Question

9.38

An airplane propeller is 1.86 mm in length (from tip to tip) with mass 116 kgkg and is rotating at 2900 rpmrpm (rev/minrev/min) about an axis through its center. You can model the propeller as a slender rod.

Part A

What is its rotational kinetic energy?

Part 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 ?

Answer 1

Part A. Rotational kinetic energy is given by:

KE = (1/2)*I*w^2

I = moment of inertia of propeller = (1/12)*m*L^2

w = angular speed of propeller = 2900 rpm = 2900*2*pi/60 = 303.7 rad/sec

m = mass of propeller = 116 kg

L = length of propeller = 1.86 m

So,

KE = (1/2)*(1/12)*m*L^2*w^2

KE = (1/24)*116*1.86^2*303.7^2

KE = 1.54*10^6 J

Part B.

Given that mass of propeller is reduced to 75.0% of original mass, So

m1 = 0.75*m

Now kinetic energy and size needs to remain same, So

KE1 = KE

(1/24)*m1*L^2*w1^2 = (1/24)*m*L^2*w^2

w1 = w*sqrt (m/m1)

w1 = 2900*sqrt (m/(0.75*m))

w1 = 2900*sqrt (4/3)

w1 = 3348.63 rpm

w1 = 3350 rpm = new angular velocity required

Let me know if you've any query.