Question

Two uniform rods A and B, each of length l and mass $m_A$ and $m_B$, respectively are rigidly joined end to end. The combination is pivoted at the lighter end P as shown such that it can freely rotate about point P in a vertical plane. A small object of mass m moving horizontally hits the lower end of combination and sticks to it, the moment of inertia of the system about P is

• A. $m{l}^2+\frac{m_A}{3}{l}^2+m_B\frac{l^2}{12}$
• B. $2m{l}^2+\frac{m_A}{3}{l}^2+m_B{\left(\frac{l}{2}\right)}^2$
• C. $4m{l}^2+\frac{m_A}{3}{l}^2+m_B[\frac{l^2}{12}+{(\frac{l}{2}+l)}^2]$
• D. $4m{l}^2+\frac{m_A}{3}{l}^2+m_B[\frac{l^2}{2}+{(\frac{l}{12}+l)}^2]$

Answer 1

Answer: (C)

$4m{l}^2+\frac{m_A}{3}{l}^2+m_B[\frac{l^2}{12}+{(\frac{l}{2}+l)}^2]$

The moment of inertia of small object of mass m about rotational axis would be $m(2l{)}^2=4m{l}^2$
Moment of inertia of rod B about rotational axis using parallel axis theorem would be $m_B\left[\frac{l^2}{12}+(\frac{l}{2}+l{)}^2\right]$
Moment of inertia of rod A about rotational axis would be $\frac{1}{3}{m}_A{l}^2$. Thus correct is option is C.