Radius of gyration (rectangle)

[madmax2006]

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 =
I_cm = 1/12(M(a² + b²))
I_cm = 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

 

[kuruman]

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.

 

[kerimek]

Dear student,

Thank you for your question. The radius of gyration is a measure of how far the mass of an object is distributed from its axis of rotation. It is often represented by the symbol "k" and is calculated using the formula k = √(I/m), where I is the moment of inertia and m is the mass of the object.

In the case of a rectangular frame, the moment of inertia can be calculated using the formula I = 1/12*M*(a^2 + b^2), where M is the total mass of the frame and a and b are the dimensions of the frame.

Therefore, the radius of gyration of the frame about O can be calculated as follows:

k = √(I/m) = √(1/12*M*(a^2 + b^2)/M) = √(1/12*(a^2 + b^2)) = √(1/12*(1.8 lb)^2) = 0.75 ft

I hope this helps. Let me know if you have any further questions.

Sincerely,