# Inertia of unequal masses on pivoted rod

1. Nov 24, 2007

### motalha

1. The problem statement, all variables and given/known data

A rod of mass m_r and length 2L allowed to pivot freely about its center (or central axis). A small sphere of mass m_1 is attached to the left end of the rod, and a small sphere of mass m_2 is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to .

What is the moment of inertia of this assembly about the axis through which it is pivoted?
Express the moment of inertia in terms of m_r, m_2 , m_1 , and L . Keep in mind that the length of the rod is 2L, not L .

3. The attempt at a solution

My attempt at the solution was to sum up the intertia of sphere with m_1, sphere of m_2 and the rod. I got:

(2/5*m_2*L^2) + (2/5*m_1*L^2) + (1/12*m_r*2L^2)
I think there is something wrong with the inertia of the rod. Any help at solving this would be great, thanks.

2. Nov 24, 2007

### motalha

Attached a diagram applicable to the question.

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3. Nov 24, 2007

### Staff: Mentor

Each term has problems: For the two masses, where do you get the 2/5 factor? (Remember these are to be treated as point masses.) For the rod, realize that its length is 2L, so you should have (2L)^2.

4. Nov 24, 2007

### motalha

Thanks I forgot to regard the masses as point particles and thought of them as spheres using 2/5. Appreciate the help and thanks again.