someone spinns with an angular velocity of 4.55rad/s with arms held out. his rotational inertia is 2.7kgm^2. whn the he pulls his arms in his angular velocity increases to 7.60rad/s. what is his rotary ek?

To find the rotational kinetic energy (Ek) of the person, we can use the formula:

Ek = (1/2) * I * ω^2

Where:
Ek is the rotational kinetic energy
I is the rotational inertia
ω is the angular velocity

In the given scenario, the initial angular velocity (ω1) is 4.55 rad/s, and the initial rotational inertia (I1) is 2.7 kg·m^2. The final angular velocity (ω2) is 7.60 rad/s.

1. Calculate the initial rotational kinetic energy (Ek1):
Ek1 = (1/2) * I1 * ω1^2

Substitute the given values:
Ek1 = (1/2) * 2.7 kg·m^2 * (4.55 rad/s)^2

Calculate the value of Ek1.

2. Calculate the final rotational kinetic energy (Ek2):
Ek2 = (1/2) * I2 * ω2^2

When the person pulls their arms in, the rotational inertia decreases (I2 < I1). However, the question does not provide the value of I2. Therefore, we can't determine the final rotational kinetic energy (Ek2) without this information.

Hence, we can calculate the initial rotational kinetic energy (Ek1) using the given values, but we need the value of the final rotational inertia to determine the final rotational kinetic energy (Ek2).