Propeller Rotation Problem - Please check

[JWDavid]

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
$$\Theta$$_f = $$\Theta$$_i + $$\omega$$t + $$\alpha$$t²/2
$$\omega$$_f = $$\omega$$_i + $$\alpha$$t
W = $$\Delta$$KE = I$$\omega$$²/2
P_avg = $$\Delta$$P/2 = $$\tau$$$$\omega$$/2
P_ins = $$\tau$$$$\omega$$

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) $$\Theta$$_f = $$\Theta$$_i + $$\omega$$t + $$\alpha$$t²/2
5*2$$\pi$$ = 0 + 0 + 46.2t²/2

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

$$\omega$$_f = $$\omega$$_i + $$\alpha$$t
= 0 + 46.2*1.166 = 53.9 rad/sec

c) W = $$\Delta$$KE = I$$\omega$$²/2
= 42.2*53.9/2 = 1.14kJ

d) P_avg = $$\Delta$$P/2 = $$\tau$$$$\omega$$/2 = 1950*53.9/2 = 52.6 kW

e) P_ins = $$\tau$$$$\omega$$ = 1950*53.9 = 10.5 kW/s

 

[alphysicist]

Hi JWDavid,

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
$$\Theta$$_f = $$\Theta$$_i + $$\omega$$t + $$\alpha$$t²/2
$$\omega$$_f = $$\omega$$_i + $$\alpha$$t
W = $$\Delta$$KE = I$$\omega$$²/2
P_avg = $$\Delta$$P/2 = $$\tau$$$$\omega$$/2
P_ins = $$\tau$$$$\omega$$

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) $$\Theta$$_f = $$\Theta$$_i + $$\omega$$t + $$\alpha$$t²/2
5*2$$\pi$$ = 0 + 0 + 46.2t²/2

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

$$\omega$$_f = $$\omega$$_i + $$\alpha$$t
= 0 + 46.2*1.166 = 53.9 rad/sec

c) W = $$\Delta$$KE = I$$\omega$$²/2
= 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) P_avg = $$\Delta$$P/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:

$$P_{\rm ave} = \frac{\mbox{work}}{\Delta t}$$

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) P_ins = $$\tau$$$$\omega$$ = 1950*53.9 = 10.5 kW/s

 

[JWDavid]

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.

 

[alphysicist]

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!