# Moment of Inertia for 3 masses, on an axis

1. Nov 26, 2012

### matxjos

1. The problem statement, all variables and given/known data
[refer to picture, thanks]
The "L-shaped" figure rotates on the axis which which intercepts through 9 kilogram and 2.5 kilogram masses. Find the moment of inertia of the object for this type of rotation. Disregard the masses of connecting bars.

2. Relevant equations
MR^2

3. The attempt at a solution
Not really sure on what to do. I am assuming we treat the objects like point masses but I'm not really sure on that either. Do you take the 1.2 kg object and only use this with its radius to each object? I tried but the answer wasnt in the answer options.

The 2 objects on the axis are confusing me. Where is the center of mass the object is rotating on.

Thanks for any help!

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2. Nov 26, 2012

### SHISHKABOB

moment of inertia is

$I = \sum MR^{2}$

where the R is the distance from each mass component to the axis of rotation

if we assume that each mass is a point mass, then they aren't going to be doing any rotating if they are on the axis of rotation

so the problem here is to find the distance from the 1.2 kg mass to the axis of rotation

3. Nov 26, 2012

### matxjos

Hi, thanks for the response.

That is what I initially thought, but what I got isnt an answer choice.

I calculated that the 1.2kg mass is 1.788854383m from the axis of rotation.

so 1.2*1.78^2 = 3.84 kg*m^2

the options are: 50, 19, 11, 0.96, 0.72, 29, 60, 20, 0.60
in kg*m^2

4. Nov 26, 2012

### SteamKing

Staff Emeritus
The distance that you calculated for the 1.2 kg mass from the axis of rotation is wrong.

5. Nov 26, 2012

### matxjos

Okay. I assume you have to use geometry to figure that out.

Tried to use cosine.

Can you tell me how to figure out that distance?

6. Nov 26, 2012

### matxjos

Nvm, I figured it out. Was using cosine instead of sine.

Thanks for the help, questions done.

:rofl: