Question

A propeller consists of two blades each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate.

a) What is the angular momentum of the propeller at ????=10s;????=20s?
b) What is the torque on the propeller?

Answer 1

Answer:

Explanation:

a)

$l$ = length of each blade = 3 m

$m$ = mass of each blade = 120 kg

$L$ = length of the rod

$M$ = mass of the rod

Length of the rod is given as

$L = 2 l = 2 (3) = 6 m$

$w_{0}$ = Angular velocity at t = 0, = 0 rad/s

$w_{30}$ = Angular velocity at t = 30, = 1200 rpm = 125.66  rad/s

$\Delta t$ = time interval = 30 - 0 = 30 s

angular acceleration is given as

$\alpha = \frac{w_{30} - w_{0}}{\Delta t} = \frac{125.66 - 0}{30} = 4.2 rad/s^{2}$

Angular velocity at t = 10 s is given as

$w_{10} = w_{0} + \alpha t\\w_{10} = 0 + (4.2) (10)\\w_{10} = 42 rad/s$

Angular velocity at t = 20 s is given as

$w_{20} = w_{0} + \alpha t\\w_{20} = 0 + (4.2) (20)\\w_{10} = 84 rad/s$

Mass of the rod is given as

$M = 2m = 2(120) = 240 kg$

Moment of inertia of the propeller is given as

$I = \frac{ML^{2} }{12} =  \frac{(240)(6)^{2} }{12} = 720 kgm^{2}$

Angular momentum at t = 10 s is given as

$L_{10} = I w_{10} = (720) (42) = 30240 kgm^{2}/s$

Angular momentum at t = 20 s is given as

$L_{20} = I w_{20} = (720) (84) = 60480 kgm^{2}/s$

b)

Torque on propeller is given as

$\tau = I \alpha \\\tau = (720) (4.2)\\\tau = 3024 Nm$

Answer 2

Answer:

(a) 4800 kg m^2/s^2 , 9600 kg m^2/s^2

(b) 480 Nm

Explanation:

length of each blade = 3 m

total length, L = 2 x 3 = 6 m

mass of each blade = 120 kg

total mass, m = 2 x 120 = 240 kg

initial angular velocity, ωo = 0 rad/ s

final angular velocity, ω = 1200 rpm = 1200 / 60 = 20 rps = 125.6 rad/s

t = 30 s

Let the angular acceleration is α.

α = (ω - ωo)/t = 120 / 30 = 4 rad/s^2

(a) moment of inertia. I = mL^2 / 12

I = 240 x 6 / 12 = 120 kg m^2

Let ω' be the angular velocity at the end of 10 s

use first equation of motion

ω' = 0 + αt

ω' = 4 x 10 = 40 rad/s

Angular momentum at t = 10 s

L = I x ω = 120 x 40 = 4800 kg m^2/s^2

Let ω'' be the angular velocity at the end of 20 s

use first equation of motion

ω'' = 0 + αt

ω'' = 4 x 20 = 80 rad/s

Angular momentum at t = 20 s

L = I x ω = 120 x 80 = 9600 kg m^2/s^2

(b) Torque, τ = I x α

τ = 120 x 4 = 480 Nm