Center Gravity and Moment of Inertia

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a system of four objects arranged at the corners of a rectangle. Participants clarify the formulas needed for moments of inertia about the x-axis, y-axis, and an axis perpendicular to the page. The center of mass (CoM) is determined as a prerequisite for these calculations, with emphasis on using the correct distances from the origin to each mass. Pythagorean theorem is suggested as a helpful tool for finding these distances. The conversation concludes with participants confirming their understanding of the calculations involved.
ymehuuh
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Homework Statement


Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. The mass values are given below.
M1 (kg) M2 (kg) M3 (kg) M4 (kg)
3.50 1.50 3.90 1.70
p8-29alt.gif

(a) Find the moment of inertia of the system about the x axis.

(b) Find the moment of inertia of the system about the y axis.

(c) Find the moment of inertia of the system about an axis through O and perpendicular to the page.

Homework Equations



Center of Gravity: sumM1*X1+M1*X2.../M1+M2...

The Attempt at a Solution



I found the center of gravity for the x-axis to be .037736...how do I find the moment of inertia with that information?
 

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You need to find the x and y CoM.

Knowing that then the moment I about the CoM is Σmr2.

The axes are simpler in that for

I_x = Σm_i*y_i2

I_y = Σm_i*x_i2
 
Oh, I reversed the two and that's why I got it wrong. Thanks.
What would I use for part c?
 
ymehuuh said:
Oh, I reversed the two and that's why I got it wrong. Thanks.
What would I use for part c?

What are the distances to each corner from O to each mass?

I_o = Σmr2

It's made a little easier by Pythagoras, so be sure and thank him.
 
lowlypion said:
what are the distances to each corner from o to each mass?

I_o = Σmr2

it's made a little easier by pythagoras, so be sure and thank him.

perfect! Thanks!
 

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