Merry-go-round physics problems

I missed these two questions on my last test and would like to know how to do them. (Final next week)

5 boys, each of mass 40kg, are riding on the edge of a merry-go-round (I=(1/2)MR^2) with an angular velocity of w. If they each walk to the center, the angular velocity increases by a factor of 1.4. What is the mass of the merry-go-round?

A girl of mass 30kg runs and jumps onto the edge of a merry-go-round of radius 2m. If the merry-go-round has a mass of 500kg, what is the slowest that she could run and still obtain an angular velocity of 0.2 rad/s after she jumps on?

Any help is appreciated, thanks.

•Find the initial angular momentum (Li = Iω) ..
Boys at rim treated as point masses,Ib = 40R² each
Li = {½MR² + 5*40R²}ω

•Find final ang. mom. (Lf)
Boys have zero mom. of inertia at centre (R = 0)
Lf = {½MR²}*1.4ω

•Equate Li = Lf (cons. of ang. mom.) .. R²,ω cancel out .. find M(kg)
(1000kg)


2)
•Calculate mom. of inertia for m.g.r, I(m) = ½MR².... M=500, R=2
•Calculate final ang. mom.(Lf) for (m.g.r + girl as point mass) = {I(m) + mR²}ω
•Calc. initial ang.mom.for girl(Lg) as she jumps on ...
ang.mom. = R*(linear mom.) = R.mv
•Equate Lg = Lf (cons. of ang. mom.) to find v (m/s)
(3.73m/s)

200 . R^2 = loss of MI when the boys go to the centre

(1/2 MR^2 + 200R^2) w = (1/2 MR^2 ) . 1.4 w

M + 400 = 1.4 M

M = 400 / 0.4 = 1000 kg

2) m . v . r angular momentum of girl

MI = 30 . 2^2 + 1/2 . 500 . 2^2 = 1120 kgm^2

Angular momentum wanted = 1120 . 0.2 = 224 = 30 . 2 . v

v = 3.73 m/s