Point masses $m, 4 m$, and $2 m$ are connected by a thin rod of length $L$ and negligible mass with the $4 m$ mass midway between the $m$ and $2 m$ masses, as shown. If this composite object is rotated around an axis perpendicular to the rod and through a point midway between the $4 \mathrm{~m}$ and $2 \mathrm{~m}$ masses, what is its moment of inertia?
$\frac{5}{2} m L^{2}$
$\frac{15}{16} m L^{2}$
$\frac{7}{16} m L^{2}$
$7 m L^{2}$

Step 1 ：The moment of inertia of a point mass is given by \(I = m r^2\), where \(m\) is the mass and \(r\) is the distance from the axis of rotation. In this case, we have three point masses at different distances from the axis of rotation. We can calculate the moment of inertia of each mass and then add them up to get the total moment of inertia.

Step 2 ：The \(4m\) mass is at a distance of \(\frac{L}{4}\) from the axis of rotation, the \(2m\) mass is at a distance of \(\frac{L}{2}\) from the axis of rotation, and the \(m\) mass is at a distance of \(\frac{3L}{4}\) from the axis of rotation.

Step 3 ：Calculate the moment of inertia for each mass: \(I_{4m} = 4m * (\frac{L}{4})^2 = \frac{mL^2}{4}\), \(I_{2m} = 2m * (\frac{L}{2})^2 = \frac{mL^2}{2}\), and \(I_{m} = m * (\frac{3L}{4})^2 = \frac{9mL^2}{16}\).

Step 4 ：Add up the moments of inertia to get the total moment of inertia: \(I_{total} = I_{4m} + I_{2m} + I_{m} = \frac{mL^2}{4} + \frac{mL^2}{2} + \frac{9mL^2}{16} = \frac{21mL^2}{16}\).

Step 5 ：Final Answer: The moment of inertia of the composite object is \(\boxed{\frac{21}{16} m L^{2}}\). However, this option is not listed in the given choices. There might be a mistake in the problem or in my calculations.