Ans.

Here the mass of the uniform rod M=5 lbs,
length of the rod is l=2 ft.
The rod is rotating about one end with frequency \( n=\displaystyle{\frac{60}{60}} \) =1 rev./sec.
The angular velocity is \( \omega=2n\pi \)

We know that the moment of inertia of the rod about on end is

\( I=\displaystyle{\frac{1}{3}Ml^2} \)
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\( or,\ I=\frac{1}{3}\times{5}\times{4}=6.66\ lb\cdot{ft}^2 \)

The kinetic energy K.E.= \( \frac{1}{2}I{\omega}^2 \), [ To know in details ( CLICK HERE ) ]

or, \( K.E.=\frac{1}{2}I{(2n\pi)}^2 \)

or, \( K.E.= 2I{\pi}^2n^2 \)

or, K.E.= \( 2\times{6.6}\times{\frac{22}{7}}^2\times{1}\\=4.07\ ft\cdot{lb} \)