The rotational speed of a threeblade propeller increases fro

The rotational speed of a three-blade propeller increases from 300 rpm to 2500 rpm in three seconds. Each blade weighs 30 lb and is 4-ft long. Assuming that the mass per unit length of the blade is uniform and that the angular acceleration is constant, determine the required torque.

Solution

Mass of each blade, M = 30 lb = 13.61 kg

length of each blade, L=4 ft = 1.22 m

We can consider each blade as a bar of mass M, length L and rotating about any corner.

So, the moment of inertia of a blade is = 1/3 ML² = 1/3 x 13.61x1.22² = 6.75 kg-m²

So, total moment of inertia of the entire system = 3xmoment of inertia of each blade,I = 3x6.75 = 20.26 kg-m²

Initial angular velocity, W1 = 300 rpm = (2x3.14x300)/60 rad/s = 31.4 rad/s

Final angular velocity, W2 = 2500 rpm = (2x3.14x2500)/60 rad/s = 261.67 rad/s

So, angular acceleration,A = (W2-W1)/t = (261.67-31.4)/3 = 76.76 rad/s²

Torque required = IA = 20.26x76.76 Nm = 1555.1 Nm