Question

Assuming the expression for the moment of inertia of a ring about its transverse symmetry axis, obtain the expression for its moment of inertia about 

(1) a diameter 

(2) a tangential axis in its plane. Also deduce the expressions for the corresponding radii of gyration.

Answer 1

Let M be the mass of a thin ring of radius R. Let /CM be the moment of inertia (MI) of the ring about its transverse symmetry axis. Then, 

I_CM = MR … (1)

(1) MI about a diameter : Let x- and y-axes be along two perpendicular diameters of the ring as shown in below figure. Let I_x , I_y and I_z be the moments of inertia of the ring about the x, y and z axes, respectively.

Both I_x and I_y represent the moment of inertia of the ring about its diameter and, by symmetry, the MI of the ring about any diameter is the same.

∴ I_x = I_y ….. (2)

Also, I_z being the MI of the ring about its transverse symmetry axis,

Radius of gyration : The radius of gyration of the ring for rotation about its diameter is

(2) MI about a tangent in its plane: Let I be its MI about an axis in plane of the ring, i.e., parallel to a diameter, and tangent to it. Here, h = R and

I_CM = I_x = \(\frac{1}2\) MR² .

By the theorem of parallel axis,

=  \(\frac{1}2\) MR² + MR² = \(\frac{3}2\) MR² ....(7)

Radius of gyration : The radius of gyration of the ring for rotation about a tangent its olane is