What Is the Radius of Gyration of a Rectangular Frame About Its Center?

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SUMMARY

The radius of gyration for a rectangular frame about its center of mass (point O) can be calculated using the formula R = √(I/m), where I is the moment of inertia and m is the total mass. In this discussion, the moment of inertia was initially calculated as Icm = 101.4 kg ft², leading to a radius of gyration of 3.75 ft when the total mass is 7.2 kg. However, the calculation of the moment of inertia was questioned due to the assumption of massless rods, indicating that only the masses at the corners should be considered for accurate results.

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madmax2006
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I'm trying to figure out the radius of gyration of the frame about O.

Homework Statement



A rectangular frame is put together with massless rods having identical 1.8 lb weights placed at the corners as shown in the figure. The frame is pivoted about an axis passing through O, the center of mass of the system, perpendicular to the paper.
a) Find the angular acceleration of the frame about O
b) Find the radius of gyration of the frame about O

Homework Equations


Angular acceleration of the fram about O =
Icm = 1/12(M(a2 + b2))
Icm = 101.4kg ft2

I can't find an equation for b) and I can't find anything about "radius of gyration" in my book. Is there another name for it?

I found R = [tex]\sqrt{I/m}[/tex] On wiki & a few other sites..

The Attempt at a Solution



[tex]\sqrt{101.4kg ft^2 / 7.2kg}[/tex]

=

3.75ft
 
Last edited:
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You have the correct expression for the radius of gyration. I know of no other name for it.

Your calculation of the moment of inertia looks suspicious, but to be 100% sure, we have to see a picture. I say "suspicious" because the rods are said to be massless so the factor of 1/12 should not be there. Just find the total moment of inertia of the four masses at the corners because that's what the problem appears to want.
 

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