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Q7.26(a):
(a) Prove the theorem of perpendicular axes. (Hint : Square of the distance of a point $(x, y)$ in the $x–y$ plane from an axis through the origin and perpendicular to the plane is $x^2+y^2$).

Solution :

1. Theoretical Statement of the Theorem

The Theorem of Perpendicular Axes states that the moment of inertia of a planar laminar body about an axis perpendicular to its plane ($I_z$) is equal to the sum of the moments of inertia of the body about two mutually perpendicular axes lying in the plane of the body ($I_x$ and $I_y$), such that all three axes are concurrent at a point (the origin $O$).

Mathematically, the theorem is expressed as: $I_z = I_x + I_y$

2. Geometric Construction and Variables

x y O P(x, y)

3. Derivation Steps

Consider a planar laminar object of mass $M$ lying in the $xy$-plane. Let there be a particle of mass $m_i$ located at coordinates $(x_i, y_i)$ relative to the origin $O$.

Step A: Defining Moments of Inertia

  • The distance of the particle from the $x$-axis is $y_i$. Thus, the moment of inertia about the $x$-axis is: $I_x = \sum m_i y_i^2$ [Per definition of Moment of Inertia $I = \sum mr^2$]
  • The distance of the particle from the $y$-axis is $x_i$. Thus, the moment of inertia about the $y$-axis is: $I_y = \sum m_i x_i^2$

Step B: Moment of Inertia about the $z$-axis

The $z$-axis passes through the origin and is perpendicular to the $xy$-plane. The distance $r_i$ of the particle from the $z$-axis (the origin) is given by the Pythagorean theorem: $r_i^2 = x_i^2 + y_i^2$ [By the Pythagorean Theorem in the $xy$-plane]

The moment of inertia about the $z$-axis is: $I_z = \sum m_i r_i^2$

Step C: Algebraic Substitution

Substitute the expression for $r_i^2$ into the equation for $I_z$: $I_z = \sum m_i (x_i^2 + y_i^2)$

Distribute the summation: $I_z = \sum m_i x_i^2 + \sum m_i y_i^2$

Substituting the definitions derived in Step A: $I_z = I_y + I_x$

4. Analytical Justification

Axis Distance of $m_i$ Moment of Inertia Contribution
$x$-axis $y_i$ $m_i y_i^2$
$y$-axis $x_i$ $m_i x_i^2$
$z$-axis $r_i = \sqrt{x_i^2 + y_i^2}$ $m_i (x_i^2 + y_i^2)$

Final Solution: $I_z = I_x + I_y$. The theorem holds that the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia about two mutually perpendicular axes in that plane.


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