Question

The moment of inertia of a sphere of mass M and radius R about an axis passing through the centre is \[\dfrac{2}{5}M{{R}^{2}}\]. The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is
A. \[\dfrac{7}{5}R\]
B. \[\dfrac{3}{5}R\]
C. \[\left( \sqrt{\dfrac{7}{5}} \right)R\]
D. \[\left( \sqrt{\dfrac{3}{5}} \right)R\]

Answer 1

Hint: First we will be using the theorem of parallel axis to find the moment of inertia about the axis parallel to an axis passing through the centre of the sphere and tangent to the sphere. Then, according to the definition of radius of gyration we will find the radius of gyration for the sphere about that axis.
Formula used:
Theorem of parallel axis.
${{I}_{R}}={{I}_{COM}}+M{{R}^{2}}$

Complete answer:
First, we will use the theorem of parallel axis to find the moment of inertia along the axis parallel to the axis passing through the centre and tangent to the sphere.

The theorem of parallel axis states that the moment of inertia of a body about any axis parallel to the axis passing through the centre of mass of the body is equal to the sum of the moment of inertia of body about the axis passing through the centre of mass and the product of the mass of the body and square of the distance between both the axes. So, here we have the moment of inertia about the axis passing through the centre and the mass of the body is given as M. The distance between both the axes will be equal to the radius of the sphere, which is R.

\[I=\dfrac{2}{5}M{{R}^{2}}+M{{R}^{2}}=\dfrac{7}{5}M{{R}^{2}}\]

The radius of gyration is defined as the radial distance from the axis to a point at which the mass of the body must be concentrated for it to have the same moment of inertia.
Let us take G to be the radius of gyration the moment of inertia of a point mass will then be $M{{G}^{2}}$. Equating both we get G as
 $\begin{align}
  & M{{G}^{2}}=\dfrac{7}{5}M{{R}^{2}} \\
 & {{G}^{2}}=\dfrac{7}{5}{{R}^{2}} \\
 & G=\sqrt{\dfrac{7}{5}}R \\
\end{align}$

So, the correct answer is “Option C”.

Note:
We can also take a moment of inertia of many small elements and then find the complete moment of inertia about the given axis, but that process will be very long, tedious, and time-consuming. So, we will use the theorem of parallel axis directly.