Energy in Rotational Motion?

Question

An airplane propeller is 2.08 m in length (from tip to tip) and has a mass of 117 kg. When the airplane's engine is first started, it applies a constant torque of 1950 N*m to the propeller, which starts from rest.

Part A: What is the angular acceleration of the propeller? Treat the propeller as a slender rod (I =(1/3)ML^2).

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?

Answer 1

First, the I=1/3 M*L^2 is for a slender rod attached at one end, yet you describe the propeller as "tip-to-tip". It seems that the propeller should be attached to the shaft at the center making I=1/12*M*L^2. I will use your inertia model as if it is one blade and the length is shaft to tip at 2.08 m. I will show you a method using variables that can be calculated from the information given

T=I*α
where α is the angular acceleration, I is the moment of inertia, and T is the torque

so α=T/I
w=w0+α*t
Where w is the angular velocity
since w0 = 0
w=T*t/I

The angle traversed is th, in this case, since the propeller started from rest, th0=0

th=T*t^2/(2*I)

Part 1
5 revolutions is 10*pi radians

t=sqrt(20*pi*I/T)

and w=sqrt(T*20*pi/I)


Part 2

at the time of 5 revolutions the work can be expressed in two ways

Work=T*th
or
T*10*pi

and by looking at the rotational kinetic energy

.5*I*w^2

from above
w^2=T*20*pi/I

so the energy is
.5*I*T*20*pi/I
or
T*10*pi same answer, which is good

Part 3

Power is T*w

Average power is T*w(average)

The instantaneous power of the propeller at 5 revolutions is
T*sqrt(T*20*pi/I)

The average power is found by computing the average angular velocity over the time period and since torque was constant, multiply average w by torque.

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