Explanation

(a) $m = 0.6\ kg$
[image:7_image176.emf]
Mass per unit length $= \frac{0.6}{100}\ kgcm^{- 1}$
Mass of part $AB,\ m_{1} = \frac{0.6}{100} \times 20 = \frac{0.6}{5}\ kg$
Mass of part $BC,\ m_{2} = \frac{0.6}{100} \times 80$
Moment of inertia $= \frac{0.6 \times 4}{5} = \frac{2.4}{5}\ kg$
$I = m_{1}\left( \frac{AB}{2} \right)^{2} + m_{2}\left( \frac{BC}{2} \right)^{2}$
$= \frac{0.6}{5} \times \left( \frac{20}{2} \times 10^{- 2} \right)^{2} + \frac{24}{5} \times \left( \frac{80}{2} \times 10^{- 2} \right)^{2}$
$= \frac{0.6}{5} \times 10^{- 2} + \frac{2.4}{5} \times \left( 4 \times 10^{- 1} \right)^{2}$
$= \frac{0.6}{5} \times 10^{- 2} + \frac{2.4}{5} \times 16 \times 10^{- 2}$
$= \left( \frac{0.6 + 38.4}{5} \right) \times 10^{- 2}$
$= 7.8 \times 10^{- 2}kg$-$m^{2} = 0.078\ kg$-$m^{2}$