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A 27 kg disk with a radius of 1.3m is spinning at an angular speed of 15 rad/s. What is the rotational kinetic energy of the disk?

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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

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