An airplane propeller is 2.18 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2600 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy?

1.4*10^6 J

Explanation:

Given that

Length of the propeller, l = 2.18 m

Mass of the propeller, m = 97 kg

Speed of the propeller, w = 2600 rpm

The formula for finding rotational Kinetic energy, K is = ½Iw²

Where, I is the moment of Inertia, and is given as 1/12 * m * l²

I = 1/12 * 97 * 2.18²

I = 8.083 * 4.7524

I = 38.41 kgm²

w = 2600 rpm, converting to rad/s, we have

w = 2600 * 2π rad/s

w = 272.31 rad/s

Now, Kinetic Energy, K is

K = ½Iw²

K = ½ * 38.41 * 272.31²

K = 19.205 * 74152.7361

K = 1424103.3 J

K = 1.4 MJ or 1.4*10^6 J

Thus, the rotational Kinetic Energy is 1.4*10^6 J

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation