Question

The radius of gyration of a body about an axis, at a distance of 0.4m from its center of mass is 0.5m. Find its radius of gyration about a parallel axis passing through its center of mass.

Answer 1

From the parallel axis theorem for moment of inertia (MI), we know that the MI of a body about an axis which is parallel to another axis passing through the center of mass (CM) is same as the sum of MI about the axis passing through CM and mass of the body times square of the distance between two axes.

Formulae used:
Moment of inertia of any body is given by:
$I = M{R^2}$..........................(1)
Where,
I is the moment of inertia(MI) of a body about some axis,
R is the radius of gyration about that axis.

Moment of inertia of a body about an axis parallel to the axis passing through center of mass(CM) is given by:
${I_P} = {I_{CM}} + M{h^2}$......................(2)
Where,
${I_P}$ denotes the MI of the body about a parallel axis of an axis passing through CM,
${I_{CM}}$ denotes the MI of the body about an axis passing through CM,
h is the distance between two axes.

Step by step solution:
Given:
Radius of gyration about the parallel axis is ${R_P} = 0.5m$.
Distance between the two axes is $h = 0.4m$.

To find: The radius of gyration ${R_{CM}}$about the axis passing through its CM.

Step 1
Assume the radius of gyration about the parallel axis and the axis passing through CM to be ${R_P}$and ${R_{CM}}$. Now, using eq.(1) substitute these radii in eq.(2) to get an expression for ${R_{CM}}$as:
${I_P} = {I_{CM}} + M{h^2} \\\ \Rightarrow MR_P^2 = MR_{CM}^2 + M{h^2} \\\ \Rightarrow R_{CM}^2 = R_P^2 - {h^2} \\\ \therefore {R_{CM}} = \sqrt {R_P^2 - {h^2}} ……...(3) \\\$

Step 2
Substitute the values of ${R_P}$ and h in eq.(3) to get the value of ${R_{CM}}$ as:
${R_{CM}} = \sqrt {{{(0.5m)}^2} - {{(0.4m)}^2}} \\\ \therefore {R_{CM}} = \sqrt {0.25 - 0.16} \;m = 0.3m \\\$

Final answer:
The radius of gyration about a parallel axis passing through CM is 0.3 m.

Note: Radius of gyration is the radial distance of an imaginary point from the axis such that if all the mass of the whole body is assumed to be at that point then the moment of inertia (MI) of that point will be the same as that of the actual object. This position of the point hence the radius of gyration doesn’t depend on the total mass of the object but depends on how the mass is distributed throughout the object.