Pivoting Bar and Kinetic Energy

[DanielB]

Homework Statement

A thin, uniform 12.0kg bar that is 2.00m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds.

What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energy of all these segments.)

Homework Equations

K = 1/2 integral of v^2 dm
dm = (M/L)*dx

The Attempt at a Solution

I was confused on how to integrate this particular integral as velocity is not a constant and is clearly not in terms of dm. I attempted to integrate in terms of dx rewriting the integral as:

K= 1/2 * (2Pi/t)^2 * M/L integral x^2 dx

I don't think that is the correct direction but I am generally confused with the integration.
[EDIT]: Template was incorrect

 

[gneill]

I don't see where your velocity takes into account the 5.00 revolutions it does in t=3.00 seconds.