Question

A rod of length $L$ is composed of a uniform length $\frac{1}{2}L$ of wood whose mass is $m_w$ and a uniform length $\frac{1}{2}L$ of brass whose mass is $m_b$.

(i) Find $I$ for the rod about an axis perpendicular to the rod and through its center. (ii) Repeat for a perpendicular axis through the wood end.

• A. (i) $I = \frac{1}{12} (m_w + m_b) L^2$; (ii) $I = \frac{1}{12} (m_w + 7m_b) L^2$
• B. (i) $I = \frac{1}{12} (m_w + m_b) L^2$; (ii) $I = \frac{1}{12} (m_w + m_b) L^2$
• C. (i) $I = \frac{1}{12} (m_w + m_b) L^2$; (ii) $I = \frac{1}{12} (m_w + 7m_b) L^2$ and $I = \frac{1}{12} (m_w + m_b) L^2$
• D. (i) $I = \frac{1}{12} (m_w + 7m_b) L^2$; (ii) $I = \frac{1}{12} (m_w + m_b) L^2$

Answer 1

Answer: (A), (B), (C)

(i) $I = \frac{1}{12} (m_w + m_b) L^2$; (ii) The possible answers for part (ii) are:

• $\frac{1}{12} (m_w + 7m_b) L^2$ (Axis at an outer wood end of the rod)
• $\frac{1}{12} (m_w + m_b) L^2$ (Axis at the wood end adjacent to brass, i.e., center)

Answer 2

For part (i), the center of the rod is at $L/2$. For both wood and brass segments, the axis is at one of their ends. The total moment of inertia is the sum of individual moments of inertia about this common axis. For part (ii), "wood end" can be interpreted as an outer extremity of the rod made of wood, or the end of the wood segment adjacent to brass (the center). The former leads to $I = \frac{1}{12} (m_w + 7m_b) L^2$ using the parallel axis theorem for the brass segment. The latter leads to $I = \frac{1}{12} (m_w + m_b) L^2$, same as part (i).