Question 506: Centre of Mass and Rotational Motion

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 of the spheres and perpendicular to the rod will be

Accepted Answer

From parallel axis theorem, $\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2$ Let the moment of inertia of sphere 1 be
$$ I_1=\frac{2}{5} M(R)^2+M(4 R)^2 $$
and,
Let the moment of inertia of sphere 2 be
$$ I_2=\frac{2}{5} M(R)^2 $$
Moment of inertia of the $\operatorname{rod} I_3=0$
$	herefore \quad$ Moment of inertia of the system,
$$ \begin{aligned} \mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \ \mathrm{I} & =\frac{2}{5} \mathrm{M}(\mathrm{R})^2+\mathrm{M}(4 \mathrm{R})^2+\frac{2}{5} \mathrm{M}(\mathrm{R})^2 \ & =\frac{4}{5} \mathrm{M}(\mathrm{R})^2+16 \mathrm{MR}^2 \ & =\frac{4}{5} \mathrm{MR}^2+\frac{80}{5} \mathrm{MR}^2=\frac{84}{5} \mathrm{MR}^2 \end{aligned} $$

Answer

$\frac{21}{5} \mathrm{MR}^2$

Answer

$\frac{84}{5} \mathrm{MR}^2$

Answer

$\frac{42}{5} \mathrm{MR}^2$

Answer

$\frac{5}{21} \mathrm{MR}^2$

Centre of Mass and Rotational Motion

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 of the spheres and perpendicular to the rod will be

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