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

AI Thread Summary
The discussion focuses on calculating the radius of gyration for a rectangular frame with massless rods and weights at the corners, pivoted about its center of mass. The angular acceleration and moment of inertia are critical to solving the problem, with the moment of inertia given as Icm = 101.4 kg ft². The radius of gyration is expressed as R = √(I/m), leading to a calculated value of 3.75 ft. However, concerns are raised about the moment of inertia calculation, suggesting it should only consider the weights at the corners since the rods are massless. Accurate assessment of the moment of inertia is essential for determining the radius of gyration correctly.
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 = \sqrt{I/m} On wiki & a few other sites..

The Attempt at a Solution



\sqrt{101.4kg ft^2 / 7.2kg}

=

3.75ft
 
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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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