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
Show Hint
Don't forget the Parallel Axis Theorem for the second sphere! It's not just a point mass; it's a rotating sphere.
Step 1: Understanding the Concept:
The total moment of inertia of a system is the scalar sum of the moments of inertia of its individual components about the specified axis.
For objects not rotating about their own center of mass, we must use the Parallel Axis Theorem. Step 2: Key Formula or Approach:
1. Moment of inertia of a solid sphere about its central diameter: $I_{cm} = \frac{2}{5}MR^2$. (Assuming standard solid spheres as is convention when not specified otherwise).
2. Parallel Axis Theorem: $I = I_{cm} + Md^2$, where $d$ is the perpendicular distance from the center of mass to the axis of rotation.
3. Total system inertia: $I_{\text{total}} = I_1 + I_2$. Step 3: Detailed Explanation:
The axis of rotation passes through the center of the first sphere (Sphere 1) and is perpendicular to the connecting rod. Sphere 1:
Since the axis passes right through its center of mass, its moment of inertia is simply its standard formula:
\[ I_1 = \frac{2}{5}MR^2 \] Sphere 2:
The axis is parallel to its central diameter but shifted by a distance $d$.
Looking at the diagram, the distance $d$ between the center of Sphere 1 and the center of Sphere 2 is explicitly given by the dimension line as $4R$.
Using the Parallel Axis Theorem for Sphere 2:
\[ I_2 = I_{cm,2} + M d^2 \]
\[ I_2 = \frac{2}{5}MR^2 + M(4R)^2 \]
\[ I_2 = \frac{2}{5}MR^2 + 16MR^2 \]
Find a common denominator to add:
\[ I_2 = \frac{2}{5}MR^2 + \frac{80}{5}MR^2 = \frac{82}{5}MR^2 \] Total System:
Add the inertias of both spheres (the rod is massless, so $I_{\text{rod}} = 0$):
\[ I_{\text{total}} = I_1 + I_2 \]
\[ I_{\text{total}} = \frac{2}{5}MR^2 + \frac{82}{5}MR^2 \]
\[ I_{\text{total}} = \frac{84}{5}MR^2 \] Step 4: Final Answer:
The moment of inertia of the system is $\frac{84}{5}\text{MR}^2$.
