Question

Moment of inertia of a rod about an axis passing through a point at distance \(\ell/4\) from the center and perpendicular to the rod is: (The uniform rod has mass \(m\) and length \(\ell\)).

• A. \(\dfrac{7m\ell^2}{48}\)
• B. \(\dfrac{3m\ell^2}{48}\)
• C. \(\dfrac{4m\ell^2}{48}\)
• D. \(\dfrac{m\ell^2}{48}\)

Answer 1

The Correct Option is A

Concept: Use the parallel axis theorem. \[ I = I_{cm} + md^2 \] where \(d\) is distance from center of mass. \includegraphics[width=0.5\linewidth]{25p ans.png} Step 1: Moment of inertia about center For a rod about axis perpendicular to length: \[ I_{cm} = \frac{m\ell^2}{12} \] Step 2: Distance from center \[ d = \frac{\ell}{4} \] Step 3: Apply parallel axis theorem \[ I = \frac{m\ell^2}{12} + m\left(\frac{\ell}{4}\right)^2 \] \[ I = \frac{m\ell^2}{12} + \frac{m\ell^2}{16} \] Step 4: Simplify LCM of \(12\) and \(16\) is \(48\) \[ I = \frac{4m\ell^2}{48} + \frac{3m\ell^2}{48} \] \[ I = \frac{7m\ell^2}{48} \]