Chapter 10 - Section 10.3 - Rotational Inertia and the Analog of Newton’s Law - Example - Page 181: 10.4

$.29 \ kgm^2$

Recall, the moment of inertia is equal to the sum of the moments of inertia of every aspect of the object, including the masses. Thus, we find: $I = m(\frac{1}{4}L)^2 + m(\frac{3}{4}L)^2 = \frac{5mL^2}{8}= \frac{5(.64)(.85)^2}{8}= .29 \ kgm^2$