Airplane Propeller: Rotational Kinetic Energy & Angular Speed

[azila]

Homework Statement

An airplane propeller is 1.85 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2300 rpm (rev/min) about an axis through its center. YOu can model the propeller as a slender rod.

What is its rotational kinetic energy?

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 enrgy. What would its angular speed have to be, in rpm?

Homework Equations

I = 1/12ML^2
after substituting, K = 1/2Iomega^2

The Attempt at a Solution

Ok, for the first part, I solved for inertia and I got inertia as 165.99. THen I plugged that into the kinetic equation and with the velocity is rad/sec which was 240.86, and got the kinetic energy to be 4.81 x 10^6, which is not the answer.

For the second part, I first found the mass without 75% of it to be 24.25 kg. So, I solved for Inertia again and got 6.92. Used that value in the kinetic equation to solve for omega and then converted that to rpm to get 112592 rpm which looks really weird and is not the answer.

Any help?? Thanks in advance...

 

[Doc Al]

Ok, for the first part, I solved for inertia and I got inertia as 165.99.

Redo this calculation.

 

[azila]

thanks, I got it..