Rotational motion of a uniform solid disk

  • Thread starter Momentum09
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  • #1
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Homework Statement



A uniform solid disk of radius 7.1m and mass 30.3kg is free to rotate on a fricionless pivot through a point on its rim. [picture: http://www.wellesley.edu/Physics/phyllisflemingphysics/107_p_angular_images/figure_9.gif] [Broken]
If the disk is released from rest in the position shown, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle?
What is the speed of the lowest point on the disk in the dashed position?


Homework Equations



mgh = 1/2 Iw^2 + 1/2mv^2.


The Attempt at a Solution



From the above equation, I was able to subsitute w = v/r and solved for v...v = (4/3gh)^1/2. I don't know if this velocity pertains to the one about the center of mass or the lowest point. Please help! Thank you!
 
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Answers and Replies

  • #2
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I got the speed of the center of mass. Now I just need to somehow link that to the speed of the lowest point.
 
  • #3
andrevdh
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You should not include the second term in

mgh = 1/2 Iw^2 + 1/2mv^2

that is it should just be

mgh = 1/2 Iw^2

since the disc is not translating, it is just rotating. The second term is for translational kinetic energy.

If one takes the bottom position as zero potential energy level then the potential energy at the top will be just

mgR

since the centre of mass dropped a distance equal to the radius below the top position.

What do you get the angular speed of the disc at the dashed position?
 
Last edited:
  • #4
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I got w = 2.349 rad/s
 
  • #5
andrevdh
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I (1.36 rad/s) do not get the same answer as you. Please show your calculations so that we can compare notes.
 
Last edited:
  • #6
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I got the answer; there was something wrong with my calculation.
Thank you for your help! :)
 

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