Question

g A four bladed propeller on a cargo aircraft has a moment of inertia of 40 kg m2. If the prop goes from rest to 400 rpm in 14 sec, findA) the torque required andB) the number of revolutions turned as the prop achieves operating speed.

Answer 1

a. T = 119.68 N.m

b. r = 140 rev

Step-by-step explanation:

first we know that:

∑T = Iα

where ∑T is the sumatory of the torques, I is the moment of inertia and α is the angular aceleration.

so, if the prop goes from rest to 400 rpm in 14 seconds we can find the α of the system:

1. change the 400 rpm to radians as:

W = 400*2π/60

W = 41.888 rad /s

2. Then, using the next equation, we find the α as:

w = αt

solving for α

α =
`(w)/(t)`

note: t is the time, so:

α =
`(41.888)/(14)`

α = 2.992 rad/s^2

Now using the first equation, we get:

T = Iα

T = 40(2.992)

T = 119.68 N.m

On the other hand, for know the the number of revolutions turned as the prop achieves operating speed, we use the following equation:

θ = wt +
`(1)/(2)`α
`t^2`

Where w = 41.888 rad /s, α = 2.992 rad/s^2, t is the time and θ give as the number of radians that the prop made in the fisrt 14 seconds, so:

θ = (41.888)(14)+
`(1)/(2)`(2.992)(14
`)^2`

θ = 879.648 rad

and that divided by 2π give us the number of revolutions r, so:

r = 140 rev