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 piate 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) (b) Find the moment of inertia of a ring of radius a and mass M about an axis perpendicular to the plane of the ring and passing through its edge. (7 marks)

4. (a)(i) **Perpendicular Axes Theorem:**<br /><br />The perpendicular axes theorem states that for a planar body lying in the XY-plane, 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 (X and Y) lying in the plane of the body.<br /><br />Mathematically, it can be expressed as:<br />\[ I_z = I_x + I_y \]<br /><br />**Proof:**<br /><br />Consider a planar object lying in the XY-plane. Let \( dm \) be an infinitesimal mass element at a point with coordinates \((x, y)\). The moment of inertia of this mass element about the Z-axis is given by:<br />\[ dI_z = dm \cdot (x^2 + y^2) \]<br /><br />The total moment of inertia about the Z-axis is obtained by integrating over the entire body:<br />\[ I_z = \int (x^2 + y^2) \, dm \]<br /><br />This can be split into two separate integrals:<br />\[ I_z = \int x^2 \, dm + \int y^2 \, dm \]<br /><br />These integrals represent the moments of inertia about the X-axis and Y-axis, respectively:<br />\[ I_x = \int y^2 \, dm \]<br />\[ I_y = \int x^2 \, dm \]<br /><br />Thus, we have:<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 \(a\), the moment of inertia about an axis through its center and parallel to one of its sides is \(\frac{1}{12}Ma^2\).<br /><br />Using the perpendicular axes theorem:<br />\[ I_z = I_x + I_y \]<br /><br />Since the plate is symmetric, \(I_x = I_y = \frac{1}{12}Ma^2\).<br /><br />Therefore:<br />\[ I_z = \frac{1}{12}Ma^2 + \frac{1}{12}Ma^2 = \frac{1}{6}Ma^2 \]<br /><br />(b) **Moment of Inertia of a Ring:**<br /><br />To find the moment of inertia of a ring of radius \(a\) and mass \(M\) about an axis perpendicular to the plane of the ring and passing through its edge, we use the parallel axis theorem.<br /><br />The moment of inertia of the ring about an axis through its center and perpendicular to its plane is:<br />\[ I_{\text{center}} = Ma^2 \]<br /><br />Using the parallel axis theorem, the moment of inertia about an axis at the edge is:<br />\[ I_{\text{edge}} = I_{\text{center}} + Md^2 \]<br /><br />where \(d = a\) (the distance from the center to the edge).<br /><br />Thus:<br />\[ I_{\text{edge}} = Ma^2 + Ma^2 = 2Ma^2 \]