# The mechanics of ice skating

by DiracPool
Tags: mechanics, skating
 P: 534 Hello, I'm trying to bone up on my conservation of angular momentum skills as well as my ice skating skills so I can be like my hero, Michio Kaku. Unfortunately my ice skating skills are better than my physics skills, so I thought y'all might be able to help. Here's the question, if angular momentum, L, equals the moment of inertia of a body, I, multiplied by its angular velocity, ω, then does $L=mr^2(2πf)$? Now, if that's true, then does $r=\sqrt{L/m2πf}$? And, accordingly, $f=L/m2πr^2$? I just attempted to derive these myself so I don't know if I'm missing something here. Plugging in some values, then, if I weighed 100 kg and started spinning at 1 cycle per second with my arms extended at a 1 meter radius, then would my angular momentum be 628.32 joule-seconds? Now say we were to conserve this figure as I varied my "moment" during my spin by moving my arms inward and outward of my torso. Say I brought my arms in so that my radius was .5 meters instead of 1 meter. Would my rotation rate then be 4 cycles per second? Finally, if I decided I wanted to rotate at a comfortable 2 cycles per second, would I need to move my arms to a position whereby my radius was 0.7 meters? Am I calculating these figures correctly? Thanks for your help. Also, I do have one follow up question once I get all of this checked out. Attached Thumbnails
 P: 1,395 The moment of inertia is $mr^2$ for a point mass, but in general you have to integrate over the entire volume of the body. (divide body in small parts, sum mr^2 for these parts) It would be easier to put someone on a turntable and try to measure the angular momentum for different positions.
 Quote by willem2 The moment of inertia is $mr^2$ for a point mass, but in general you have to integrate over the entire volume of the body. (divide body in small parts, sum mr^2 for these parts) It would be easier to put someone on a turntable and try to measure the angular momentum for different positions.