Question

The radius of gyration of a disc about its axis passing through its centre and perpendicular to its plane is
$$\begin{array}{l}\mathrm{A}.{\displaystyle \frac{R}{\sqrt{2}}}\\ \mathrm{B}.{\displaystyle \frac{R}{2}}\\ \mathrm{C}.\sqrt{2}R\\ \mathrm{D}.2R\end{array}$$

Answer 1

Hint: The radius of gyration of a body is equal to the square root of the ratio of moment of inertia and the total mass. We need the value of the moment of inertia of a circular disc from which the radius of gyration can be calculated easily.
Formula used:
The moment of inertia of a body is given as
$I=M{R}^2...(\mathrm{i})$
Here I represent the moment of inertia, M is the total mass of the body and R is the distance of this mass from the axis of rotation.
The moment of inertia of a circular disc about its axis passing through its centre and perpendicular to its plane is given as
$I={\displaystyle \frac{M{R}^2}{2}}...(\mathrm{i}\mathrm{i})$
The radius of gyration is given as
$K=\sqrt{{\displaystyle \frac{I}{M}}}...(\mathrm{i}\mathrm{i}\mathrm{i})$
Here K represents the radius of gyration.

Complete step by step solution:
The moment of inertia of a body can be expressed in terms of the radius of gyration. The radius of gyration of a body is defined as the distance at which we will obtain the same moment of inertia as that obtained from the mass distribution of the body as if all the mass of the body is concentrated. It is denoted by K and moment of inertia I is given in terms of K in the following way.
$\begin{array}{l}I=M{K}^2\\ \Rightarrow K=\sqrt{{\displaystyle \frac{I}{M}}}\end{array}$
We are given a circular disc. Let the radius of this circular disc be R while the mass of the disc be M. We know that the moment of inertia of a circular disc is given as
$I={\displaystyle \frac{M{R}^2}{2}}$
Now we can calculate the radius of gyration of the circular disc easily by using equation (iii) in the following way.
$K=\sqrt{{\displaystyle \frac{I}{M}}}=\sqrt{{\displaystyle \frac{{\displaystyle \frac{M{R}^2}{2}}}{M}}}={\displaystyle \frac{R}{\sqrt{2}}}$
This is the radius of gyration of the disc and hence the correct answer is option A.

Note: It should be noted that the value of moment of inertia is different for bodies of different shapes. The formulas for moment of inertia for other shapes like a rod and cylinder are given as follows:
$$\begin{array}{l}{I}_{rod}={\displaystyle \frac{1}{12}}M{r}^2\\ I_{cylinder}={\displaystyle \frac{1}{2}}M{r}^2\end{array}$$