Kinetic energy

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1. KINETIC ENERGY The kinetic energy, T, of a mass point m whose absolute velocity is v is given by: 21T 2 m= v (1.1) Conversely, the kinetic energy of a 3-D rigid body is given by: 2 V 1TV 2 dρ= ∫ v (1.2) where v is the absolute velocity vector of the local infinitesimal volume d V of the body. The general expression of the eq.(1.2) can be modified depending on the type of motion of the rigid body. 1.1 Transaltional motion In the case of a translating body ev ery point P has the same velocity v. Therefore, also the center of gravity G has the same velocity vector: v = v. In this case, being vector v independent on the position of the volume G d V, it is possible to write: 2 V 1TV 2 dρ= ∫v (1.3) However, the mass m of the rigid body is given by: V Vmd ρ= ∫ (1.4) Therefore, we have: () 22 G 11 1 1Tv 22 2 2mm m m== × = =vv v 2v (1.5) That is, the kinetic energy of a translating body can be evaluated consideri ng its global mass and the square value of the barycenter absolute velocity, v . G 1.2 Rotational motion For a rigid body subjected to a rotational motion about a fixed axis, passing through point O, the velocity of a generic point P, at which the infinitesimal volume d V is considered, is given by: ( )P PO=∧ −v ω (1.6) where vector ω is the body angular velocity. This vector can be also expressed as: x yz=ω +ω + ωijkω (1.7) where ω , ω , ω , are the angular velocity components about the axes x, y, z. x y z Therefore, vector v can be expressed as:P () () (P det xyz y z z x x y zy xz y xyz = ω ω ω = ω −ω + ω −ω + ω −ω ij k vi j k )x (1.8) That is: () () ( )P vv vyz z x x y x yzy xz y x= ω −ω + ω −ω + ω −ω = + +vi j k i j k z (1.9) Then, the kinetic energy can be written as: () () 22 VV V 11 1TV V v v v22 2 xy zddρρ ρ== × = + +∫∫ ∫ vv v 2 2 Vd (1.10) Substituting…