Answer:
Part a)
[tex]I_{end} = \frac{mL^2}{3}[/tex]
Part b)
[tex]I_{edge} = \frac{2ma^2}{3}[/tex]
Explanation:
As we know that by parallel axis theorem we will have
[tex]I_p = I_{cm} + Md^2[/tex]
Part a)
here we know that for a stick the moment of inertia for an axis passing through its COM is given as
[tex]I = \frac{mL^2}{12}[/tex]
now if we need to find the inertia from its end then we will have
[tex]I_{end} = I_{cm} + Md^2[/tex]
[tex]I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2[/tex]
[tex]I_{end} = \frac{mL^2}{3}[/tex]
Part b)
here we know that for a cube the moment of inertia for an axis passing through its COM is given as
[tex]I = \frac{ma^2}{6}[/tex]
now if we need to find the inertia about an axis passing through its edge
[tex]I_{edge} = I_{cm} + Md^2[/tex]
[tex]I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2[/tex]
[tex]I_{edge} = \frac{2ma^2}{3}[/tex]