A rigid body is a system of particles in which the distances between the particles do not vary. To describe the motion of a rigid body we use two systems of coordinates, a space-fixed system X, Y, Z, and a moving system x, y, z, which is rigidly fixed in the body and participates in its motion.
Let the origin of the body-fixed system be the body’s center of mass (CM). The orientation of the axes of that system relative to the axes of the space-fixed system is given by three independent angles. The vector R points from the origin of the spaced-fixed system to the CM of the body. Thus a rigid body is a mechanical system with six degrees of freedom.
Let r denote the position of an arbitrary point
P in the body-fixed system. In the space fixed system its position is
given by r + R, and its velocity is
v = d(R + r)/dt = dR/dt + dr/dt = V + Ω × r.
Here V is the velocity of the CM and Ω is the angular velocity of the body. The direction of Ω is along the axis of rotation and Ω = dφ/dt.
The kinetic energy of the body is
T = ½Σmivi2 = ½Σmi(V + Ω × r)2.
T = ½MV2 + ½Σmi(Ω2ri2 - (Ω×ri)2), M = Σmi, Σmiri = 0.
We find T = TCM + Trot, i.e. the kinetic energy is the sum of the kinetic energy of the motion of the CM and the kinetic energy of the rotation about the CM.
In component form we
Trot = ½ΣijIijΩiΩj.
Iij = Σkmk[Σl(xk)l2δij - (xk)i(xk)j] is the inertia tensor.
The Ωi are the components of Ω along the axis of the body fixed system. For a continuous system Σkmk --> ∫dm = ∫ρdV.
By appropriate choice of the orientation of the
body-fixed coordinate system the inertia tensor can be reduced to diagonal form.
The directions of the axes xi are then called the
axes of inertia and the diagonal components of the tensor are
then called the principal
moments of inertia.
Then Trot = ½[I1Ω12 + I2Ω22 + I3Ω32]
|Asymmetrical top:||I1 ≠ I2, I1 ≠ I3, I2 ≠ I3|
|Symmetrical top:||I1 = I2 ≠ I3|
|Spherical top:||I1 = I2 = I3|
The total angular momentum of a rigid body about a
point is Ltot = Σmiri×(V +
Ω × ri)
= MR×V + Σmiri×(
Ω × ri).
Ltot = angular momentum of the CM about the point plus angular momentum about the CM.
Let L denote the angular momentum about the CM of the body.
L = Σkmkrk×(V + Ω × rk) = Σkmk[rk2Ω - rk(rk∙Ω)],
which in component form yields
Li = ΣjΣkmk[Σl(xk)l2δij - (xk)i(xk)j]Ωj = ΣjIijΩj.
If x1, x2, and x3 are
the principal axes of inertia, then
L1 = I1Ω1, L2 = I2Ω2, L3 = I3Ω3.
The equations of motion of a rigid body are dP/dt
where P = total momentum and F = external forces, and dL/dt
= τ ,
where L = angular momentum about CM and τ = total torque produced
by external forces.
Let A be an arbitrary vector and dA/dt its rate of change with respect to the space fixed axes, d'A/dt its rate of change with respect to the body fixed axes.
dA/dt = d'A/dt + Ω × A.
dP/dt = d'P/dt + Ω × P = F and dL/dt = d'L/dt + Ω × L = τ.
Let the body fixed axes be the principal axes of inertia of the body. Then
Li = IiΩi and d'Li/dt = IidΩi/dt,
and we have Euler’s equations:
I1dΩ1/dt + (I3 - I2)Ω2Ω3 = τ1
I2dΩ1/dt + (I1 - I3)Ω3Ω1 = τ2
I3dΩ1/dt + (I2 - I3)Ω1Ω2 = τ3
Center of mass:
xCM = Σmixi/M, yCM = Σmiyi/M, yCM = Σmiyi/M.
M = Σmi.
MvCM = Σmivi = Σpi = ptot.
MaCM = Σmiai = ΣFi = Ftot.MvCM = Σmivi = Σpi = ptot.
For an object to be in static equilibrium, the net force and the net torque about any axis must be zero.