Work and Rotational Kinetic Energy

  • Thread starter G-reg
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  • #1
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Homework Statement


A 38.0 kg wheel, essentially a thin hoop with radius 1.13 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
(a) How much work must be done to stop it?


Homework Equations


change in KE = (1/2)Iwi^2 - (1/2)Iwf^2
I = (1/2)mr^2


The Attempt at a Solution


(1/2)[(1/2)(38)(1.13)^2](28pi/3) = 355.686 kg(m)^2/s

I don't see where the time comes into play though..can anyone help?
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
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(1/2)[(1/2)(38)(1.13)^2](28pi/3)
It should be
(1/2)[(1/2)(38)(1.13)^2](28pi/3)^2

To find the work done, time is not necessary.
 
  • #3
44
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Oh ok, I get it now!
Thank you so much!
 

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