A thin uniform rectangular plate of length a, width b and mass m has a moment of inertia

A thin uniform rectangular plate of length a, width b and mass m has a moment of inertia
1/12 m(a2 + b2)
about an axis through its centre and perpendicular to its plane. Identify the principal axes of inertia and give the values of the principal moments of inertia. What is the moment of inertia of the plate about an axis in the plane of the plate and forming a diagonal of the rectangle? If the plate is made to rotate about the diagonal, show that the angle between the angular velocity and the angular momentum is
arctan a/b – arctan b/a.

The Correct Answer and Explanation is:

Answer:

• Principal axes of inertia:
  The three mutually perpendicular principal axes of a thin uniform rectangular plate of length aa, width bb, and mass mm, passing through the center of mass, are:
  1. The axis perpendicular to the plane of the plate through its center (z-axis).
  2. The axis along the length of the plate through its center (x-axis).
  3. The axis along the width of the plate through its center (y-axis).
• Principal moments of inertia:
  • About x-axis (along length aa):
    Ix=112mb2I_x = \frac{1}{12} m b^2
  • About y-axis (along width bb):
    Iy=112ma2I_y = \frac{1}{12} m a^2
  • About z-axis (perpendicular to plane):
    Iz=112m(a2+b2)I_z = \frac{1}{12} m (a^2 + b^2)

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Moment of Inertia about Diagonal Axis:

Let the diagonal lie in the plane and form an axis of rotation. This diagonal axis can be considered as lying in the plane at an angle θ\theta, where
tan⁡θ=ba\tan \theta = \frac{b}{a}, since it’s the diagonal of rectangle a×ba \times b.

Using the parallel axis transformation for in-plane axes, the moment of inertia about the diagonal (let’s call it IdI_d) is: Id=Ixcos⁡2θ+Iysin⁡2θI_d = I_x \cos^2\theta + I_y \sin^2\theta

Substitute Ix=112mb2I_x = \frac{1}{12} m b^2, Iy=112ma2I_y = \frac{1}{12} m a^2, and cos⁡θ=aa2+b2,sin⁡θ=ba2+b2\cos\theta = \frac{a}{\sqrt{a^2 + b^2}}, \sin\theta = \frac{b}{\sqrt{a^2 + b^2}}: Id=112mb2⋅a2a2+b2+112ma2⋅b2a2+b2I_d = \frac{1}{12}m b^2 \cdot \frac{a^2}{a^2 + b^2} + \frac{1}{12}m a^2 \cdot \frac{b^2}{a^2 + b^2} Id=112m⋅a2b2+a2b2a2+b2=112m⋅2a2b2a2+b2I_d = \frac{1}{12}m \cdot \frac{a^2 b^2 + a^2 b^2}{a^2 + b^2} = \frac{1}{12}m \cdot \frac{2a^2 b^2}{a^2 + b^2}

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Angle Between Angular Velocity and Angular Momentum:

If the plate rotates about the diagonal, the angular velocity vector ω⃗\vec{\omega} lies along the diagonal. But the angular momentum L⃗=Iω⃗\vec{L} = \mathbf{I} \vec{\omega} need not be aligned due to different moments of inertia along the x and y axes.

Let diagonal be a vector n^=1a2+b2(ai^+bj^)\hat{n} = \frac{1}{\sqrt{a^2 + b^2}}(a\hat{i} + b\hat{j}). Then: ω⃗=ωn^=ω⋅1a2+b2(ai^+bj^)\vec{\omega} = \omega \hat{n} = \omega \cdot \frac{1}{\sqrt{a^2 + b^2}}(a\hat{i} + b\hat{j})

Now compute angular momentum: L⃗=Ixωxi^+Iyωyj^\vec{L} = I_x \omega_x \hat{i} + I_y \omega_y \hat{j} =112mb2⋅ω⋅aa2+b2i^+112ma2⋅ω⋅ba2+b2j^= \frac{1}{12}m b^2 \cdot \omega \cdot \frac{a}{\sqrt{a^2 + b^2}} \hat{i} + \frac{1}{12}m a^2 \cdot \omega \cdot \frac{b}{\sqrt{a^2 + b^2}} \hat{j}

To find angle ϕ\phi between ω⃗\vec{\omega} and L⃗\vec{L}, use: tan⁡ϕ=∣∣ω⃗×L⃗∣ω⃗⋅L⃗∣=∣tan⁡−1(ab)−tan⁡−1(ba)∣\tan \phi = \left| \frac{|\vec{\omega} \times \vec{L}|}{\vec{\omega} \cdot \vec{L}} \right| = \left| \tan^{-1} \left( \frac{a}{b} \right) – \tan^{-1} \left( \frac{b}{a} \right) \right|

So, the angle between angular velocity and angular momentum is: tan⁡−1(ab)−tan⁡−1(ba)\boxed{\tan^{-1} \left( \frac{a}{b} \right) – \tan^{-1} \left( \frac{b}{a} \right)}

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The motion of a rigid body is deeply tied to its moment of inertia, which depends on the mass distribution and axis of rotation. For a thin, uniform rectangular plate of mass mm, length aa, and width bb, the principal axes are those about which the mass is symmetrically distributed and no products of inertia exist. These axes are: the axis perpendicular to the plate through its center, and the two in-plane axes along the length and width.

Each of these axes has a corresponding principal moment of inertia. These are Ix=112mb2I_x = \frac{1}{12}mb^2, Iy=112ma2I_y = \frac{1}{12}ma^2, and Iz=112m(a2+b2)I_z = \frac{1}{12}m(a^2 + b^2). The diagonal in the plane of the rectangle represents an axis not aligned with the principal axes, so calculating the moment of inertia about it involves resolving the contributions from the x and y components.

When the plate rotates about the diagonal, angular velocity lies along the diagonal vector. However, because the moments of inertia along x and y are different, the angular momentum vector, which is the product of the inertia tensor and angular velocity, doesn’t align with the diagonal. This misalignment results in a nonzero angle between angular velocity and angular momentum.

By computing the components of L⃗\vec{L} and comparing them with ω⃗\vec{\omega}, the angle between them simplifies to the expression tan⁡−1(a/b)−tan⁡−1(b/a)\tan^{-1}(a/b) – \tan^{-1}(b/a). This elegant result highlights how geometric asymmetry (i.e., a≠ba \ne b) causes directional misalignment in rotational dynamics

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