Moment of inertia of a rod about an axis

In summary, the moment of inertia of a rod about an axis through its center can be found by using the formula I = (1/8) M L^2. However, the given formula for mass per unit length, lambda = lambda (sub zero) times X, may be confusing as it is not a constant value. In order to integrate properly, the absolute value of X must be used. The equation for mass is also given, but it can be combined with the moment of inertia equation to solve for lambda (sub zero).
  • #1
Supernerd2004
20
0
Find the moment of inertia of a rod about an axis through its center if the mass per unit length is lambda = lambda (sub zero) time X.
Answer (I = (1/8) M L^2)

This problem is totally throwing me off. Normally lambda is equal to (M/L) so I am not sure what this new formula is doing to the problem. I tried substituting and came up with (1/32)ML^3 which is horribly off from the answer. My idea was to add the x term from the new lambda giving me x^3. I then brought lambda (sub zero) out of the integral and integrated from (-L/2) to positive (L/2). Which brings us back to the wrong answer. Any ideas as to where I’m going wrong would be much appreciated!

Thanks

Dan
 
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  • #2
I found this thread on the internet, but I guess it still doesn't clear things up.

"Okay, I'll do the first one for you.

The line density is given as [tex]\lambda=\lambda_{0}x[/tex]

It's no wonder that you are stuck with this exercise, because this expression is by itself meaningless.
The correct expression must be:

[tex]\lambda=\lambda_{0}|x|[/tex], -L/2<=x<=L/2
i.e the absolute value of x, rather than x itself.
(I assume you gave us ALL the information present in the exercise?)
We gain:

[tex]I=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|{x}^{2}dx= \lambda_{0}\frac{2}{4}(\frac{L}{2})^{4}[/tex]

The mass M of the rod is readily calculated:

[tex]M=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|dx=\lambda _{0}(\frac{L}{2})^{2}[/tex]

Combining the expressions yiels the desired result"

I guess at this point I just need a little help understanding the integration that took place. The absolute value of X is a little confusing. Thanks again in advance!

Here is the problem again in more readable text:

Find the moment of inertia of a rod about an axis through its center if the mass per unit length is [tex]\lambda=\lambda_o x[/tex].

Answer: [tex]I=\frac 1 8 ML^2[/tex]

Dan
 
  • #3
It just means that the density is given as 0 at the axis and as [tex]\lambda_0 x[/tex] at a distance x away from the axis.

You have to use the absolute value because you are integrating x from -L/2 to L/2, but you are concerned about the distance. The distance to the point -L/2 on the rod is L/2 so it follows that if x is -L/2 and you want the distance to x you want the absolute value.
 
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  • #4
Oh, so the density of the rod is not constant. That part makes sense, but why was an equation for the mass given? The other thread explains that combining the equations gives the result, but it just isn't making sense yet. Thanks so much for your help,

Dan
 
  • #5
It depends on how you interpret the problem. It says the mass per unit length is [tex]\lambda=\lambda_0 x[/tex]. I assumed (as did the site where you got the solution from) that x is the distance from the axis to any point on the rod. So if you are looking for the density a distance L/4 away then it would be [tex]\lambda_0\frac{L}{4}[/tex]

You need to combine your equations because the calculated moment of inertia [tex]I=\lambda_{0}\frac{2}{4}(\frac{L}{2})^{4}[/tex] is still in terms of [tex]\lambda_0[/tex]. You can solve for [tex]\lambda_0[/tex] in terms of M using the second integration that gave [tex]M=\lambda _{0}(\frac{L}{2})^{2}[/tex]
 
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  • #6
Supernerd2004 said:
I guess at this point I just need a little help understanding the integration that took place. The absolute value of X is a little confusing. Thanks again in advance!
Since the integrand is symmetric with respect to the origin, just integrate the positive side and double it.

Here is the problem again in more readable text:

Find the moment of inertia of a rod about an axis through its center if the mass per unit length is [tex]\lambda=\lambda_o x[/tex].
I find that statement odd. If [itex]\lambda_o[/itex] has units of [itex]\lambda[/itex], then [itex]\lambda_o x[/itex] cannot. I presume it should read (I haven't checked it) something like: [tex]\lambda=\lambda_o |x|/(L/2)[/tex]
 

1. What is the moment of inertia of a rod about an axis?

The moment of inertia of a rod about an axis is a measure of its resistance to rotational motion. It is a physical quantity that describes how mass is distributed around an 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 factors affect the moment of inertia of a rod?

The moment of inertia of a rod is affected by its mass and distribution of mass along its length. A rod with more mass or a greater concentration of mass towards its ends will have a larger moment of inertia.

4. How does the moment of inertia of a rod about its center of mass compare to the moment of inertia about other points?

The moment of inertia of a rod about its center of mass is the lowest possible value. It increases as the distance from the axis of rotation increases, reaching its maximum value at the ends of the rod.

5. What are the practical applications of understanding the moment of inertia of a rod?

Understanding the moment of inertia of a rod is important in various fields, including engineering, physics, and mechanics. It is used in designing structures and machines that involve rotational motion, such as bridges, cars, and bicycles. It also helps in analyzing and predicting the behavior of objects in rotational motion.

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