Question

The moment of inertia of a hollow sphere of mass m and radius r about an axis passing through its centre of mass is , then its radius of gyration about an axis which is parallel to first axis and at a distance 3r will be

Answer 1

The moment of inertia 'I' of hollow sphere is $\frac{2}{3} mr^{2}$ .

The radius of gyration about an axis parallel and at a distance of '3r' is $\sqrt{\frac{29k^{2} }{3} }$.

Explanation:

Given:

     Mass of hollow sphere is 'm'.

     Radius of hollow sphere is 'r'.

Solution,

      we know that moment of inerta 'I' of hollow sphere is $\frac{2}{3} MR^{2}$.

     So, by putting given values.

   

                                         ⇒ I = $\frac{2}{3} mr^{2}$

    Moment of inertia 'I''of axis which is parallel and at a distance of '3r' from centre axis is,

                                         ⇒ I' = $\frac{2}{3} mr^{2}$ + $m\times(3r)^{2}$

                                                = $\frac{2}{3} mr^{2}$ + $9mr^{2}$

                                         ⇒ I' = $\frac{29}{3} mr^{2}$

Radius of gyration 'k' is,

                                          ⇒ $mk^{2}$ = $\frac{29}{3} mr^{2}$

                                          ⇒ k = $\sqrt{\frac{29}{3}r^{2} }$

                                           

Answer 2

Answer:

root 29k2/3 is correct