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Rod in rotational motion. Not sure how to account for potential energy.

  • Thread starter shanest
  • Start date
  • #1
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I've also attached the picture of this problem at the end of the post.

A uniform rod of mass 4:5 kg is 10mlong. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 10 m from the center of
mass of the rod. Initially the rod makes an
angle of 60± with the horizontal. The rod is
released from rest at an angle of 60± with the
horizontal, as shown in the ¯gure below
The acceleration of gravity is 9:8 m=s2 :
Hint: The moment of inertia of the rod
about its center-of-mass is Icm = (1/12)m*l^2

What is the angular speed of the rod at
the instant the rod is in a horizontal position?
Answer in units of rad=s.




I tried doing mgh = 1/2 I * omega^2, but it doesn't feel right at all. Especially since I was unsure of whether I could use the height of the center of mass of the rod for the h in mgh. Can I? There's gotta be a better way to solve this problem...
 

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  • #2
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I suppose you could do it with angular kinematic formulas, but I think energy would work just fine here. Energy would definitely be the way I would try it.

Edit: find the potential of the COM at rest, then convert that into angular velocity when it's horizontal?
 
Last edited:
  • #3
144
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Yes you need to use the height difference in the 2 positions for the CM. What is the height of CM above the horizontal? And what is the the rods moment of inertia about the point O? Once you know these two, the problem should be easily solved using energy conservation, as the way you tried, which is correct.
 

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