A System Consists Of Three Identical Spheres Of Radius ‘r’ And Mass ‘m’ Placed With Their Centres Forming The Vertices Of An Equilateral Triangle Of Side ‘a’. Calculate The Moment Of Inertia Of The About An Axis Passing Through The Centre Of Gravity Of The System And Perpendicular To The Place Of The Centres.

Share:

Ans.

Fig. 1

Three spheres of radius ‘r‘ and mass ‘m‘ placed with their centres at the vertices of an equilateral triangle ABC. Here G is the centre of gravity.

From Fig. 1 we can write,

\( AO=\sqrt{a^2-{\left(\frac{a}{2}\right)}^2}\\=\frac{\sqrt{3}}{2}a \)

\( AG=\frac{2}{3}AD\\=\frac{2}{3}\times\frac{\sqrt{3}}{2}a\\=\frac{a}{\sqrt{3}} \)

We know that the moment of inertia of a sphere about its diameter \( \frac{2}{5}mr^2 \), [ To know the derivation of moment of inertia of a sphere about its diameter, (CLICK HERE) ].

By applying the theorem of parallel axes, the moment of inertia of each sphere about an axis passing through the centre of gravity G and perpendicular to the plane of the triangle is

\( \frac{2}{5}mr^2+m{\left(\frac{a}{\sqrt{3}}\right)}^2 \)

\( =\frac{2}{5}mr^2+\frac{1}{3}ma^2 \)

\( =m\left(\frac{2}{5}r^2+\frac{1}{3}a^2 \right) \)

So the moment of inertia of the system of three spheres about the axis passing through the centre of gravity G and perpendicular to the plane of the triangle is

\( 3m\left(\frac{2}{5}r^2+\frac{1}{3}a^2 \right) \)

Share:
Tags: