Answer:
part a.
[tex]I = \frac{1}{3}ML^2[/tex]
part b.
I = [tex]\frac{2}{3}Ma^2[/tex]
Explanation:
Part a.
For the center of mass the moment of inertia is:
[tex]I_{cm}=\frac{1}{12}ML^2[/tex]
Using the parallel axis theorem, we get:
[tex]I = I_{cm}+Md^2[/tex]
I = [tex]\frac{1}{12}ML^2+Md^2[/tex]
Additionally,
d = L/2
so:
[tex]I =\frac{1}{12}ML^2+M\frac{L^2}{4}[/tex]
[tex]I = \frac{1}{3}ML^2[/tex]
part b.
For the center of mass the moment of inertia is:
[tex]I_{cm}=\frac{1}{6}Ma^2[/tex]
Using the parallel axis theorem, we get:
[tex]I = I_{cm}+Md^2[/tex]
I = [tex]\frac{1}{6}Ma^2+Md^2[/tex]
Adittionally, in this case:
d = [tex]\sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2}=\sqrt{\frac{a^2}{2}}=\frac{a}{\sqrt{2}}[/tex]
so:
[tex]I =\frac{1}{6}Ma^2+M(\frac{a}{\sqrt{2}})^2[/tex]
[tex]I =\frac{1}{6}Ma^2+M\frac{a^2}{2}[/tex]
I = [tex]\frac{2}{3}Ma^2[/tex]