# Energy in rotational motion

1. Oct 2, 2007

### Edwardo_Elric

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

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

3. The attempt at a solution
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)

2. Oct 2, 2007

### andrevdh

There is another block at the centre of the rod.

3. Oct 2, 2007

### 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

4. Oct 2, 2007

### learningphysics

you should have (1/6)^2

5. Oct 2, 2007

### Edwardo_Elric

oh yeah thanks LP