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## Main Question or Discussion Point

I want to ask why a rod's rotational inertia with rotational axis at its end is not

1/2 mR

1/2 mR

^{2}but 1/3 mR^{2}?- Thread starter destroyer130
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I want to ask why a rod's rotational inertia with rotational axis at its end is not

1/2 mR^{2} but 1/3 mR^{2}?

1/2 mR

- #2

cepheid

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- #3

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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

- #5

K^2

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destroyer, imagine I replace the rod by N+1 point masses located at distance r

[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.

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