# Moment of Inertia of a Rod: Translating to Off-Center Axis

• whozum
In summary, the moment of inertia for a rod centered at the origin and rotating along the x-axis can be represented by the integral \int_{-L/2}^{L/2} r^2 \frac{M}{L} dr, where M/L denotes mass density. When the rod is not rotating about the center but instead about an axis at L/5 from its edge, the integral becomes \int_{-L/5}^{4L/5} r^2 \frac{M}{L} dr and evaluates to \frac{M}{3L}\left(\frac{4L}{5}\right)^3 - \left(\frac{-L}{5}\right)^3. This can also be
whozum
Moment of inertia for a rod along the x-axis centered at the origin is :

$$\int_{-L/2}^{L/2} r^2 \frac{M}{L} dr$$

M/L denotes mass density, this evaluates to $\frac{1}{12}mR^2$ as we'd expect, but my questoin is, if the rod was not rotating about an axis through its center, but an axis at L/5 from its edge, would the integral directly translate to:

$$\int_{-L/5}^{4L/5} r^2 \frac{M}{L} dr$$ ?

In which case it would evaluate to

$$\frac{M}{3L}\left(\frac{4L}{5}\right)^3 - \left(\frac{-L}{5}\right)^3$$ expanded?

Yep, that's it. Parallel axis theorem gives the same thing--(13/75)ML^2

I thought parallel axis was only for the edge?

far as i know, it works whenever you are measuring with respect to the CM. So if u know the inertia about an axis thru the CM, then the parallel axis theorem gives the inertia about an axis that is parallel to the axis through the CM. If they are perp. distance A apart, then

I = I_cm + mA^2

Is this the version you use?

Yeah actually your definition is more correct, I knew it as the same except add "at the edge" to "about an axis parallel to the axis thru the CM" with

what does the edge have to do with anything?

No idea, that's what I remember it as for some reason.

Thanks.

## 1. What is the moment of inertia of a rod?

The moment of inertia of a rod is a measure of its resistance to rotational motion around an axis. It is affected by the mass distribution of the rod and the location of the axis of rotation.

## 2. How is the moment of inertia of a rod calculated?

The moment of inertia of a rod can be calculated using the formula I = 1/12 * m * L^2, where m is the mass of the rod and L is the length of the rod.

## 3. What is the significance of an off-center axis when calculating the moment of inertia of a rod?

An off-center axis means that the axis of rotation is not located at the center of mass of the rod. This can affect the moment of inertia calculation and result in a different value compared to when the axis is at the center of mass.

## 4. How does the moment of inertia of a rod change when the axis is moved off-center?

The moment of inertia of a rod increases when the axis is moved off-center, as the mass distribution is farther away from the axis and therefore has a greater effect on the rotational motion.

## 5. What are some real-world applications of understanding the moment of inertia of a rod with an off-center axis?

Understanding the moment of inertia of a rod with an off-center axis is important in engineering and design, particularly in structures that rotate or move around a specific axis. It is also relevant in sports, such as in the design of golf clubs or the physics of figure skating jumps.

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