What is the moment of inertia for a rod with an attached mass at the end?

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Homework Help Overview

The problem involves calculating the moment of inertia for a system consisting of a rod pivoted at one end with an additional mass attached at the free end. The subject area pertains to rotational dynamics and the application of the parallel axis theorem.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the use of the parallel axis theorem and the need to calculate the moment of inertia for both the rod and the attached mass. There is consideration of how to determine the distance from the pivot point and the contributions of each component to the total moment of inertia.

Discussion Status

The discussion is active with multiple perspectives being shared. Some participants suggest different approaches to the problem, while others express uncertainty about the effectiveness of the parallel axis theorem in this context. There is no explicit consensus, but various insights are being explored.

Contextual Notes

Participants note the importance of understanding the contributions of both the rod and the point mass to the overall moment of inertia, and there is mention of constraints related to the problem setup and the need for clarity on definitions and assumptions.

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Homework Statement


Consider a rod of length L and mass m which
is pivoted at one end. The moment of inertia
of the rod about an end is Irod = 1/3 mL2. An object with mass m is attached to the free end of the rod.

Find the moment of inertia of the system
with respect to the pivot point. Consider
the mass at the end of the rod to be a point
particle.

I tried a lot of different things and now I only get one more try. I need help. Thanks in advance

Homework Equations


Irod = 1/3 mL2.
I=Icm+mr2

The Attempt at a Solution


I think you have to use the Parrallel axis theorem ( I=Icm+mr2). But I am stuck here.
 

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Moment of inertia is the sum of all the masses times the square of their distance from the pivot. Someone has already summed all the masses in the rod itself to get 1/3 mL². You just have to add the one additional contribution from the point particle at the end of the rod.
 
I guess you could use parallel axis theorem. You would need to find where the centre of mass is (to determine r in your equation). And you would need to find the inertia around the centre of mass. So you still have a similar problem, so I don't think parallel axis theorem helps much.

I think you should try thinking differently about the problem. You want to find the moment of inertia around the pivot point. You are told the moment of inertia due to the rod and you can work out the moment of inertia due to the point particle.

So once you have these, you effectively have two moments of inertia which contribute to the total moment of inertia. Do you know how to calculate the total moment of inertia? (Hint, its a simple formula).

Edit: Sorry, I started writing before Delphi51 made the post. I don't mean to be treading on toes. (if that's the right expression...)
 
I appreciate it, Bruce! I think it was you who bailed me out of a mistake this morning.
And great for the OP to get two different views.
 
yep, that's a lucky OP'er. I am pretty tired right now. I think I'll go to sleep, or I am bound to make a mistake soon!
 
thank you
 

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