The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.
What is the moment of inertia of the propeller?
How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.

Solution :

Given :

Length of the propeller rods, L =0.30 m

Mass of each, M = 0.040 kg

Moment of inertia of one propeller rod is given by  

[tex]$I=\frac{1}{3}\times M \times L^2$[/tex]

Therefore, total moment of inertia is

[tex]$I=3 \times \frac{1}{3}\times M \times L^2$[/tex]

[tex]$I=M\times L^2$[/tex]

[tex]$I=0.04\times (0.3)^2$[/tex]

  [tex]$0.0036 \ kg \ m^2$[/tex]

Now energy required is given by

[tex]$E=\frac{1}{2}\times I \times \omega^2 $[/tex]

where, angular speed, ω = 5800 rpm

[tex]$\omega = 5800 \times \frac{2 \pi}{60} $[/tex]

 = 607.4 rad/s

Therefore energy,

[tex]$E=\frac{1}{2}\times 0.0036 \times (607.4)^2 $[/tex]

   = 664.1 J

The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J

Given that the mass of the propellers is m = 0.040kg,

and their length is L = 0.30m

The moment of inertia of a rod with the rotation axis at one end is given by :

[tex]I = \frac{1}{3}m L^2[/tex]

so for 3 propellers:

[tex]I=3\times\frac{1}{3}\times(0.04)\times(0.3)^2[/tex]

I = 0.04 × 0.09

I = 0.0036 kgm²

Now, the frequency is given f = 5800 rpm

so anguar speed, ω = 5800×(2π/60)

ω = 607 rad/s

Energy required:

E = ¹/₂Iω²

E = 0.5 × 0.0036 × (607)² J

E = 663.21 J

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