JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 1)
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :
_10th_January_Evening_Slot_en_1_1.png)
$${{17} \over {15}}$$ MR2
$${{137} \over {15}}$$ MR2
$${{209} \over {15}}$$ MR2
$${{152} \over {15}}$$ MR2
Explanation
Position of the axis of rotation for Solid Sphere :
About its diametric
axis which passes
through its centre
of mass
_10th_January_Evening_Slot_en_1_2.png)
Moment of Inertia (I) = $${2 \over 5}M{R^2}$$_10th_January_Evening_Slot_en_1_3.png)
_10th_January_Evening_Slot_en_1_4.png)
Moment of Inertia (I) = $\frac{\mathrm{ML}^2}{12}$_10th_January_Evening_Slot_en_1_5.png)
Moment of Inertia (I) = $${2 \over 5}M{R^2}$$
Using parallel axis theorem, moment of inertia about the given axis in the figure below will be
$\begin{aligned} & I_1=\frac{2}{5} M R^2+M(2 R)^2 \\\\ & I_1=\frac{22}{5} M R^2\end{aligned}$
Considering both spheres at equal distance from the axis, moment of inertia due to both spheres about this axis will be
$$
2 I_1=2 \times 22 / 5 M R^2
$$ = $${{44} \over 5}$$MR2
Moment of inertia of rod
Position of the axis of rotation : About an axis passing through centre of mass and perpendicular to its length
Moment of Inertia (I) = $\frac{\mathrm{ML}^2}{12}$
Here, given that L = 2R
$\therefore I_2=\frac{1}{12} M(2 R)^2=\frac{1}{3} M R^2$
So, total moment of inertia of the system is
$$ \begin{aligned} I & =2 I_1+I_2=2 \times \frac{22}{5} M R^2+\frac{1}{3} M R^2 \\\\ \Rightarrow I & =\left(\frac{44}{5}+\frac{1}{3}\right) M R^2=\frac{137}{15} M R^2 \end{aligned} $$
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