Look, I have a big doubt with this, but isn't it 30 feet ??, because, how come can it dropeed 10 feet !!!, if it is onle 3 feet long, well, anyway, lets work with 30 feet.
Look, the angular velocity changes, because, the distance from the center of mass of the bar to the normal, you know, the vertical line es different when it drops. BUT IF THE BAR IS ROTATING AROUND ITS CENTER OF MASS, THEN THE ANGULAR VELOCITY WOULD NOT CHANGE, but I don't think that's the case in this problem, because the bar is making and angle with the horizon (30) degrees, I imagine, the bar, is rotating around one of it's ends.
The angular momentum (L) is the same at the beginning of the movement, and after it :
I*w = L = constant
I = m*r^2
where "r", is the distance from the center of the bar to the normal.
When it drops, the distance from the center of mass to the vertical will be :
r^2 = (5/2)^2 + (15/2)^2 = 62.5 = r^2
r = 7.9
20*[15*sqr(3)/4]^2*3 = 20*w*[sqr(62.5) ]^2
now, sqr(100), is the distance from the center of the bar to the normal
then :
3*20*42.18 = 20*62.5*w
w = 2.02 rad/s
Now, that will be the value of angular velocity after the bar drops 10 feet, if the bar is 30 feet long.
If you make a graphic of the problem, you will realize, the distances, and why it has to be 30 feet long
2007-03-01 11:19:57
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answer #1
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answered by anakin_louix 6
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Is it held at one end?? The bar is 3 feet long. How can it drop 10 feet?? If it isn't held at one end, then Steve is right. Dropping something that is rotating won't affect the angular velocity??
2007-03-01 11:19:21
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answer #2
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answered by hello 6
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I don't understand what axis the bar is rotating about or how dropping it could affect the angular velocity.......?????????
2007-03-01 11:02:28
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answer #3
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answered by Steve 7
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