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There are two main descriptions of motion: dynamics and kinematics.Dynamics is general, since the momenta, forces and energy of the particles are taken into account. In this instance, sometimes the term dynamics refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler–Lagrange equations), and sometimes to the solutions to those equations.
The problem of compatibility in continuum mechanics involves the determination of allowable single-valued continuous fields on simply connected bodies. More precisely, the problem may be stated in the following manner. [5] Figure 1. Motion of a continuum body. Consider the deformation of a body shown in Figure 1.
The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields = Applying the trigonometric identity
[4] [5] [6] A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system. Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined.
An analytic solution to an inverse kinematics problem is a closed-form expression that takes the end-effector pose as input and gives joint positions as output, = (). Analytical inverse kinematics solvers can be significantly faster than numerical solvers and provide more than one solution, but only a finite number of solutions, for a given end ...
The second called inverse kinematics uses the position and orientation of the end-effector to compute the joint parameters values. Remarkably, while the forward kinematics of a serial chain is a direct calculation of a single matrix equation, the forward kinematics of a parallel chain requires the simultaneous solution of multiple matrix ...
We may write down the Lagrangian in terms of the position coordinates as they are, but it is an established procedure to convert the two-body problem into a one-body problem as follows. Introduce the Jacobi coordinates ; the separation of the bodies r = r 2 − r 1 and the location of the center of mass R = ( m 1 r 1 + m 2 r 2 )/( m 1 + m 2 ) .
Torque-free precessions are non-trivial solution for the situation where the torque on the right hand side is zero. When I is not constant in the external reference frame (i.e. the body is moving and its inertia tensor is not constantly diagonal) then I cannot be pulled through the derivative operator acting on L.