Question 298: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

Position of the axis of rotation for Solid Sphere :
About its diametric
 axis which passes
through its centre
of mass
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
$$
2 I_1=2 	imes 22 / 5 M R^2
$$ = $${{44} \over 5}$$MR
2
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
$	herefore I_2=\frac{1}{12} M(2 R)^2=\frac{1}{3} M R^2$
So, total moment of inertia of the system is
$$
\begin{aligned}
I & =2 I_1+I_2=2 	imes \frac{22}{5} M R^2+\frac{1}{3} M R^2 \\
\Rightarrow I & =\left(\frac{44}{5}+\frac{1}{3}ight) M R^2=\frac{137}{15} M R^2
\end{aligned}
$$

Answer

$${{17} \over {15}}$$ MR
2

Answer

$${{137} \over {15}}$$ MR
2

Answer

$${{209} \over {15}}$$ MR
2

Answer

$${{152} \over {15}}$$ MR
2

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: