A rigid body consists of three thin uniform rods, each of mass m and length 2a, held mutually perpendicular at their midpoints. Choose a coordinate system with axes along the rods. (a) Find the angular momentum and kinetic energy of the body if it rotates with angular velocity cc about an axis passing through the origin and the point (1, 1, 1). (b) Show that the moment of inertia is the same for any axis passing through the origin. (e) Show that the moment of inertia of a uniform square lamina is that given in Example 9.1.1 for any axis passing through the center of the lamina and lying in the plane of the lamina.
The correct answer and explanation is :
Let’s solve the problem in parts:
Given:
A rigid body is made of three thin uniform rods, each of mass m and length 2a, mutually perpendicular at their midpoints. The coordinate axes are aligned with the rods: x, y, z.
Let’s denote angular velocity vector ω, and compute:
(a) Angular Momentum and Kinetic Energy
The body rotates about an axis through the origin and the point (1, 1, 1), which means the angular velocity vector is:
[
\vec{\omega} = \omega \cdot \frac{1}{\sqrt{3}}(1, 1, 1)
]
Step 1: Moment of Inertia Tensor
Each rod has length 2a and mass m, centered at the origin. The moment of inertia of a thin rod of length L about an axis perpendicular to its length and through its center is:
[
I = \frac{1}{12} m L^2 = \frac{1}{12} m (2a)^2 = \frac{1}{3}ma^2
]
So we calculate the moment of inertia tensor:
- Rod along x: contributes to ( I_{yy} ) and ( I_{zz} )
- Rod along y: contributes to ( I_{xx} ) and ( I_{zz} )
- Rod along z: contributes to ( I_{xx} ) and ( I_{yy} )
Hence, total moment of inertia tensor ( I ) is diagonal:
[
I = \begin{bmatrix}
\frac{2}{3}ma^2 & 0 & 0 \
0 & \frac{2}{3}ma^2 & 0 \
0 & 0 & \frac{2}{3}ma^2 \
\end{bmatrix}
]
Step 2: Angular Momentum
[
\vec{L} = I \vec{\omega} = \frac{2}{3}ma^2 \cdot \vec{\omega} = \frac{2}{3}ma^2 \cdot \omega \cdot \frac{1}{\sqrt{3}} (1, 1, 1)
\Rightarrow \vec{L} = \frac{2}{3\sqrt{3}}ma^2 \omega (1, 1, 1)
]
Step 3: Kinetic Energy
[
T = \frac{1}{2} \vec{\omega} \cdot \vec{L}
= \frac{1}{2} \left( \omega \cdot \frac{1}{\sqrt{3}} (1, 1, 1) \cdot \frac{2}{3\sqrt{3}}ma^2 \omega (1, 1, 1) \right)
= \frac{1}{2} \cdot \omega^2 \cdot \frac{2}{9}ma^2 \cdot 3
= \frac{1}{3}ma^2 \omega^2
]
(b) Moment of Inertia is Same for Any Axis Through Origin
Since the inertia tensor is isotropic (diagonal with equal elements), it behaves like a scalar multiple of identity. Hence:
For any unit vector ( \hat{n} ), the moment of inertia about that axis is:
[
I_n = \hat{n}^T I \hat{n} = \frac{2}{3}ma^2 (\hat{n}_x^2 + \hat{n}_y^2 + \hat{n}_z^2) = \frac{2}{3}ma^2
]
Because ( \hat{n} ) is a unit vector, the squared components sum to 1.
(e) Moment of Inertia of Square Lamina in Plane
From Example 9.1.1, the moment of inertia of a square lamina (mass m, side 2a) about an axis in the plane and through the center (e.g., x-axis) is:
[
I = \frac{1}{3}ma^2
]
Due to symmetry of the square, rotation about x or y (in-plane, through center) yields the same result.
This is verified by integrating over the area:
[
I_x = \int y^2 \, dm = \int_{-a}^a \int_{-a}^a y^2 \cdot \frac{m}{4a^2} \, dy \, dx = \frac{1}{3}ma^2
]
Conclusion
- (a): Angular momentum is ( \vec{L} = \frac{2}{3\sqrt{3}}ma^2 \omega (1,1,1) ), and kinetic energy is ( \frac{1}{3}ma^2\omega^2 ).
- (b): Moment of inertia is the same (( \frac{2}{3}ma^2 )) for any axis through the origin due to symmetry.
- (e): The square lamina’s in-plane central axis inertia is ( \frac{1}{3}ma^2 ), same as derived in the referenced example.