Question

Three solid spheres of mass
Mand radius R are shown in
the figure. The moment
of inertia of the system about
XX' axis will be :-
(1) 7/2 MR²
(2) 14/5 MR²
(3)16/5MR²
(4)21/5 MR²
​

Answer 1

Given

Three solid spheres of radius R are shown as follows

To find

Moment of inertia of the system through XX'

Explanation

Let us name the spheres as sphere 1,sphere 2,sphere 3

From the figure

Sphere 1:

XX' is passing through the diameter

$\sf moment\: of\: inertia\:, I_1 = \dfrac{2}{5}MR^2$

Now

Sphere 2 and Sphere 3:

XX' is passing about the tangent

$\sf I_2=\dfrac{7}{5}MR^2 \\ \sf I_3 =\dfrac{7}{5}MR^2$

$\sf moment \:of \:inertia \:of\: system ,I\\ \sf= I_1+I_2+I_3$

$I=\dfrac{2}{5}MR^2+\dfrac{7}{5}MR^2+\dfrac{7}{5}MR^2$

$\sf Taking\: MR^2 \: common$

$=>I =(\dfrac{2}{5}+\dfrac{7}{5}+\dfrac{7}{5})MR^2$

$=>I=\dfrac{2+7+7}{5}MR^2$

$\boxed {I =\dfrac{16}{5}MR^2}$

Option 3

Additional information:

→If u don't know the formula of moment of inertia of solid sphere about tangent u can refer here

→We know moment of inertia of solid sphere about diameter

$I =\dfrac{2}{5}MR^2$

→As axis about tangent is parallel to the axis about diameter,

axis about tangent is parallel to the axis about diameter,

Using parallel axis theorem

$I =I_c+Md^2$

here d=R

$I=\dfrac{2}{5}MR^2+MR^2$

$I=\dfrac{2+5}{5}MR^2$

$I=\dfrac{7}{5}MR^2$

Answer 2

Solution:-Moment of inertia of sphere at centre=2mr^2/5

from parallel axis theorem,we can know that

M.O.I at tangent of hemisphere=2mr^2/5 +mr^2=7mr^2/5

two sphere have the same tangent,

total MOI=sum of (MOI at centre+2×MOI at tangent)

Required moment of inertia=(2/5 +7/5 +7/5)mR^2=16mr^2/5