Meri go round: Angular momentum

[kalupahana]

Homework Statement

A meri go round is rotating it vertical axis through center with a diameter 4.5m. It's moment of inertia is 1750 kg m² and angular velocity is 0.70 rad/s. 4 men with mass 65 kg instantly enter the the edge of the meri go round. Calculate the angular velocity of meri go round after that instant.

Homework Equations

Conservation of angular momentum
I₁ω₁ = I₂ω₂

I = ma²

The Attempt at a Solution

I₁ = 1750 kg m²
ω₁ = 0.70 rad/s

The mass of meri go round
1750 = ma²
1750 = m(4.5/2)²
(1750 x 4)/ (4.5)² = m

The mass of meri go round after 4 men get in

m = (1750 x 4)/ (4.5)² + 65 x 4
m = [7000 + (260 x (4.5)²]/(4.5)²
m = 12265/(4.5)²

Then
I₂ = 12265 kg m²

Applying conservation of angular momentum

1750 x 0.70 = 12265 ω₂
ω₂ = 1.099 rad/sBut the answer is given as 0.004 rad/s.
Please help me to reach to this answer

 

[cantgetaname]

The mistake you've made is 'I=ma^2' which is only valid for a point particle.
The ride could be any complex shape you can't say anything about.

So moment of inertia after 4 men go on is: 1750 + 4*(65)*((4.5/2)^2)

the men are assumed to be point masses at a distance a/2 so their moments of inertia is m(a/2)^2 each.
This can be just added to the moment of inertia of the ride, as the axis of rotation stays same throughout the problem.