Question

the radius of gyration of a hollow sphere of mass m and radius r about an axis which is at a distance 2R from surface and parallel to its diameter will be...​

Answer 1

Given:

Hollow sphere of mass m and radius r .

To find:

Radius of gyration at a distance 2r from the surface and parallel to the diameter.

Calculation:

First of all, the moment of inertia along the geometric axis of the sphere is :

$\boxed{MI = \dfrac{2}{3} m {r}^{2} }$

Now, the new axis is at a distance of 2r + r = 3r from the centre of the sphere.

Since the new axis is parallel to the initial axis, we can apply PARALLEL AXES THEOREM :

New moment of Inertia :

$\therefore \: MI_{new} = MI + m {(3r)}^{2}$

$\implies \: MI_{new} = \dfrac{2}{3}m {r}^{2} + 9m {r}^{2}$

$\implies \: MI_{new} =( \dfrac{2 + 27}{3})m {r}^{2}$

$\implies \: MI_{new} = \dfrac{29}{3}m {r}^{2}$

Now, radius of gyration is defined as the radius of a ring which will have the same moment of inertia, let radius of gyration be x :

$\therefore \: MI_{new} = m {x}^{2}$

$\implies \: m {x}^{2} = \dfrac{29}{3}m {r}^{2}$

$\implies \: {x}^{2} = \dfrac{29}{3} {r}^{2}$

$\implies \: x =r \times \sqrt{ \dfrac{29}{3}}$

So radius of gyration is:

$\boxed{ \bf\: x =r \times \sqrt{ \dfrac{29}{3}} }$