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Step-by-Step Solution
Step 1: Identify the Relevant Physical Situation
The rod of length $l$ is massless, with two masses attached at its ends: one mass $m$ and the other $\frac{m}{2}$. This rod is suspended about its center of mass by a thin wire with torsional constant $k$. When the rod is rotated by an angle $\theta_0$ and released, it executes rotational motion about the center of mass.
Step 2: Determine the Position of the Center of Mass
Let us choose coordinates so that the heavier mass $m$ is at the left end (position 0) and the lighter mass $\frac{m}{2}$ is at the right end (position $l$). The rod itself is massless. The center of mass $X_{CM}$ of the system is given by:
$X_{CM} \;=\; \dfrac{m \times 0 + \left(\tfrac{m}{2}\right)\times l}{m + \tfrac{m}{2}}
\;=\; \dfrac{\tfrac{m}{2}\,l}{\tfrac{3m}{2}}
\;=\; \dfrac{l}{3}.$
This means the pivot (center of mass) is at a distance $\tfrac{l}{3}$ from the heavier mass and $\tfrac{2l}{3}$ from the lighter mass.
Step 3: Calculate the Moment of Inertia about the Center of Mass
The rod is massless, so only the two masses contribute to the moment of inertia. Denoting the pivot at the center of mass, the distances of each mass from the pivot are:
Heavier mass $m$: distance $\tfrac{l}{3}$.
Lighter mass $\tfrac{m}{2}$: distance $\tfrac{2l}{3}$.
The moment of inertia $I$ about the center of mass is then:
$I \;=\; m\left(\tfrac{l}{3}\right)^2 \;+\; \tfrac{m}{2}\,\bigl(\tfrac{2l}{3}\bigr)^2
\;=\; m\,\dfrac{l^2}{9} \;+\; \tfrac{m}{2}\,\dfrac{4l^2}{9}
\;=\; \dfrac{m\,l^2}{9} \;+\; \dfrac{2m\,l^2}{9}
\;=\; \dfrac{3m\,l^2}{9}
\;=\; \dfrac{m\,l^2}{3}.
Step 4: Use Energy Conservation to Find the Angular Velocity at the Mean Position
When the rod is rotated by $\theta_0$ and released from rest, all energy is initially stored as torsional potential energy in the wire. At the mean position (i.e., $\theta = 0$), the potential energy is fully converted into rotational kinetic energy. The torsional potential energy is:
$U_{\text{torsion}} \;=\; \tfrac{1}{2} \, k \,\theta_0^{2}.
The rotational kinetic energy at the mean position is:
$K_{\text{rot}} \;=\; \tfrac{1}{2}\,I\,\omega^{2},
where $\omega$ is the angular velocity at the mean position. By energy conservation:
$\tfrac{1}{2}\,k\,\theta_0^{2} \;=\; \tfrac{1}{2}\,\bigl(\tfrac{m\,l^2}{3}\bigr)\,\omega^{2}.
Canceling the factor $\tfrac{1}{2}$ on both sides and solving for $\omega^2$ gives:
$k\,\theta_0^{2} \;=\; \tfrac{m\,l^2}{3}\,\omega^{2}
\;\;\Longrightarrow\;\;
\omega^{2} \;=\; \dfrac{3\,k}{m\,l^{2}}\,\theta_0^{2}.
Step 5: Find the Tension in the Rod at the Mean Position
As the rod passes through the mean position, each mass moves in a circular path around the pivot. The tension in the rod on each end provides the necessary centripetal force. Consider first the heavier mass $m$ at distance $\tfrac{l}{3}$ from the pivot. The centripetal force (which equals the tension in that segment of the rod) is:
$T \;=\; m \,\bigl(\tfrac{l}{3}\bigr)\,\omega^{2}.
Substitute $\omega^{2} = \dfrac{3\,k}{m\,l^{2}}\,\theta_0^{2}$:
$T
\;=\; m \,\bigl(\tfrac{l}{3}\bigr)\,\dfrac{3\,k}{m\,l^{2}}\;\theta_0^{2}
\;=\; \dfrac{k\,\theta_0^{2}}{l}.
Thus, when the system passes through the mean position, the tension in the rod (on the heavier mass side) is:
$T \;=\; \dfrac{k\,\theta_0^{2}}{l}.
This matches the given correct answer.
Final Answer
$\displaystyle \dfrac{k\,\theta_0^{2}}{l}$
