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

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SUMMARY

The moment of inertia for a rod of length L and mass m, pivoted at one end, is calculated using the formula Irod = 1/3 mL². When a point mass m is attached to the free end of the rod, the total moment of inertia can be determined by applying the parallel axis theorem, which states I = Icm + mr². The total moment of inertia is the sum of the moment of inertia of the rod and the additional contribution from the point mass. This approach allows for the accurate calculation of the system's moment of inertia with respect to the pivot point.

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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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