Question

The moment of inertia of a solid sphere an axis passing through the center of gravity is $\dfrac{2}{5}M{R^2}$ then its radius of gyration about a parallel axis at a distance $2R$ from first axis is
A. $5R$
B. $\sqrt {\dfrac{{22}}{5}} R$
C. $\dfrac{5}{2}R$
D. $\sqrt {\dfrac{{12}}{5}} R$

Answer 1

Hint- For solving this numerical firstly we will see the terms like radius of gyration and moment of inertia of the sphere. And then we will use the parallel axis theorem formula to get the answer. Parallel axis theorem states that the moment of inertia of a body about its parallel axis is equal to the sum of its moment inertia about its axis and the product of the square of distance between them and its mass.

Step-By-Step answer:
Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia the same as the body's actual distribution of mass, if the total mass of the body were concentrated.
The moment of inertia is the reluctance to change the state of motion (rotation) of an object rotating around a given axis
Therefore, the moment of inertia of a solid sphere about its diameter (axis) is expressed as follows;
$I = \left( {\dfrac{2}{5}} \right) \times m \times {r^2}$
Where;
$m$ = mass of the solid sphere
$r$ - Radius of the sphere.
By parallel axis theorem, the moment of inertia at 2R is
$I = \left( {\dfrac{2}{5}M{R^2}} \right) + M{\left( {2R} \right)^2}$
$I = \left( {\dfrac{{22}}{5}M{R^2}} \right)$
The radius of gyration is

$M{k^2} = \left( {\dfrac{2}{5}M{R^2}} \right)$
Therefore
$k = \sqrt {\dfrac{{22}}{5}} R$
Hence the correct option is B .

Note- The parallel axis theorem is used for finding the moment of inertia of the area of a rigid body whose axis is parallel to the axis of the known moment body and it is through the center of gravity of the object. Using the parallel axis theorem keeps in mind that the reference axis must pass through the center of mass of the object.