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Question 5–2
190 Chapter 5. Kinetics of Rigid Bodies where ¯IR is the moment of inertia tensor of the rod relative to the center of mass and FωR is the angular velocity of the rod in reference frame F. Now since the {er,eθ,Ez} is a principle-axis basis, we have that ¯IR = ¯I rrer ⊗er + ¯Iθθeθ ⊗eθ + ¯IzzEz ⊗Ez (5.58) Furthermore, using the expression for FωR as given in Eq.
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LECTURE NOTES FOR COURSE EML 4220 - Anil V. Rao
vdol.mae.ufl.edu2 Chapter 1. Response of Single Degree-of-Freedom Systems to Initial Conditions ℓ0 c P g m x K O Q Figure 1–1 Block of mass m sliding without friction along a horizontal surface con- nected to a linear spring and a linear viscous damper.
Chapter 5 Kinetics of Rigid Bodies - Anil V. Rao
vdol.mae.ufl.eduFigure 5-1 Free Body Diagram of Disk Now from the geometry we have that ... matic quantities in an inertial reference frame: • The acceleration of the collar • The acceleration of the center of mass of the rod • The rate of change of angular momentum of the rod relative to the center of
Question 3–12
vdol.mae.ufl.edu105 Question 3–12 A particle of mass m is attached to a linear spring with spring constant K and unstretched length r0 as shown in Fig. P3-12. The spring is attached at its other end at point P to the free end of a rigid massless arm of length l.The arm is hinged at its other end and rotates in a circular path at a constant angular
Chapter 3 Kinetics of Particles - Anil V. Rao
vdol.mae.ufl.eduChapter 3 Kinetics of Particles Question 3–1 A particle of mass m moves in the vertical plane along a track in the form of a circle as shown in Fig. P3-1. The equation for the track is r = r0 cosθ Knowing that gravity acts downward and assuming the initial conditions θ(t = 0) = 0 and θ(t˙ = 0) = θ˙0, determine (a) the differential equation of motion for the particle and (b) the force ...
Chapter 2 Kinematics
vdol.mae.ufl.edu4 Chapter 2. Kinematics Finally, corresponding to reference frame B, we choose the following coordinate system: Origin at point O er = Along OP ez = Out of Page eθ = ez ×er Then, the position of the particle can be desribed in terms of the basis {er,eθ,ez} as r = rer. (2.10) Now, in order to compute the velocity of the particle, it is ...
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