Figure skater spinning (rate of rotation)

  • Thread starter kbyws37
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  • #1
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A figure skater is spinning at a rate of 1.40 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to 68.0% of its original value. What is her new rate of rotation?



I am having trouble starting this problem.
I know that
(I)(alpha) = (I)(omega)(t)

however, the problem doesn't mention time..
 

Answers and Replies

  • #2
Doc Al
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What physical quantity is conserved in this problem?
 
  • #3
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Angular momentum is conserved.
So...
(I1)(omega1) = (I2)(omega2)
however i am confused as to how to find omega when i do not have time
 
  • #4
Doc Al
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Where do you see time mentioned in that equation? What makes you think you need it? (It doesn't matter how fast or how slowly the skater pulls her arms in, since all you care about is the end result.)
 
  • #5
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sorry i am struggling in physics so i don't really understand.

so
if she reduces her inertia by 68% of its original value....
it would be 0.68*omega

I don't know where to start
 
  • #6
Doc Al
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kbyws37 said:
so
if she reduces her inertia by 68% of its original value....
Express that mathematically. If her original rotational inertia is I1, what's I2? Note that the precise instructions were:
reducing her rotational inertia to 68.0% of its original value​
That's "to", not "by".
 
  • #7
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ok thanks. i got it

1.40 rev/s = (0.68)I2
= 2.06 rev/s
 
  • #8
Doc Al
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Good! I'll rewrite it more systematically.
Conservation of angular momentum says:
(I1)(omega1) = (I2)(omega2)

Given:
omega1 = 1.4 rev/s
I2 = 0.68 I1

So:
(I1)(1.40 rev/s) = (0.68 I1)(omega2)
(1.40 rev/s) = (0.68)(omega2)
omega2 = (1.40 rev/s)/(0.68)
 

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