Describe how you could use part (b) of Table 9.2 to derive the result in part (d).

Solution 13DQ Step 1: Consider a slender rod of length L and mass M. The moment of inertia of the slender rod about an axis passing through one end, 2 I = ML Consider two rods having lengths a and b are joined. 2 2 Moment of inertia of this system through its center of mass, I = 1/12 M (a + b cm Where, M - Mass of the system 2 According to parallel axes theorem equation, I = I + Md cm Where, d - distance of the center of mass from the origin. d = (a/2) + (b/2) 2 d = (a + b ) / 4 2 2 2 2 Then, I = M/12 (a + b ) + M/4 (a + b ) I = (a + b ) {(M/12) + (M/4)} = (a + b ) M/3 2 2 2 I = M (a + b ) If the axis moved to b , then, I = Ma , (b =0) 2