Question

All are cylindrical rods having radius of cross-section $R$ and mass of each rod $\dfrac{M}{4}$. Find the moment of inertia about $yy'$ axis:

• A. $I=\dfrac{1}{16}MR^2+\dfrac{1}{6}M\ell^2$
• B. $I=\dfrac{5}{16}MR^2+M\ell^2$
• C. $I=\dfrac{16}{5}MR^2+\dfrac{1}{6}M\ell^2$
• D. $I=\dfrac{3}{8}MR^2+\dfrac{M\ell^2}{6}$

Hint:

For composite bodies, always break the system into simple parts, find MOI of each about the required axis, and then add them carefully using the parallel axis theorem.

Answer 1

The Correct Option is D

To find the moment of inertia of the given system of cylindrical rods about the axis \(yy'\), we need to use the parallel axis theorem and the formula for the moment of inertia of a cylindrical rod.

Each rod has a mass of \(\frac{M}{4}\) and is aligned in a square configuration with side length \(\ell\).

We will calculate the moment of inertia for each rod separately and then sum them up:

1. The moment of inertia of a cylindrical rod of mass \(m\) and length \(L\) about an axis through its center is: \(I_{\text{center}} = \frac{1}{12} m L^2\).
2. Using the parallel axis theorem, the moment of inertia about a parallel axis at a distance \(d\) from the center is: \(I = I_{\text{center}} + m d^2\).
3. For rods along the top and bottom (parallel to the axis but not passing through it), the distance from the centroid to the \(yy'\) axis is \(\frac{\ell}{2}\). Hence, the moment of inertia of each of these rods is: \(I_{\text{top}} = I_{\text{bottom}} = \frac{1}{12} \cdot \frac{M}{4} \cdot \ell^2 + \frac{M}{4} \left(\frac{\ell}{2}\right)^2\).
4. Simplifying the above expression: \(I_{\text{top}} = I_{\text{bottom}} = \frac{1}{48} M \ell^2 + \frac{M \ell^2}{16} = \frac{1}{48} M \ell^2 + \frac{3}{48} M \ell^2 = \frac{1}{12} M \ell^2\).
5. For rods along the sides (perpendicular to \(yy'\)), the moment of inertia about its center remains: \(I_{\text{side}} = \frac{1}{12} \cdot \frac{M}{4} \cdot R^2 + \frac{M}{4} \left(\frac{\ell}{2}\right)^2 = \frac{1}{48} M R^2 + \frac{M \ell^2}{16}\).
6. Simplifying further: \(I_{\text{side}} = \frac{1}{48} M R^2 + \frac{3}{48} M \ell^2 = \frac{1}{48} M R^2 + \frac{1}{16} M \ell^2\).

Now summing up the moments of inertia of all four rods:

\(I = 2(I_{\text{side}}) + 2(I_{\text{top}}) = 2\left(\frac{1}{48} M R^2 + \frac{1}{16} M \ell^2\right) + 2\left(\frac{1}{12} M \ell^2\right)\)

\(= \frac{1}{24} M R^2 + \frac{1}{8} M \ell^2 + \frac{1}{6} M \ell^2\)

\(= \frac{1}{24} M R^2 + \frac{4}{24} M \ell^2 + \frac{4}{24} M \ell^2\)

\(= \frac{1}{24} M R^2 + \frac{8}{24} M \ell^2 = \frac{3}{8} M \ell^2\)

Thus, the correct answer is:

\(I = \frac{3}{8} MR^2 + \frac{M \ell^2}{6}\)

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