Three solid spheres each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about YY' will be
Show Hint
Always separate composite bodies into individual elements relative to the target axis. Recognizing that the lower spheres are shifted by exactly $d = R$ allows you to immediately assign them a value of $\frac{7}{5}MR^2$ using the standard parallel axis shift, turning the calculation into simple fraction addition.
Step 1: Understanding the Question:
The system consists of three identical solid spheres, each with mass $M$ and radius $R$, grouped together. We need to compute the total moment of inertia of this configuration about the vertical axis of symmetry $YY'$.
Step 2: Key Formula or Approach:
1. The moment of inertia of a single uniform solid sphere about its central diameter axis is:
$$I_{\text{cm}} = \frac{2}{5}MR^2$$
2. For an axis shifted away from the center of mass, we use the Parallel Axis Theorem:
$$I = I_{\text{cm}} + Md^2$$
where $d$ is the perpendicular distance between the parallel axes.
Step 3: Detailed Explanation:
Let's analyze each sphere individually relative to the rotation axis $YY'$:
Top Sphere: The axis $YY'$ passes exactly through its center of mass. Therefore, its moment of inertia is simply:
$$I_1 = \frac{2}{5}MR^2$$
Bottom Two Spheres: Look at the geometry of the arrangement. The centers of the two lower spheres are each located at a horizontal distance of exactly one radius $R$ away from the vertical axis $YY'$.
Applying the parallel axis theorem for one of these lower spheres ($d = R$):
$$I_2 = I_{\text{cm}} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$
Since the bottom two spheres are placed symmetrically, the third sphere shares this exact value:
$$I_3 = \frac{7}{5}MR^2$$
Now, add the individual moments of inertia together to find the total system value $I_{\text{total}}$:
$$I_{\text{total}} = I_1 + I_2 + I_3$$
$$I_{\text{total}} = \frac{2}{5}MR^2 + \frac{7}{5}MR^2 + \frac{7}{5}MR^2$$
$$I_{\text{total}} = \frac{2 + 7 + 7}{5}MR^2 = \frac{16}{5}MR^2$$
Step 4: Final Answer:
The total moment of inertia of the system is $\frac{16}{5} MR^2$, which corresponds to option (A).
