Rotational Motion of a spindle

  • Thread starter Bones
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  • #1
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Homework Statement


A uniform disk turns at 9.5 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed. What is the angular velocity of the combination?




Homework Equations





The Attempt at a Solution



I1w1= 1/2mr^2*9.5rev/sec
I2w2= 1/12ml^2*w2
That's as far as I got...is it even right??
 

Answers and Replies

  • #2
tiny-tim
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A uniform disk turns at 9.5 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed. What is the angular velocity of the combination?

I1w1= 1/2mr^2*9.5rev/sec
I2w2= 1/12ml^2*w2
That's as far as I got...is it even right??

Hi Bones! :smile:

(have an omega: ω :wink:)

Sort-of … the Is are right …

but you need conservation of angular momentum,

so for the "after" side, you'll need the total I, = I1 + I2, so that you can find (I1 + I22. :smile:
 

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