Moment Of Inertia Of Non-uniform Rod

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Homework Help Overview

The discussion revolves around finding the moment of inertia of a non-uniform rod with an additional mass attached at a certain position along its length. The context includes the analysis of a physical pendulum and the calculation of its oscillation period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the moment of inertia for both the uniform rod and the additional point mass, questioning how to combine their effects. There are inquiries about the center of mass of the system and how it changes with the added mass.

Discussion Status

Some participants have offered guidance on treating the additional mass as a point mass and combining the moments of inertia. There is ongoing exploration of how to express the center of mass as a function of the position of the added mass.

Contextual Notes

Participants are considering the implications of the mass distribution along the rod and how it affects the calculations, particularly in relation to the definitions of moment of inertia and center of mass.

BOAS
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Hello,

I am trying to find the moment of inertia of a uniform rod, that has a mass added to it at some position along it's length, which is equal to the mass of the rod itself, and the axis of rotation is at one end.

1. Homework Statement


A uniform, \mathrm{1.00m} stick hangs from a horizontal axis at one end and oscillates as a physical pendulum with period T_{0}. A small object of mass equal to that of the stick can be clamped to the stick at a distance y below the axis. The system then has a period T.

Find the ratio \frac{T}{T_{0}}

Homework Equations

The Attempt at a Solution


[/B]
I know that the moment of inertia of a uniform rod with the axis about one end is equal to;

I = \frac{1}{3}ML^{2}

The period of a physical pendulum is given by;

T = 2 \pi \sqrt{\frac{I}{mgd}} where d is the distance from the pivot to the center of gravity. So, for the initial case, we have a period of

T_{0} = 2 \pi \sqrt{\frac{\frac{1}{3}ML^{2}}{mg\frac{L}{2}}} = 2 \pi \sqrt{\frac{2L}{3g}}

I am stuck here though, because I don't know how to find a moment of inertia for the second case. I understand how to calculate moments of inertia of non-uniform rods using integration, but I don't have a function for the linear density...

Thanks for any help you can give!
 
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BOAS said:
I am stuck here though, because I don't know how to find a moment of inertia for the second case.
What's the second case? Where the mass is stuck to the rod? What's the moment of inertia of a point mass? Just add the separate moments of inertia to find the total.
 
Doc Al said:
What's the second case? Where the mass is stuck to the rod? What's the moment of inertia of a point mass? Just add the separate moments of inertia to find the total.

Thanks, I was making that far more complicated in my head than it needed to be.

How do I describe the center of mass of such an object? I need to find an expression that describes the center of mass as a function of y, the masses distance along the rod.
 
BOAS said:
How do I describe the center of mass of such an object? I need to find an expression that describes the center of mass as a function of y, the masses distance along the rod.

You treat it as a point mass i.e. the mass itself is dimensionless so it has no radius and so on to consider. So you use your basic definition of I for a point mass.

What will change is your center of gravity of the new system.
 
BOAS said:
How do I describe the center of mass of such an object? I need to find an expression that describes the center of mass as a function of y, the masses distance along the rod.
You can find the center of mass of each separately (which is trivial, of course). Then treat them as two point masses for the purpose of find the center of mass of the system.
 
Doc Al said:
You can find the center of mass of each separately (which is trivial, of course). Then treat them as two point masses for the purpose of find the center of mass of the system.

So I have my uniform rod, and I say that the left end is at 0 and the right is at L. I treat the rod as a point mass at L/2. I then consider another point mass at position y, where y is between 0 and L somewhere along the length of the rod.

So the center of mass of the system is CM = \frac{M\frac{L}{2} + My}{2M}
 
BOAS said:
So I have my uniform rod, and I say that the left end is at 0 and the right is at L. I treat the rod as a point mass at L/2. I then consider another point mass at position y, where y is between 0 and L somewhere along the length of the rod.

So the center of mass of the system is CM = \frac{M\frac{L}{2} + My}{2M}
Exactly.
 
Thank you.
 

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