An airplane propeller is 2.88 m in length (from tip to tip) and has a mass of 10

Question

An airplane propeller is 2.88 m in length (from tip to tip) and has a mass of 107 kg . When the airplane's engine is first started, it applies a constant torque of 1930 Nm to the propeller, which starts from rest.

1-What is the angular acceleration of the propeller? Treat the propeller as a slender rod.

Part B

What is the propeller's angular speed after making 5.00 rev ?

Part C

How much work is done by the engine during the first 5.00 rev ?

Part D

What is the average power output of the engine during the first 5.00 rev ?

Part E

What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.00 rev?

Explanation / Answer

given that

m = 107 kg

L = 2.88 m

torque t = 1930 N*m

treat the propeller as a slender rod

moment of inertia of rod I = mL^2/12 = = 107*(2.88)^2 / 12

I = 73.95 kg*m^2

(a) we know that

torque T = I*alpha

alpha = T/l = 1930/73.95

angular acceleration alpha =26.09 rad/s^2

(b)

w = sqrt(2*alpha*theta)

theta = 2*pi*rev = 2*3.14*5 = 31.4 rad/rev

w = sqrt (2*26.09*31.4)

w = 40.77 rad/s

(c)

KE = 1/2*l*w^2 = 1/2*73.95*(40.77)^2

KE = 61019.09 J = 61.01 kJ

this is the amount of work done by the engine.

(d)

average power output P = work / time

time t = w/alpta = 40.77/26.09

t = 1.56 s

so P = 61.01/1.56 = 39.10 kW

(e)

instantaneous power output P = T*w

P = 1930*40.77 = 78686.1 W

P = 78.68 kW

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