13.12: Kinetic Energy of Rotating Rigid Body
( \newcommand{\kernel}{\mathrm{null}\,}\)
An important observable is the kinetic energy of rotation of a rigid body. Consider a rigid body composed of N particles of mass mα where α=1,2,3,…N. If the body rotates with an instantaneous angular velocity ω about some fixed point, with respect to the body coordinate system, and this point has an instantaneous translational velocity V with respect to the fixed (inertial) coordinate system, see Figure 13.3.1, then the instantaneous velocity vα of the αth particle in the fixed frame of reference is given by
vα=V+v′′α+ω×r′α
However, for a rigid body, the velocity of a body-fixed point with respect to the body is zero, that is v′′α=0, thus
vα=V+ω×r′α
The total kinetic energy is given by
T=N∑α12mαvα⋅vα=N∑α12mα(V+ω×r′α)⋅(V+ω×r′α)=12N∑αmαV2+N∑imαV⋅ω×r′α+12N∑αmα(ω×r′α)⋅(ω×r′α)
This is a general expression for the kinetic energy that is valid for any choice of the origin from which the body-fixed vectors r′α are measured. However, if the origin is chosen to be the center of mass, then, and only then, the middle term cancels. That is, since V⋅ω is independent of the specific particle, then
N∑αmαV⋅ω×r′α=V⋅ω×(N∑αmαr′α)
But the definition of the center of mass is
∑αmαr′=MR
and R=0 in the body-fixed frame if the selected point in the body is the center of mass. Thus, when using the center of mass frame, the middle term of Equation 13.12.3 is zero. Therefore, for the center of mass frame, the kinetic energy separates into two terms in the body-fixed frame
T=Ttrans+Trot
where
Ttrans=12N∑αmαV2
Trot=12N∑αmi(ω×r′α)⋅(ω×r′α)
The vector identity
(A×B)⋅(A×B)=A2B2−(A⋅B)2
can be used to simplify Trot
Trot=12N∑αmα[ω2r′2α−(ω⋅r′α)2]
The rotational kinetic energy Trot can be expressed in terms of components of ω and r′α in the body-fixed frame. Also the following formulae are greatly simplified if r′α=(xα,yα,zα) in the rotating body-fixed frame is written in the form r′α=(xα,1,xα,2,xα,3) where the axes are defined by the numbers 1,2,3 rather than x,y,z. In this notation the rotational kinetic energy is written as
Trot=12N∑αmα[(∑iω2i)(∑kx2α,k)−(∑iωixα,i)(∑jωjxα,j)]
Assume the Kronecker delta relation
ωi=3∑jωjδij
where δij=1 if i=j and δij=0 if i≠j.
Then the kinetic energy can be written more compactly
Trot=12N∑αmα[(∑iω2i)(∑kx2α,k)−(∑iωixα,i)(∑jωjxα,j)]=12N∑α3∑i,jmα[(ωiωjδij)(3∑kx2α,k)−(ωixα,i)(ωjxα,j)]=123∑i,jωiωj[N∑αmα[δij(3∑kx2α,k)−xα,ixα,j]]
The term in the outer square brackets is the inertia tensor defined in equation (13.4.1) for a discrete body. The inertia tensor components for a continuous body are given by equation (13.4.2).
Thus the rotational component of the kinetic energy can be written in terms of the inertia tensor as
Trot=123∑i,jIijωiωj
Note that when the inertia tensor is diagonal, then the evaluation of the kinetic energy simplifies to
Trot=123∑iIiiω2i
which is the familiar relation in terms of the scalar moment of inertia I discussed in elementary mechanics.
Equation ??? also can be factored in terms of the angular momentum L.
Trot=12∑i,jIijωiωj=12∑iωi∑jIijωj=12∑iωiLi
As mentioned earlier, tensor algebra is an elegant and compact way of expressing such matrix operations. Thus it is possible to express the rotational kinetic energy as
Trot=12(ω1 ω2 ω3)⋅(I11I12I13I21I22I23I31I32I33)⋅(ω1ω2ω3)
Trot≡T=12ω⋅{I}⋅ω
where the rotational energy T is a scalar. Using equation (13.11.1) the rotational component of the kinetic energy also can be written as
Trot≡T=12ω⋅L
which is the same as given by ???. It is interesting to realize that even though L={I}⋅ω is the inner product of a tensor and a vector, it is a vector as illustrated by the fact that the inner product Trot=12ω⋅L=12ω⋅({I}⋅ω) is a scalar. Note that the translational kinetic energy Ttrans must be added to the rotational kinetic energy Trot to get the total kinetic energy as given by Equation ???.