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Answer:
the rotational kinetic energy of the disk is 5,133.375 J
Explanation:
Given;
mass of the disk, m = 27 kg
radius of the disk, r = 1.3 m
angular speed, ω = 15 rad/s
The rotational kinetic energy of the disk is calculated as;
[tex]K.E_{rot} = \frac{1}{2}I \omega^2\\\\ where;\\I \ is \ moment \ of \ inertia\\\\K.E_{rot} = \frac{1}{2} \times (mr^2) \times \omega ^2\\\\ K.E_{rot} = \frac{1}{2} \times (27\times 1.3^2) \times \ 15^2\\\\K.E_{rot} = 5,133.375 \ J[/tex]
Therefore, the rotational kinetic energy of the disk is 5,133.375 J