Merry-go-round angular speed physics

[beocom6000yello]

A merry-go-round of radius R=2.0 m has a moment of inertia I=250 kg m2, and is rotating at 10 rpm. A child whose mass is 25 kg jumps onto the edge of the merry-go-round, by jumping directly toward the center at 6.0 m/s. The new angular speed (in rpm) of the merry-go-round is approximately
A) 7.1
B) 8.5
C) 9.2
D) 10
E) 6.4

this is what i did: the moment of a thick disk is I = 1/2 MR^2... I THINK you could use that to find the mass of the merry-goat-round...but...if he jumps directly towards the center...does that have any impact on the rotational speed? I don't think it does.

i would think the answer would be 10 but my friend got a different solution...is there anyway that someone can help telling what they got?

 

[Doc Al]

A merry-go-round of radius R=2.0 m has a moment of inertia I=250 kg m2, and is rotating at 10 rpm. A child whose mass is 25 kg jumps onto the edge of the merry-go-round, by jumping directly toward the center at 6.0 m/s. The new angular speed (in rpm) of the merry-go-round is approximately
A) 7.1
B) 8.5
C) 9.2
D) 10
E) 6.4

this is what i did: the moment of a thick disk is I = 1/2 MR^2... I THINK you could use that to find the mass of the merry-goat-round...but...if he jumps directly towards the center...does that have any impact on the rotational speed? I don't think it does.

i would think the answer would be 10 but my friend got a different solution...is there anyway that someone can help telling what they got?

You don't need to find the mass of the merry-go-round. Consider angular momentum.

 

[Shooting Star]

i would think the answer would be 10 but my friend got a different solution...is there anyway that someone can help telling what they got?

It's been a few days, but I'm just curious to know whether you at all tried to find the right answer, after following the advice given by Doc Al, seeing that you labeled the question urgent.