A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.
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Answer:
Explanation:
a )
Each blade resembles a rod with its axis positioned near one end.
The moment of inertia for one blade is:
= 1/3 x m l²
where m stands for the mass of the blade
l represents the length of each blade.
Total moment of inertia for 3 blades is:
= 3 x x m l²
ml²
2 )
Details provided include:
m = 5500 kg
l = 45 m
Substituting these values produces:
moment of inertia of one blade:
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia for 3 blades:
= 3 x 37.125 x 10⁵ kg.m²
= 111.375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I denotes the moment of inertia of the turbine
ω symbolizes angular velocity
ω = 2π f
f indicates the rotational frequency of the blades
d )
We have I = 111.375 x 10⁵ kg.m² (Calculated)
f = 11 rpm (revolutions per minute)
= 11 / 60 revolutions per second
ω = 2π f
= 2π x 11 / 60 rad / s
Calculating angular momentum yields
= I x ω
111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹.