Show that the parallel axis theorem applied to a rod whose axis of rotation is perpendicular to is length gives the same answer as direct integration.
Solution
In the first case the axis of rotation is through the centre of mass. The moment of inertia is calculated as follows were
$\lambda$ is the linear mass density in kg/m:
$$ I_{\rm cm}=\int^{L/2}_{-L/2} \lambda x^2dx = \frac{1}{3} \frac{m}{L} \left[ x^3 \right]^{L/2}_{-L/2}= \frac{1}{3} \frac{m}{L} \left[ \left(\frac{L}{2}\right)^3 - \left(-\frac{L}{2}\right)^3\right] $$
$$ I_{\rm cm}= \frac{1}{12}mL^2$$
Now integrating the case where the axis is on the end:
$$ I_{\rm end}=\int^{L}_{0} \lambda x^2dx = \frac{1}{3} \frac{m}{L} \left[ x^3 \right]^{L}_{0}= \frac{1}{3} \frac{m}{L} \left[ L^3 - 0^3\right] $$
$$ I_{\rm end}= \frac{1}{3}mL^2$$
Moment of Inertia about end using Parallel Axis Theorem
$$ I_{\rm end}=I_{\rm cm} +m\left( \frac{L}{2}\right)^2$$
$$ = \frac{1}{12}mL^2 + \frac{1}{4}mL^2$$
$$ = \frac{1}{12}mL^2 + \frac{3}{12}mL^2$$
$$ I_{\rm end}= \frac{1}{3}mL^2$$
This is the same result we got by direct integration.
N. Alberding, 2013