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 January 8th, 2017, 01:31 PM #11 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 i am also struggling with a very similar topic, i am trying to calculate the radius of gyration however for the experimental value it says to use the outside radius of the disc but for the experimental calculations it says use the roller radius, why would that be? and also how do i calculate the radius of gyration? i have transposed a formula to get k = √(2ghr)-r^2/(v^2) -r^2 but how do i use this to find what i need?
 January 9th, 2017, 01:13 PM #12 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. I would have thought using $I = k^2 A$ would be more straight forward. Rearranging would give, $\displaystyle k = \sqrt{\dfrac{I}{A}}$ Substituting moment of inertia of a solid disk (about x,y plane), $\displaystyle k = \sqrt{\dfrac{m r^2}{4A}}$ EDIT: Experimental value? Experimental calculations? I can only guess that maybe you are required to find $k$ theoretically, then compare this result with experimental values. Or, it is all to be done theoretically, and you are to observe the differences that appear when changing point that you are measuring radius from. Last edited by Joppy; January 9th, 2017 at 01:17 PM.
 January 10th, 2017, 02:47 PM #13 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 Yes we must calculate both a theoretical value and the experimental value

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