Rotating Bar- physics?

Question

A thin, uniform 12.0-kg bar that is 2.00 m 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.)

Answer 1

the angular speed of the bar is
w=5/(3*2*pi) rdians per second

The kinetic energy of the element dl
=.5*m*w^2*l^2
where l is the distance fro mthe pivot point to the derivative element

the total mass is 12kg
the mass of the derivative element is
12/dl
so integrate
dke/dl=.5*12/dl*(5/(3*2*pi))^2*l^2 from 0 to 2m. The boundry condition that ke=0 at l=0

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