Chemical Engineering - Apllied Mechanics

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: 2 1 T 2m =v (1.1) Conversely, the kinetic energy of a 3-D rigid body is given by: 2 V 1 TV 2d ρ = ∫ 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 every 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 1 TV 2d ρ = ∫ v (1.3) However, the mass m of the rigid body is given by: V V md ρ = ∫ (1.4) Therefore, we have: () 22 G 11 1 1 Tv 22 2 2mm m m ==×= =vvv 2v (1.5) That is, the kinetic energy of a translating body can be evaluated considering 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 =∧ − vω (1.6) where vector ω is the body angular velocity. This vector can be also expressed as: x yz =ω+ω+ ωijk ω (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 −ωijk vijk )x (1.8) That is: () () ( ) P vv v yz 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 1 TV Vvvv 22 2 xyz dd ρρ ρ ==×=++ ∫∫ ∫ vvv 22 V d (1.10) Substituting eq.(1.9) into eq.(1.10) we obtain: () () 222 1 TJ J J J J J 2 xx x yy y zz zxy x y yz y z zx z x =ω+ω+ω+ωω+ωω+ω ω (1.11) If the principal axes of the body are used to evaluate the mass moments of inertia eq.(1.11) becomes: () 22 1 TJ J J 2 2 xxx yy y zzz =ω+ω+ ω (1.12) 1.3 Roto-translational motion If the rigid body is subjected to a roto-translating motion the velocity of a generic point P, at which the infinitesimal volume d V is considered, is given by: ( ) PO PO =+∧− vvω (1.13) where vector ω is the body angular speed and v is the absolute velocity of a generic point O of the body. This involves the use of a translating mobile reference system with origin at O. The square value of the velocity v is given by: O P ( ) ( ) 2 2 PPP OO vPO2 P 2 ⎡⎤⎡⎤ =× = + ∧ − + × ∧ − ⎣⎦⎣⎦ vvv v ωω O (1.14) Then, the kinetic energy can be written as: () () 2222 O 11 TvJJJ J J J 22 xx x yy y zz zxy x y yz y z zx z x m =+ω+ω+ω+ωω+ωω+ωω + ( ) { } O +Gm O ⎡ ⎤ ×∧− ⎣ ⎦ v ω (1.15) where G is the center of mass of the body. () () 222 1 TJ J J J J J 2 xx x yy y zz zxy x y yz y z zx z x =ω+ω+ω+ωω+ωω+ω ω (1.16) If the principal axes of the body are used to evaluate the mass moments of inertia and point O coincides with the center of mass G, eq.(16) becomes: () 222 G 11 TvJJJ 22 2 xxx yy y zzz m = + ω+ ω+ ω (1.17)