I What is the purpose of radius of gyration

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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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