4. (a)(i)State and prove the perpendicular axes theorem for the moment of inertia. (7 marks) (ii) Find the moment of inertia of a square plate of mass M and side a about an axis perpendicular to the plate and passing through its centre. (Given that the moment of inertia of the plate about its side is (1)/(3)Ma^2 (6 marks)

(a)(i) **Perpendicular Axes Theorem:**<br /><br />The perpendicular axes theorem is applicable to planar bodies (2D objects). It states that if a planar object lies in the XY-plane, then the moment of inertia about an axis perpendicular to this plane (the Z-axis) is equal to the sum of the moments of inertia about the two perpendicular axes lying in the plane (the X and Y axes).<br /><br />Mathematically, it can be expressed as:<br /><br />\[ I_z = I_x + I_y \]<br /><br />**Proof:**<br /><br />Consider a planar body lying in the XY-plane. Let \( dm \) be an infinitesimal mass element at a point with coordinates \((x, y)\).<br /><br />1. The moment of inertia about the Z-axis is given by:<br /> \[ I_z = \int (x^2 + y^2) \, dm \]<br /><br />2. The moment of inertia about the X-axis is:<br /> \[ I_x = \int y^2 \, dm \]<br /><br />3. The moment of inertia about the Y-axis is:<br /> \[ I_y = \int x^2 \, dm \]<br /><br />4. Adding \( I_x \) and \( I_y \):<br /> \[ I_x + I_y = \int y^2 \, dm + \int x^2 \, dm = \int (y^2 + x^2) \, dm \]<br /><br />5. From the expressions above, we see that:<br /> \[ I_z = I_x + I_y \]<br /><br />This completes the proof of the perpendicular axes theorem.<br /><br />(ii) **Moment of Inertia of a Square Plate:**<br /><br />Given that the moment of inertia of the plate about its side is \(\frac{1}{3}Ma^2\), we need to find the moment of inertia about an axis perpendicular to the plate and passing through its center.<br /><br />For a square plate of side \(a\) and mass \(M\), the moment of inertia about an axis through its center and parallel to one of its sides is given as \(\frac{1}{3}Ma^2\).<br /><br />Since the plate is symmetric, the moment of inertia about the other axis parallel to the other side and through the center is also \(\frac{1}{3}Ma^2\).<br /><br />Using the perpendicular axes theorem for the square plate:<br /><br />\[ I_z = I_x + I_y = \frac{1}{3}Ma^2 + \frac{1}{3}Ma^2 = \frac{2}{3}Ma^2 \]<br /><br />Thus, the moment of inertia of the square plate about an axis perpendicular to the plate and passing through its center is \(\frac{2}{3}Ma^2\).