Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
The pilot might be correct (I think), because, if the gravity of the planet is strong, then the planet’s gravity will pull the spaceship into its orbit, so the engines don’t need to be on for the ship to get pushed toward the planet.
You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3.
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= 4.5 g/cm3
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