# Inertia of a uniform rod

i've been wondering why the I for a uniform rod held at one end is $$\frac{1}{3}ML^2$$

i know that it has something to do with the parallel axis theorm, but after finiding the first I, being: $$M(\frac{L}{2})^2$$
I am confused as how to find the I of the meter stick rotating about its parallel axis in the center. Since the center of mass is at the center, and it is being pivioted about the center, it seems there would be no I at all.

can anyone help me out?

Well one thing you have incorrect is that the Moment of Inertia of a Long Rod pivoted at the center is (1/12) ML^2. The reason why its (1/3) ML^2 is because of the parallel axis theorem. You have:

I = (1/12) ML^2 + MD^2.
D is this case is (L/2). Therefore, if you add (1/12) + (1/4), you should get (1/3).

Doc Al
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cair0 said:
I am confused as how to find the I of the meter stick rotating about its parallel axis in the center. Since the center of mass is at the center, and it is being pivioted about the center, it seems there would be no I at all.
Rotational inertia depends on the distribution of all the mass, not just the location of the center of mass. To find the rotational inertia of an object about an axis, you must integrate the contributions of each mass element. The rotational inertia of a small piece of mass Δm is ΔmR^2, where R is the distance from the axis. Doing the integration for a stick rotating about its center will give you (1/12)ML^2, as harsh noted.

im not entirely sure how to integrate R^2 theough dm

i mean in a uniform rod, M = Lp, so dm = p dL, but then integrating that, i get pR^3 /3 where am i going wrong here?

arildno
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cair0 said:
im not entirely sure how to integrate R^2 theough dm

i mean in a uniform rod, M = Lp, so dm = p dL, but then integrating that, i get pR^3 /3 where am i going wrong here?

dm = p dL:
Wrong!
dm = p dR, where 0<=R<=L

Hence, we have:
$$I=\int_{0}^{L}pdR=\frac{p}{3}L^{3}=\frac{M}{3}L^{2}$$

When doing integrations, you should ALWAYS INTRODUCE A DUMMY VARIABLE OF INTEGRATION!!!!
This example is mathematically meaningless, but, unfortunately, it's common in physics texts:
$$F(t)=\int_{0}^{t}f(t)dt$$
Never use this "convention"!

When integrating a function f, depending on time (for example) from 0 to an arbitrary time value t, write instead:
$$F(t)=\int_{0}^{t}f(\tau)d\tau$$

Here, $$\tau$$ is called a dummy variable of integration.

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