(g) Find the moment of inertia of a solid circular cylinder of radius a and mass m about an axis on the surface of the cylinder and parallel to the axis of the cylinder. (Given that the moment of inertia of the cylinder about its axis is (1)/(2)Ma^2 )

To find the moment of inertia of a solid circular cylinder of radius \(a\) and mass \(m\) about an axis on the surface of the cylinder and parallel to its axis, we use the **parallel axis theorem**.<br /><br />The **parallel axis theorem** states:<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 (the axis passing through the center of the cylinder),<br />- \(M\) is the mass of the object,<br />- \(d\) is the perpendicular distance between the two axes.<br /><br />### Step 1: Moment of inertia about the central axis<br />The moment of inertia of the cylinder about its central axis (through the center of mass) is given as:<br />\[<br />I_{\text{cm}} = \frac{1}{2}Ma^2<br />\]<br /><br />### Step 2: Distance between the axes<br />The axis on the surface of the cylinder is parallel to the central axis and at a distance \(d = a\) (the radius of the cylinder).<br /><br />### Step 3: Apply the parallel axis theorem<br />Using the parallel axis theorem:<br />\[<br />I = I_{\text{cm}} + Ma^2<br />\]<br />Substitute \(I_{\text{cm}} = \frac{1}{2}Ma^2\) and \(d = a\):<br />\[<br />I = \frac{1}{2}Ma^2 + Ma^2<br />\]<br />\[<br />I = \frac{3}{2}Ma^2<br />\]<br /><br />### Final Answer:<br />The moment of inertia of the cylinder about the given axis is:<br />\[<br />\boxed{\frac{3}{2}Ma^2}<br />\]