Question

Calculate the rotational energy of a segment, given mass of the segment is 2.2 kg, moment of inertia is 0.57 kg-m2, and angular velocity is 25 rad/s.

Answer 1

The rotational energy of the segment is 179.06 Joules.

Step-by-step explanation:

In order to calculate the rotational energy of a segment, we need to use the formula for rotational kinetic energy: KE_rot = 0.5 * I * w^2, where KE_rot is the rotational kinetic energy, I is the moment of inertia, and w is the angular velocity.

Given that the mass of the segment is 2.2 kg, the moment of inertia is 0.57 kg-m^2, and the angular velocity is 25 rad/s, we can substitute these values into the formula:

KE_rot = 0.5 * 0.57 kg-m^2 * (25 rad/s)^2 = 179.06 J

Therefore, the rotational energy of the segment is 179.06 Joules.

Answer 2

To solve this problem we will use the concepts given by the rotational kinetic energy which describes the energy in a body product of its moment of inertia and its angular velocity, mathematically this is given as,

`KE = (1)/(2) I\omega^2`

Where,

I = Moment of Inertia

`\omega`= Angular Velocity

Replacing with our values we have that,

`KE = (1)/(2) (0.57)(25)^2`

`KE = 178.125J T`

Therefore the rotational energy is 178.125J