(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)

To find the moment of inertia of a ring about an axis perpendicular to its plane and passing through its edge, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia \( I \) about any axis parallel to an axis through the center of mass is given by:<br /><br />\[ I = I_{\text{cm}} + Md^2 \]<br /><br />where:<br />- \( I_{\text{cm}} \) is the moment of inertia about the center of mass,<br />- \( M \) is the mass of the object,<br />- \( d \) is the distance between the two axes.<br /><br />For a ring of radius \( a \) and mass \( M \), the moment of inertia about an axis through its center and perpendicular to its plane is:<br /><br />\[ I_{\text{cm}} = Ma^2 \]<br /><br />The distance \( d \) from the center of the ring to the edge (where the new axis is located) is equal to the radius \( a \). Therefore, using the parallel axis theorem:<br /><br />\[ I = Ma^2 + M(a)^2 = Ma^2 + Ma^2 = 2Ma^2 \]<br /><br />Thus, the moment of inertia of the ring about an axis perpendicular to its plane and passing through its edge is \( 2Ma^2 \).