Answer :
Answer:
Angular velocity of merry go round is 0.64 [tex]\frac{rad}{s}[/tex].
Explanation:
Given :
Radius of merry go round [tex]R = \frac{D}{2} = 2.2[/tex] m
Angular velocity of merry go round [tex]\omega = 0.8 \frac{rad}{s}[/tex]
Momentum of inertia [tex]I = 1950[/tex] [tex]Kg m^{2}[/tex]
Mass of each people [tex]M = 50[/tex] Kg
From the formula of angular momentum conservation,
[tex]L = I\omega[/tex]
So initial angular momentum,
[tex]L_{i} = 1950 \times 0.8 = 1560[/tex]
Now we calculate the MOI after people jump on merry go round,
[tex]I_{p} =\frac{1}{2} MR^{2}[/tex]
But there are four people so we have to multiply with 4 and add merry go round MOI,
[tex]I _{tot} = 1950 + 2\times 50 \times (2.2) ^{2}[/tex]
So total MOI after people jump on is,
[tex]I _{tot} = 2434[/tex] [tex]Kg m^{2}[/tex]
From conservation of angular momentum,
[tex]L_{i} = L_{f}[/tex]
[tex]1560 = 2434 \times \omega_{f}[/tex]
So angular velocity is,
[tex]\omega _{f} = \frac{1560}{2434} = 0.64[/tex] [tex]\frac{rad}{s}[/tex]