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Trying to find the angular speed of a rod.

  • Thread starter thyrgle
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  • #1
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Homework Statement


A uniform rod (length = 2.0m) is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through the rod at a point of 0.5m from one end of the rod. If the rod is released from rest in a horizontal position, what is the angular speed of the rod as it rotates through its lowest position?


Homework Equations


I believe (although I might be wrong on some of them):
[itex]\tau = I \alpha[/itex]
[itex]K = \frac{1}{2} I \omega^{2}[/itex]
[itex]L_{z} = I \omega[/itex]
For a rod the moment of inertia is:
[itex]\frac{1}{12} M L^{2}[/itex]
And the parallel axis thereom:
[itex]I = I_{cm} + M D^{2}[/itex]

The Attempt at a Solution


So far I have used the parallel axis theorem and the moment of inertia for a rod to get the moment of inertia to be 0.5833 for this rods rotation.
I am not sure exactly what to do after that. I can figure it out if it is the moment right after release, but not sure how to do it when the rod is at the bottom.
 

Answers and Replies

  • #2
haruspex
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Energy?
 
  • #3
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I first started thinking I could take the potential energy the system has before release as (mgh) and then at the bottom i could take mgh = (1/2)I*w^2 and find omega... But I couldn't figure out how to use it right...
 
  • #4
haruspex
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That's the right method. Just need to figure out h and I. Any ideas?
 
  • #5
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I believe I have found I for the entire rod. The issue is with the height I think. I want to say the height is 1.5 but what about the 0.5 end? I try using m = 1, g = 9.8 and h = 1.5.

1.5*9.8 = 14.7
14.7 = (1/2)(0.5833333)(w^2)
29.4 = 0.5833(w^2)
50.4 = w^2
w = 7.099

But the answer is supposedly 4.1...
I then tried subtracting the small 0.5 stub:
0.5*9.8*1 (mgh)
4.9 = (1/2)(0.5833)(w^2)
9.8 = (0.5833)w^2
16.8 = w^2
w = 4.099

Subtracting the two gives 3.00 which is not right either though.
 
  • #6
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[itex]I[/itex] was found using doing the following:

I = I_cm + m*d^2

I = (1/12)m*L^2 + m*d^2

d = 0.5 since there is a 0.5 shift from the center.

I = (1/12)*1*4 + 1*(0.5^2)

I = (0.5833333)
 
  • #7
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Actually never mind I should use h = 2 not 1.5 for the first part of my original attempt and then everything else should be the same.

1.5*9.8 = 14.7
14.7 = (1/2)(0.5833333)(w^2)
29.4 = 0.5833(w^2)
50.4 = w^2
w = 8.199

0.5*9.8*1 (mgh)
4.9 = (1/2)(0.5833)(w^2)
9.8 = (0.5833)w^2
16.8 = w^2
w = 4.099

8.199 - 4.099

Thanks!

Although I am still not sure why I should use 2...
 
  • #8
haruspex
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No, 0.5m is right for h (it's the distance the CoM descends).
You should have
mgh = (1/2)m(L2/12+h22
4.9*2 = (1/3+1/4)ω2
ω ≈ 4.1
Are you saying the given answer is 8.2?
 
  • #9
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Oh, my bad. Haha the given answer was 4.1, for some reason I didn't even realize using h = 0.5 gives the answer approx. 4.1. My bad thanks again!
 

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