# Moment of inertia for physical pendulum

1. Dec 14, 2008

### BoogieBot

I have a rigid, massless strut of length 11m. A mass of 55kg is located on the free end, and another mass of 55kg is located at the midpoint. I'm having some trouble calculating the moment of inertia I for this physical pendulum.

I = $$\int r^{2} dm$$ (for continuous objects)

I = $$\sum m D$$ for all particles/objects composing the system.

I'm a little confused. Should I calculate I for each 55kg mass and add them together?

2. Dec 14, 2008

### Staff: Mentor

Exactly.

3. Dec 14, 2008

### BoogieBot

so:
55*5.5 = 302.5
55*11 = 605

so I for the whole pendulum is 907.5 kg-m^2, correct?

4. Dec 14, 2008

### Staff: Mentor

No, your formula for I of a point mass is incorrect. (I didn't notice that earlier.)

For both continuous objects and point masses, the distance must be squared. (Imagine integrating the first formula to get the second. All of the mass is at the same distance from the axis, so the integral is trivial.)

5. Dec 14, 2008

### BoogieBot

Ah, so then:

55*5.5^2 = 1663.75
55*11^2 = 6655

1663.75 + 6655 = 8318.75 kg-m^2

6. Dec 14, 2008

### Staff: Mentor

You got it.

7. Dec 14, 2008

### BoogieBot

awesome! thanks!