What is the moment of inertia for small blocks clamped to a light rod?

[Edwardo_Elric]

Homework Statement

Small blocks with mass m, are clamped at the ends and at the center of a light rod of length L. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one third of the length from one end.
Neglect the moment of inertia of the light rod.

Homework Equations

$$\sum{I} = m_{1}r_{i} + m_{2}r_{i}...$$

The Attempt at a Solution

my answer would be
1/3 at one end and 2/3 because the other farther
I = m([1/3]L)^2 + m([2/3]L)^2
I = 1/9(mL^2) + 4/9(mL^2)
I = 5/9(mL^2)

 

[andrevdh]

There is another block at the centre of the rod.

 

[Edwardo_Elric]

so the block at the center has a distance of:
since it is at the center
1/2 - 1/3 = 1/6 is its distance from the end point

I = 1/9(mL^2) + 4/9(mL^2) + 1/6(mL^2)
I = 13/18mL^2

 

[learningphysics]

so the block at the center has a distance of:
since it is at the center
1/2 - 1/3 = 1/6 is its distance from the end point

I = 1/9(mL^2) + 4/9(mL^2) + 1/6(mL^2)
I = 13/18mL^2

you should have (1/6)^2

 

[Edwardo_Elric]

oh yeah thanks LP