The impact wrench consists of a slender 1-kg rod AB which is 580 mm long, and cylindrical...

Calculate the mass moment of inertia of the section about the axle.

\(=\left[\frac{m\left(L_{A B}\right)^{2}}{12}\right]+2\left[\frac{m R^{2}}{2}+m d^{2}\right]\)

Here, length of the rod \(A B\) is \(L_{A B}\), mass of the rod and the cylinder is \(m\), radius of the cylinder is \(R\) and distance of the center of mass of the cylinder from the centroidal axis is \(d\).

Substitute \(1 \mathrm{~kg}\) for \(m, 600-(2 \times 10) \mathrm{mm}\) for \(L_{A B}, 10 \mathrm{~mm}\) for \(R\) and \(300 \mathrm{~mm}\) for \(d\).

$$ \begin{aligned} I_{a+l_{k}} &=\left[\frac{(1)(600-(2 \times 10))^{2}}{12}\right]+2\left[\frac{(1)\left(\frac{20}{2}\right)^{2}}{2}+(1)(300)^{2}\right] \\ &=28033.33+2(90050) \\ &=208133.33 \mathrm{~kg} \cdot \mathrm{mm}^{2} \times \frac{1 \mathrm{~m}^{2}}{1000000 \mathrm{~mm}^{4}} \\ &=0.208133 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{aligned} $$