Question

The engine in a small airplane is specified to have a torque of 500 N⋅m . This engine drives a 2.4- m -long, 40 kg single-blade propeller.On startup, how long does it take the propeller to reach 2000 rpm ?

Answer 1

Answer:

8.04 second

Explanation:

torque = 500 Nm

Length of blade, L = 2.4 m

mass of blade, m  40 kg

initial angular velocity, ωo = 0  rad/s

frequency, f = 2000 rpm = 2000 / 60 rps

final angular velocity, ωf = 2 x π x 2000 / 60 = 209.33 rad/s

Moment of inertia of the blade,

$I = \frac{1}{12}mL^{2}$

$I = \frac{1}{12}\times 40\times 2.4^{2}$

I = 19.2 kg m^2

Torque = Moment of inertia x angular acceleration

500 = 19.2 x α

where, α be the angular acceleration

α = 26.04 rad/s^2

Use first equation of motion

ω = ωo + αt

where t is the time taken by the propeller to reach 2000 rpm.

209.33 = 0 + 26.04 x t

t = 8.04 second

Thus, the time taken by the propeller is 8.04 second.