Question

9.38 An airplane propeller is 1.86 mm in length (from tip to tip) with mass 116...

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 ?

Homework Answers

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.

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