Propeller Rotation Problem - Please check

AI Thread Summary
The discussion revolves around calculating various parameters related to an airplane propeller's rotation, including angular acceleration, angular speed, work done, average power, and instantaneous power. The propeller, with a length of 2.08 m and mass of 117 kg, experiences a constant torque of 1950 N m, leading to an angular acceleration of 46.2 rad/sec². After 5 revolutions, the propeller reaches an angular speed of 53.9 rad/sec, with the work done during this period calculated at 1.14 kJ. Average power output is discussed, with a correction noted regarding the formula used, emphasizing the importance of accurately calculating work over time. The conversation highlights the need for careful attention to detail in physics calculations to avoid losing points on assignments.
JWDavid
Messages
23
Reaction score
0
Thanks in advance - it went smoothly so I'm hoping I'm right but... usually that's when I've made a big mistake.

Homework Statement


An airplane propeller is 2.08 m (tip to tip) and has a mass of 117 kg. When the engine is first started it applies a constant torque of 1950 N m to the propeller which starts from rest.
a) What is the angular acceleration of the propeller? (model it on a slender rod).
b) What is the propeller's angular speed after 5.00 revolutions?
c) How much work is done during the first 5.00 revolutions by the engine?
d) What is the average power during the first 5.00 revolutions?
e) What is the instantaneous power output of the engine the instant it has completed 5.00 revolutions?


Homework Equations


I think this is all of them
I = mL2/12
\Thetaf = \Thetai + \omegat + \alphat2/2
\omegaf = \omegai + \alphat
W = \DeltaKE = I\omega2/2
Pavg = \DeltaP/2 = \tau\omega/2
Pins = \tau\omega


The Attempt at a Solution


work that got me to the answers below.
a) 42.2 b) 53.9 rad/sec2 c) 1.14 kJ d) 52.6 kW e) 10.5 kW


I = mL2/12 = 117*2.082/12 = 42.2

a) \tau = 1950 N m = I\alpha
\alpha = 1950/42.2 = 46.2 rad/sec2

b) \Thetaf = \Thetai + \omegat + \alphat2/2
5*2\pi = 0 + 0 + 46.2t2/2

t = squareroot (20\pi/46.2) = 1.166 seconds

\omegaf = \omegai + \alphat
= 0 + 46.2*1.166 = 53.9 rad/sec

c) W = \DeltaKE = I\omega2/2
= 42.2*53.9/2 = 1.14kJ

d) Pavg = \DeltaP/2 = \tau\omega/2 = 1950*53.9/2 = 52.6 kW

e) Pins = \tau\omega = 1950*53.9 = 10.5 kW/s
 
Physics news on Phys.org
Hi JWDavid,

JWDavid said:
Thanks in advance - it went smoothly so I'm hoping I'm right but... usually that's when I've made a big mistake.

Homework Statement


An airplane propeller is 2.08 m (tip to tip) and has a mass of 117 kg. When the engine is first started it applies a constant torque of 1950 N m to the propeller which starts from rest.
a) What is the angular acceleration of the propeller? (model it on a slender rod).
b) What is the propeller's angular speed after 5.00 revolutions?
c) How much work is done during the first 5.00 revolutions by the engine?
d) What is the average power during the first 5.00 revolutions?
e) What is the instantaneous power output of the engine the instant it has completed 5.00 revolutions?


Homework Equations


I think this is all of them
I = mL2/12
\Thetaf = \Thetai + \omegat + \alphat2/2
\omegaf = \omegai + \alphat
W = \DeltaKE = I\omega2/2
Pavg = \DeltaP/2 = \tau\omega/2
Pins = \tau\omega


The Attempt at a Solution


work that got me to the answers below.
a) 42.2

I just wanted to point out here that you seemed to have typed in the wrong value; 42.2 is the moment of inertia; but I see you have the correct answer for the angular acceleration below.

b) 53.9 rad/sec2 c) 1.14 kJ d) 52.6 kW e) 10.5 kW


I = mL2/12 = 117*2.082/12 = 42.2

a) \tau = 1950 N m = I\alpha
\alpha = 1950/42.2 = 46.2 rad/sec2

b) \Thetaf = \Thetai + \omegat + \alphat2/2
5*2\pi = 0 + 0 + 46.2t2/2

t = squareroot (20\pi/46.2) = 1.166 seconds

\omegaf = \omegai + \alphat
= 0 + 46.2*1.166 = 53.9 rad/sec

c) W = \DeltaKE = I\omega2/2
= 42.2*53.9/2 = 1.14kJ

I don't think this is correct; it looks like you neglected to square the angular speed.

d) Pavg = \DeltaP/2 = \tau\omega/2 = 1950*53.9/2 = 52.6 kW

This also looks incorrect to me. The average power is not just the work done divided by 2; the formula is something like:

<br /> P_{\rm ave} = \frac{\mbox{work}}{\Delta t}<br />

Your book may have it written differently, of course. But the point is its the fraction (work done in some time interval)/(time interval).

e) Pins = \tau\omega = 1950*53.9 = 10.5 kW/s
 
thank you,
You are correct I did forget to square the angular speed.
But on the average power - it does work both ways. If you note I wasn't taking the work done I was taking the change in Power/2 {(torque*angular speed)-0}/2.

thanks again for your observations, I would probably have lost points on this had I not asked for help.
 
Sorry about part d; I was so focused on the dividing by 2 that I did not actually read what you wrote down, and just assumed you did the (work)/(time) route!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top