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 = ½Σm_{i}v_{i}^{2}
= ½Σm_{i}(V + Ω
× r)^{2}.
We rewrite
T = ½MV^{2} + ½Σm_{i}(Ω^{2}r_{i}^{2}
- (Ω×r_{i})^{2}), M =
Σm_{i},
Σm_{i}r_{i} = 0.
We find T = T_{CM }+ T_{rot}, 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
write
T_{rot} = ½Σ_{ij}I_{ij}Ω_{i}Ω_{j}.
where
I_{ij} = Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}
- (x_{k})_{i}(x_{k})_{j}] is the inertia
tensor.
The Ω_{i}
are the components of Ω
along the axis of the body fixed system. For a continuous system
Σ_{k}m_{k} --> ∫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 x_{i} are then called the
principal
axes of inertia and the diagonal components of the tensor are
then called the principal
moments of inertia.
Then T_{rot} = ½[I_{1}Ω_{1}^{2} + I_{2}Ω_{2}^{2}
+ I_{3}Ω_{3}^{2}]
Definitions:
Asymmetrical top: | I_{1} ≠ I_{2}, I_{1} ≠ I_{3}, I_{2} ≠ I_{3} | |
Symmetrical top: | I_{1} = I_{2} ≠ I_{3} | |
Spherical top: | I_{1} = I_{2} = I_{3} |
The total angular momentum of a rigid body about a
point is L_{tot} = Σm_{i}r_{i}×(V +
Ω × r_{i})
= MR×V + Σm_{i}r_{i}×(
Ω × r_{i}).
L_{tot} =
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 = Σ_{k}m_{k}r_{k}×(V +
Ω × r_{k})
= Σ_{k}m_{k}[r_{k}^{2}Ω - r_{k}(r_{k}∙Ω)],
which in component form yields
L_{i} = Σ_{j}Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}
- (x_{k})_{i}(x_{k})_{j}]Ω_{j} = Σ_{j}I_{ij}Ω_{j}.
If x_{1}, x_{2}, and x_{3} are
the principal axes of inertia, then
L_{1} = I_{1}Ω_{1},
L_{2} = I_{2}Ω_{2}, L_{3}
= I_{3}Ω_{3}.
When we consider the motion of a rigid body, we distinguish two situations.
(a) No point of the body is fixed by constraints. This problem is
best treated by separating the motion of the body int motion of the CM and
motion about the CM.
Then T = T_{CM }+ T_{rot}, and L(about a point) =
angular momentum of the CM about the point plus angular momentum about the CM.
(b) One point of the body is fixed by constraints. The problem is
often best treated by treating the motion as a pure rotation about an axis
through this point.
Then T =
T_{rot} = ½Σ_{ij}I_{ij}Ω_{i}Ω_{j},
L_{i} = Σ_{j}I_{ij}Ω_{j}, with I_{ij}
= Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}
- (x_{k})_{i}(x_{k})_{j}]. The origin of
the coordinate system then is the fixed point.
The parallel axes theorem (Steiner's theorem)
Let I be the moment of inertia tensor of a body of mass M, calculated in a
body fixed system with origin S at the CM. Let I' be the moment of inertia
tensor calculated in a body fixed system with parallel axes, but with origin
S' with origin S' = S + a. Then I and I' are
related by
I'_{ij} = I_{ij }+ M (a^{2} δ_{ij}
- a_{i}a_{j}).
For the diagonal elements we have I'_{ii} = I_{ii }+ M (a^{2}
- a_{i}^{2}) = I_{ii }+ M a_{⊥}^{2}.
The equations of motion of a rigid body are dP/dt
= F,
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.
Therefore
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
L_{i} = I_{i}Ω_{i} and
d'L_{i}/dt = I_{i}dΩ_{i}/dt,
and we have Euler’s equations:
I_{1}dΩ_{1}/dt + (I_{3} - I_{2})Ω_{2}Ω_{3} = τ_{1}
I_{2}dΩ_{2}/dt + (I_{1} - I_{3})Ω_{3}Ω_{1} = τ_{2}
I_{3}dΩ_{3}/dt + (I_{2} - I_{3})Ω_{1}Ω_{2} = τ_{3}
Center of mass:
x_{CM} = Σm_{i}x_{i}/M, y_{CM} = Σm_{i}y_{i}/M,
z_{CM} = Σm_{i}z_{i}/M.
M = Σm_{i}.
Mv_{CM} = Σm_{i}v_{i} = Σp_{i}
= p_{tot}.
Ma_{CM} = Σm_{i}a_{i} = ΣF_{i}
= F_{tot}.Mv_{CM} = Σm_{i}v_{i}
= Σp_{i} = p_{tot}.
For an object to be in static equilibrium, the net force and the net torque about any axis must be zero.