# Rotational Kinetic Energy of an airplane propeller

1. Dec 8, 2008

### cantgetno

1. The problem statement, all variables and given/known data
An airplane propeller is 2.02 m in length (from tip to tip) with mass 127 kg and is rotating at 2300 rpm about an axis through its center. You can model the propeller as a slender rod.
a)What is its rotational kinetic energy?
b)Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0\% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

2. Relevant equations
K=0.5 (mr^2) w^2

3. The attempt at a solution
a)
2300 RPM = 38.33 RPS = 76.66 radians/s

KE=0.5 (127) (1.01)^2 (240)^2 = 3731117.76 J

b)
new mass = 95.25kg

KE = 0.5(95.25)(1.01)^2 (w)^2

2 KE /(95.25)(1.01)^2) = (w)^2

[2 KE /(95.25)(1.01)^2)]^0.5 = w

but im sure my original KE is wrong so..

thanks for the help

Last edited: Dec 8, 2008
2. Dec 8, 2008

3. Dec 8, 2008

### cantgetno

KE=0.5 (1/12)M L^2 (w)^2 ?
giving

1/24 x 127 x 2.02^2 x 240^2 = 1243705.9
?

4. Dec 8, 2008

### LowlyPion

I get 241 for ω.

I didn't calculate the rest out, but that looks like the right method now.

5. Dec 9, 2008

### cantgetno

bingo
thanks lots and lots