Rod rotational inertia - axis at end of rod

Try using the parallel axis theorem to compute the moment of inertia for the second case, given the first case.

 

Where did you came up with 1/2 mR^2 in the first place, it is for the solid cylinder.

The rods moment of inertia about its center is 1/12 mL^2 (where L stands for length of the rod), I hope you agree with me here.
If we would like to calculate the moment of inertia of this same rod about an axis that goes through the one end of the rod we apply the parallel axis theorem:

where d stands for distance from the center of the rod to the end of the rod d = L/2.

1/12 mL^2 + md^2 = 1/12 mL^2 + m(L/2)^2 = 1/3 mL^2

 

amiras, now you need to explain the 1/12th for moment around center.

destroyer, imagine I replace the rod by N+1 point masses located at distance r_n=n(R/N) from the end, where n runs from 0 to N. The mass of each is m=M/(N+1), of course, to give you total mass of M. Moment of inertia for a point mass is mr_n². So we need to add these up.

[tex]I = \sum_{n=0}^{N} m r_n^2 = \frac{m R^2}{N^2} \sum_{n=0}^{N} n^2[/tex]

Sum of squares of consecutive integers is known as square pyramidal number.

[tex]\sum_{n=0}^{N} n^2 = \frac{2N^3 + 3N^2 + N}{6}[/tex]

Putting it all together.

[tex]I = MR^2 \frac{2N^3 + 3N^2 + N}{6 N^2 (N+1)}[/tex]

That looks ugly, but if N is large enough, only the N³ terms are important in that fraction. Since we want the rod to be broken into infinitely many point masses, the square and linear terms can be dropped. (In other words, take limit as N goes to infinity.)

[tex]I = MR^2 \frac{2N^3}{6N^3} = \frac{1}{3}MR^2[/tex]

This isn't the cleanest way to derive moment of inertia, and it gets even worse for more complex shapes, but it shows where the 1/3 comes from without any advanced mathematics.