Three children of masses and their position on the merry go round
M1 = 22kg
M2 = 28kg
M3 = 33kg
They are all initially riding at the edge of the merry go round
Then, R1 = R2 = R3 = R = 1.7m
Mass of Merry go round is
M =105kg
Radius of Merry go round.
R = 1.7m
Angular velocity of Merry go round
ωi = 22 rpm
If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf
Using conservation of angular momentum
Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round Then,
L(initial) = L(final)
Ii•ωi = If•ωf
So we need to find the initial and final moment of inertia
NOTE: merry go round is treated as a solid disk then I= ½MR²
I(initial)=½MR²+M1•R²+M2•R²+M3•R²
I(initial) = ½MR² + R²(M1 + M2 + M3)
I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)
I(initial) = 151.725 + 1.7²(83)
I(initial) = 391.595 kgm²
Final moment of inertial when R2 =0
I(final)=½MR²+M1•R²+M2•R2²+M3•R²
Since R2 = 0
I(final) = ½MR²+ M1•R² + M3•R²
I(final) = ½MR² + (M1 + M3)• R²
I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²
I(final) = 151.725 + 158.95
I(final) = 310.675 kgm²
Now, applying the conservation of angular momentum
L(initial) = L(final)
Ii•ωi = If•ωf
391.595 × 22 = 310.675 × ωf
Then,
ωf = 391.595 × 22 / 310.675
ωf = 27.73 rpm
Answer: So, the final angular momentum is 27.73 revolution per minute
