Question

Two spheres each of mass M and radius R are connected with a massless rod of length 4 R. The moment of inertia of the system about an axis passing through the centre of one ofthe spheres and perpendicular to the rod will be ______.

Options

• \[\frac {21}{5}\]MR²

• \[\frac {84}{5}\]MR²

• \[\frac {42}{5}\]MR²

• \[\frac {5}{21}\]MR²

Answer 1

Two spheres each of mass M and radius R are connected with a massless rod of length 4 R. The moment of inertia of the system about an axis passing through the centre of one ofthe spheres and perpendicular to the rod will be \[\frac {84}{5}\]MR².

Explanation:

From Parallel axis theorem, I_o = I_c + Mh²

Let the moment of inertia of sphere 1 be 

\[\mathrm{I}_1=\frac{2}{5}\mathrm{M}(\mathrm{R})^2+\mathrm{M}(4\mathrm{R})^2\]

and,

Let the moment of inertia of sphere 2 be

\[\mathrm{I}_2=\frac{2}{5}\mathrm{M}(\mathrm{R})^2\]

Moment of inertia of the rod I₃ = 0

\[\therefore\] Moment of inertia of the syystem,

I = I₁ + I₂ + I₃

I = \[\frac{2}{5}\mathrm{M}\left(\mathrm{R}\right)^{2}+\mathrm{M}\left(4\mathrm{R}\right)^{2}+\frac{2}{5}\mathrm{M}\left(\mathrm{R}\right)^{2}\]

= \[{\frac{4}{5}}\mathrm{M}\left(\mathrm{R}\right)^{2}+16\mathrm{MR}^{2}\]

= \[\frac{4}{5}\mathrm{MR}^{2}+\frac{80}{5}\mathrm{MR}^{2}=\frac{84}{5}\mathrm{MR}^{2}\]