Rotational Kinetic Energy of an airplane propeller

[cantgetno]

Homework Statement

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?

Homework Equations

K=0.5 (mr^2) w^2
(m***, radius, w angular speed)

The Attempt at a Solution

a)
2300 RPM = 38.33 RPS = 76.66 radians/s
2300RPM=240 rad/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 = wbut I am sure my original KE is wrong so..

thanks for the help

 

[LowlyPion]

What is the moment of inertia for a rotating slender rod?

That would look to be where you have been led astray.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

[cantgetno]

2300RPM=240 rad/s

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

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

 

[LowlyPion]

I get 241 for ω.

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

 

[cantgetno]

bingo
thanks lots and lots