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1、2.4 Kinematic Analysis of a Particle in 3D SpaceInvestigation into the motion of a particle is a simplest yet the most fundamental aspect in kinematics. In this section, we deal with the kinematic equations of a particle P moving in 3D space by using vector notation. Three coordinate systems will be

2、 introduced to describe the curvilinear motion of a particle: They areq Rectangular(Cartesian) Coordinate System ()q Cylindrical Coordinate System()q Spherical Coordinate System() Rectangular (Cartesian) Coordinate System () Consider a particle P moving in 3D space, in the rectangular coordinate sys

3、tem as shown in the figure, its position vector can be given as (1a) the unit vectors of three orthogonal axes. Note that they are constant since all axes have been fixed in space, i.e.,the coordinates of P with respect to.Taking derivatives once and twice with respect time gives the velocity and ac

4、celeration of point P (1b) (1c)Obviously, this is the form which we are very familiar with.Projectile Problem(A classical problem for particle kinematics)An object located at origin of a Cartesian system with its axis upward and axis horizontal as shown, is thrown through the air with an initial vel

5、ocity and an angle about horizontal at , determine the trajectory function of the object. Solution:Initial conditionsPosition: , Velocity: , Acceleration: , ( g-the gravitational acceleration) , , One more integration leads toSubstituting into the above equation finally results inCylindrical Coordin

6、ate System () In the cylindrical coordinates, the position of a particle P can be located by-the radial distance from the -axis- the angle from the -axis to the radial line-the vertical distance from the plane. Define , and as three orthogonal unit vectors in the sense of positive , and directions,

7、respectively. The position vector of P can then be expressed by (2a)In order to achieve the velocity of P, let us rotate , and about with an small angle during an infinitesimal time interval . Then the differentiation of has the form (2b)Since keeps unchanged, . While the small changes in and occur

8、as follows Thus, This will also lead toand thereby (2c)Differentiating Eq.(2c) finally results in the acceleration of P (2d)Sample ProblemDetermine , , and of a RRPR planar mechanism as shown provided that , , , , and have been known. Solution:The polar coordinate system may be used for this problem

9、. The general formulae 2(a), 2(c) and 2(d) are reduced to xyyxxyVelocity Polygon, , Velocity AnalysisSetp1: Establish the polar coordinate system with and being as the unit vectors. The velocity of A in can be expressed by Since is a constant, . ThusStep 2: Establish another polar coordinate system

10、with and being as the unit vectors. The velocity of point A in can be expressed by - the velocity of a fictitious point B on link 3, which is momentarily coincident with the point A. - the sliding velocity of A relative to B along the direction of .Step 3: The velocity of the point A evaluated by us

11、ing these two systems SHOULD BE THE SAME. ThusProjecting onto and , respectively, leads to, Acceleration Polygon Acceleration Analysis Acceleration analysis can be implemented in the same way.Evaluated in Keep in mind that is a constant such that Evaluated inThe acceleration constraint condition givesProjecting the left hand term onto and finally results inSpherical Coordinate SystemWhen a radial distance and two angles are used to specify the position of a particle

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