To solve the exercise it is necessary to apply the concepts related to Kinetic Energy and its respective definition. Angular kinetic energy means that,
[tex]KE = \frac{1}{2}I\omega^2[/tex]
Where
I = Inertia moment
\omega = Angular velocity
On the other hand the inertia of a rod is given by
[tex]I = ml^2[/tex]
Where,
m = mass
l = length
ROD A: The data for Rod A is given as,
l = 0.88m
m = 0.76Kg
[tex]\omega = 5.2 rad/s[/tex]
The inertia would be
[tex]I = ml^2[/tex]
[tex]I = 0.76Kg*0.88^2[/tex]
[tex]I = 0.588544 kgm^2[/tex]
Therefore the Kinetic Energy would be
[tex]KE = \frac{1}{2} I\omega^2[/tex]
[tex]KE = \frac{1}{2} (0.588544)(5.2)^2[/tex]
[tex]KE= 4.683J[/tex]
ROD B: The data for Rod B is given as
[tex]l = 0.88m[/tex]
[tex]m = 0.76Kg[/tex]
[tex]\omega = 5.2 rad/s[/tex]
The inertia would be
[tex]I = \frac{1}{3}ml^2 \rightarrow[/tex] Note that the moment of inertia is not about its reference system, but about its end.
[tex]I = \frac{1}{3}0.76Kg*0.88^2[/tex]
[tex]I = 0.19618 kgm^2[/tex]
Therefore the Kinetic Energy would be
[tex]KE = \frac{1}{2} I\omega^2[/tex]
[tex]KE = \frac{1}{2} (0.19618)(5.2)^2[/tex]
[tex]KE= 2.6523J[/tex]
