I What is the purpose of radius of gyration

AI Thread Summary
The discussion revolves around the confusion regarding the use of radius 'r' and radius of gyration 'Rg' in calculating the moment of inertia for a mass rolling down an incline. It clarifies that the moment of inertia can be expressed as I = mRg^2, where Rg represents the distribution of mass, while I = 1/2 * mr^2 applies specifically to a solid cylinder. Participants question whether the problem specifies the shape of the object, as this affects the appropriate formula to use. The relationship between Rg and r is also discussed, with a suggestion that Rg could be equal to r/√2, depending on the object's characteristics. Understanding the object's geometry is crucial for correctly applying the moment of inertia formula.
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Hello,

I'm given a problem of a mass rolling down an incline with mass 'm', radius 'r', and radius of gyration Rg, and I need to write the Lagrangian for the motion. I'm confused on why both r and Rg are given. Don't I just need one of the two for the moment of inertia?
Thanks!
 
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The radius of gyration gives the moment of inertia via the equivalence of the given object and an object with all its mass concentrated at a distance of the radius of gyration from the center of mass. So I = m R_g^2.

You can treat the object as a massless cylinder of the given physical radius with a hollow cylinder of mass m and radius R_g embedded within it.
 
Hi jambaugh,

Thank you for your response. The part I'm confused on is if I use Rg the moment of inertia (assume a cylinder) would be I= mRg^2, but if I use 'r', then I=1/2 *mr^2. I guess I don't understand where the question is leading...
 
Why would I = \frac{m}{2} r^2? Is R_g given to be equal to r/\sqrt{2}? Did the problem specifically say the object was a solid disk or cylinder? Maybe the object is not what you are assuming with the \frac{m}{2}r^2 formula.
 
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