Question

A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to the plane of the disc) is also equal to I, the value of r is

A. $\dfrac{{4R}}{{\sqrt {15} }}$
B. $\dfrac{{2R}}{{\sqrt {15} }}$
C. $\dfrac{R}{{2\sqrt {15} }}$
D. $\dfrac{R}{{\sqrt {15} }}$

Answer 1

Whenever a body is set into motion, it is observed that the body will not respond quickly to the rotation. This is due to the fact there is something called inertia which indicates the inability to adapt to the quick change in the state of the body. When this inertia is associated with rotation, we call it a moment of inertia.

Complete step-by-step answer:
The moment of inertia basically, gives us the distribution of mass of a body around its axis of rotation and it is the quantity that determines the torque required to produce an angular acceleration in the rotating body.
Consider the sphere of radius R melted and converted into a disc of thickness r and t.

The moment of inertia of sphere across its axis is given by,
${I_s} = \dfrac{2}{5}M{R^2}$

When this sphere is melted and converted to a disc of radius r and t, the moment of inertia of the disc across its axis XX is –
${I_{XX}} = \dfrac{{M{r^2}}}{2}$
The moment of inertia about the axis YY which is parallel to the axis XX and at a distance of r is calculated by the Parallel Axis theorem.
It states that - The moment of inertia of a body about an axis parallel to the body passing through its centre is equal to the sum of moment of inertia of body about the axis passing through the centre and product of mass of the body times the square of distance between the two axes.
${I_{YY}} = {I_{XX}} + M{r^2}$
As per the given question,
${I_s} = {I_{YY}}$
Substituting,

$${I_s} = {I_{YY}} \\\ \dfrac{2}{5}M{R^2} = {I_{XX}} + M{r^2} \\\ Solving, \\\ \dfrac{2}{5}{R^2} = {r^2} + \dfrac{{{r^2}}}{2} \\\ \to \dfrac{{3{r^2}}}{2} = \dfrac{2}{5}{R^2} \\\ \to {r^2} = \dfrac{{2 \times 2}}{{5 \times 3}}{R^2} \\\ \to r = \sqrt {\dfrac{{2 \times 2}}{{5 \times 3}}{R^2}} \\\ \to r = 2R\sqrt {\dfrac{1}{{15}}} \\\ \to r = \dfrac{{2R}}{{\sqrt {15} }} \\\$$

Hence, the correct option is Option B.

Note: There are 2 ways of compressive stress failure in columns – Crushing and Buckling. Crushing occurs for shorter columns, also called struts and Buckling occurs in longer columns. Even though you may think that their classification of long and short columns is subjective, it depends on a number known as Slenderness Ratio, which is dependent on moment of inertia.
Slenderness ratio = $\dfrac{l}{k}$
where $l$= effective length of the column and $k$= radius of gyration.
The radius of gyration, $k = \sqrt {\dfrac{I}{A}}$
where $I$is the least moment of inertia of the cross-section of the column and A is the area of cross-section.