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.

Answer 2

The time taken to reach the frequency of 2000rpm is 8.05s.

Angular acceleration:

If we calculate the angular acceleration we can calculate the time taken to reach any specific angular speed.

Given that the torque is τ = 500Nm

mass of the blade m = 40kg

length of the blade L =2.4m

Now, the moment of inertia of the blade I when the axis passes through one end is given by:

$I=\frac{1}{12}mL^2\\\\I=\frac{1}{12}40\times(2.4)^2\\\\I=19.2kgm^2$

$\tau=I\alpha\\where\;\alpha\;is\;angular\;acceleration\\\\500=19.2\alpha\\\\\alpha=26\;rad/s^2$

Now the given frequency is 2000rpm = 2000/60 rps

Then the angular speed:

$\omega=2\pi\frac{2000}{60}\;rad/s\\\\\omega=209.33\;rad/s$

So the time taken to reach this angular speed is:

$\omega=\alpha t\\\\209.33=26t\\\\t=8.05s$

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