Question

The moment of inertia of a thin rod of length 2L and mass M about an axis normal to its length and acting through the rod at a distance 3L/4 from one of its ends is given by which one of the following and why?

a)19ML^2/48
b)15ML^2/16
c)27ML^2/8
d)19ML^2/64
e)8ML^2/27

Thank you!

Answer 1

this is based on the definition of inertia
\[dI = r^2 dm\]
the small element of the rod, length dr, as shown, has mass dm. the density per unit length of the rod is M/2L so for that small element we can say dm = M/2L dr, and as dI = r^2 dm , thus dI = M/2L r^2 dr

which delivers this integral:

\[I = \frac{M}{2L} \int\limits_{-\frac{3L}{4}}^{\frac{5L}{4}} r^2 dr = \frac{M}{6L} \left| r^3 \right|^{\frac{5L}{4}}_{-\frac{3L}{4}}\]