# 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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Commented:
w = 5*2pi/60s
I = 25.0 kg*(5.00m)^2

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