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

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a rod with an attached mass at one end, pivoted at the other end. The moment of inertia for the rod is given as Irod = 1/3 mL², and the attached mass is treated as a point particle. Participants suggest using the parallel axis theorem to find the total moment of inertia, which involves adding the contributions from both the rod and the point mass. The importance of correctly identifying the pivot point and calculating the distance from it to the mass is emphasized. The conversation highlights the need for a clear understanding of the principles involved in moment of inertia calculations.
trivk96
Messages
47
Reaction score
0

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.
 

Attachments

Last edited:
Physics news on Phys.org
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
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top