Question

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?

Answer 1

the rotational kinetic energy of the disk is 5,133.375 J

Step-by-step 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;

`K.E_(rot) = (1)/(2)I \omega^2\\\\ where;\\I \ is \ moment \ of \ inertia\\\\K.E_(rot) = (1)/(2) * (mr^2) * \omega ^2\\\\ K.E_(rot) = (1)/(2) * (27* 1.3^2) * \ 15^2\\\\K.E_(rot) = 5,133.375 \ J`

Therefore, the rotational kinetic energy of the disk is 5,133.375 J