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The mechanics of ice skating

by DiracPool
Tags: mechanics, skating
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DiracPool
#1
Mar15-13, 03:20 AM
P: 611
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 [itex]L=mr^2(2πf)[/itex]?

Now, if that's true, then does [itex]r=\sqrt{L/m2πf}[/itex]?

And, accordingly, [itex]f=L/m2πr^2[/itex]?

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.
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willem2
#2
Mar15-13, 03:29 AM
P: 1,408
The moment of inertia is [itex] mr^2 [/itex] 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.
DiracPool
#3
Mar15-13, 03:47 AM
P: 611
Quote Quote by willem2 View Post
The moment of inertia is [itex] mr^2 [/itex] 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.
Would a perfectly homogenous sphere, mass distribution-wise, like a perfectly homogenous bowling ball, qualify as such a point mass?

jbriggs444
#4
Mar15-13, 06:08 AM
P: 999
The mechanics of ice skating

The point being made is that a point-like mass not located at the defined axis of rotation has a moment of inertia given by mr2 where r is its distance from the axis of rotation.

If you like, think of it as a point-like mass mounted on a massless rod that is attached to a freely spinning massless axle.

If you inflate that point-like mass to a bowling ball you have to consider not only the moment of inertia due to the distance of the center of mass from the axle but also the moment of inertia due to the distance of the distributed mass from its own center line. The total angular momentum would be mr2 + 2/5mR2 where R is the radius of the bowling ball and r is the distance of the center of the bowling ball from the axle.


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