What is the purpose of radius of gyration

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SUMMARY

The discussion centers on the purpose of the radius of gyration (Rg) in calculating the moment of inertia for a mass rolling down an incline. The moment of inertia can be expressed as I = m Rg², where Rg represents the radius of gyration, contrasting with the formula I = 1/2 * m r² for a solid cylinder. The radius of gyration allows for the simplification of complex shapes into equivalent forms, aiding in the analysis of rotational dynamics. Understanding the distinction between Rg and the physical radius (r) is crucial for accurately applying the correct moment of inertia formula.

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  • Knowledge of the properties of cylinders and disks
  • Basic principles of rotational motion
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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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