Answer :
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
Energy required for propeller:
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
Learn more about moment of inertia:
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