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

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

(b) 480 Nm

Step-by-step 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

Answer 2

Step-by-step 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 = (w_(30) - w_(0))/(\Delta t) = (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 = (ML^(2) )/(12) =  ((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`