- #1

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I am having trouble starting this problem.

I know that

(I)(alpha) = (I)(omega)(t)

however, the problem doesn't mention time..

- Thread starter kbyws37
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- #1

- 67

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I am having trouble starting this problem.

I know that

(I)(alpha) = (I)(omega)(t)

however, the problem doesn't mention time..

- #2

Doc Al

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What physical quantity is conserved in this problem?

- #3

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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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- #5

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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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Express that mathematically. If her original rotational inertia is I1, what's I2? Note that the precise instructions were:kbyws37 said:so

if she reduces her inertia by 68% of its original value....

reducing her rotational inertia __to__ 68.0% of its original value

- #7

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ok thanks. i got it

1.40 rev/s = (0.68)I2

= 2.06 rev/s

1.40 rev/s = (0.68)I2

= 2.06 rev/s

- #8

Doc Al

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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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