Question

Given the moment of inertia of a thin uniform disc about its diameter to be 1/2 MR² , where M and R are respectively the mass and radius of the disc, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer 1

Consider a thin uniform disc of mass M and radius R in the xy plane. Let Ix, ly and Iz be the moments of inertia of the disc about the x, y and z axes respectively.

Now, I_x = I_y

since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the MI of the disc about any diameter is the same.

∴ I_x = I_y= 1/4 MR² (Given)

According to the theorem of perpendicular axes,

I_z = I_x + I_y = 2(1/4MR²) = 1/2 MR²

Let I be the MI of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,

I = I_CM + Mh²

Here, I_CM = I_z = 1/2 MR² and h = R.

∴ I = 1/2 MR² + MR² = 3/2 MR²

which is the required expression.