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Linear Kinetic Energy, Angular Kinetic Energy

In figure 5.32a)
Use the expressions K = (1/2)Iw^2 and K = (1/2)mv^2 to caluclate the kinetic energy of the sled treated as a single particle) how do these two results compare.


The figure is something like this:


A sled with mass 25.0 kg rests on a horizontal sheet of frictionless ice, it is attached by a 5.00m rpore to a post set in the ice in the center of this circle with radius 5.00m. Once given a push the sled revolves uniformly in a circle around the post.
If the sled makes 5 complete revolutions every minute.


The first part seems simple:

K = (1/2)(25.0kg)(2.62m/s)^2
k = 85.8 J

I'm not really sure if I should get a different answer for K = (1/2)Iw^2, where w = omega (angular velcoity).


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BrianGEFF719
Asked:
BrianGEFF719
2 Solutions
 
ozoCommented:
w = 5*2pi/60s
I = 25.0 kg*(5.00m)^2
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aburrCommented:
"I'm not really sure if I should get a different answer for K = (1/2)Iw^2"

You should get the same answer. KE is KE. Your system does not change just because you use angular measures instead of linear measures. You should get the same answer both ways.
And you do, using OZO's formulas.
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