTENTION QUIZ The radial component of velocity of a particle moving in a circular path is always A) zero B) constant C) greater than its transverse component D) less than its transverse component The radial component of acceleration of a particle moving in a circular path is always A) negative B) directed toward the center of the path C) perpendicular to the transverse component of acceleration D) All of the above Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved End of the Lecture Let Learning Continue Dynamics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved  ... acceleration is a = (r – r ) + (r + 2r ) Dynamics, Fourteenth Edition R. C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CYLINDRICAL COORDINATES If the particle... Edition R. C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CYLINDRICAL COMPONENTS (Section 12. 8) We can express the location of P in polar coordinates as r = r ur Note... manipulation, the acceleration can be expressed as a = (r – r )ur + (r + 2r )uθ The term (r – r ) is the radial acceleration or ar The term (r + 2r ) is the transverse acceleration or aθ