Lecture Note (6): Kinetics of Spatial Mechanisms: Kinetics Equation Objectives - Kinetics equations - Problem solving procedure - Example analysis Kinetics Equation With respect to the center of gravity G of a body system, we have the following two sets of equations: Equation (6.1) for force equilibrium and Equation (6.2) for moment equilibrium. It is noted that Equation (6.1) is the vector form and can be extended to 3 scalar equations. Therefore, in total, we have 6 equations, and they describe the relation between the force / moment and linear motion / angular motion. It is further noted that the inertia tensor (moment of inertia and product of inertia) in these equations is with regard to the coordinate system which passes through the center of gravity. According to our previous discussion in lecture note 5, in the moment equation, we could also take any other point, in particular the fixed point (P). When a fixed point is chosen, the form of the moment equation remains to be the same except that G is changed to P in the subscript of items in equation (6.2), which leads to Equation (6.3). We should also note that in Equation (6.3), the inertia tensor is with respect to a coordinate system which passes through point P. F = ma G (6.1) 2 2 M G = I α ( IY I Z ) ωyωz I Y ( α Y ω Z ω ) IYZ ( ωy ωz ) I Z ( α Z + ω ωy ) 2 2 M GY = IYα Y ( I Z I ) ωzω IYZ ( α Z ω ωy ) I Z ( ωz ω ) I Y ( α + ωyωz ) (6.2) 2 2 M GZ = I Zα Z ( I IY ) ω ωy I Z ( α ω Y ωz ) I Y ( ω ωy ) IYZ ( α Y + ωzω ) 2 2 MP = Iα ( IY IZ) ωω Y Z IY( αy ω Zω) IYZ( ωy ωz) IZ( αz + ωωy) 2 2 MPY = I α Y Y ( IZ I) ωω Z IYZ( α Z ω ω Y) IZ( ω Z ω ) IY( α ωω + Y Z) (6.3) 2 2 M = I α ( I I ) ωω I ( α ωω) I ( ω ω) I ( α + ωω ) PZ Z Z Y Y Z Y Z Y Y YZ Y Z Page 1 of 16
Problem solving procedure Refer to Example 1 in Appendix A Step 1: Establish a reference coordinate system: x-y-z (A). Usually, we prefer to set up the origin of the coordinate system at the fixed point (point A in Example 1) and to choose the principal axes. Step 2: Establish a separate force diagram. Step 3: Do the analysis of known and unknown variables to confirm if there are six or less six unknowns. Step 4: Establish the force equation ( F = ma G ). There are three scalar equations. Step 5: Select a point and establish the moment equation around the coordinate system that passes through this point. Step 6: Find the expression of a G, including the center of gravity, moment and production of inertia. Example Analysis Appendix A contains two examples. In the first example, the given conditions include the torque on the vertical shaft. When there is a torque applied, there will be usually an angular acceleration; in this case, α z, α z is an unknown variable. Quite often, we may consider α z is zero, which is wrong. So in this example, the unknown variables are: A x, A y, A z, B x, B y, α z (see Appendix A). We should also notice the following ϖ = x y y ( ω, ω, ω ) = (0,0,10) α = α, α, α ) = (0,0, α ) ( x y y z In the second example (see Appendix A), we need to notice the following points: (1) We have two bodies; CD and AB. Their connection is through a pin joint. Page 2 of 16
(2) We have a separate force diagram for the rod AB (see Appendix A). There are six unknown variables: Ax, Ay, Az, Mx, My, and T. In summary, there are basically two classes of problems: Class 1: We know the external force and / or torque. We find the acceleration. Class 2: We know the acceleration. We find the external force and / or torque which maintains the acceleration. Both classes of problems find the support force and moment. - End Page 3 of 16
Appendix A There are two examples in this appendix. The first problem falls into the class 1, while the second problem falls into the class 2. Page 4 of 16
Example 1 Figure A1 Page 5 of 16
Figure A2 Page 6 of 16
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Example 2 Page 10 of 16
Example 2 Page 11 of 16
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r = (sin 40) iˆ+ (cos 40) ˆj G/ A Page 14 of 16
[(sin 40) iˆ+ (cos 40) ˆj]) Page 15 of 16
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