Angular velocity of a merry go round after a child has jumped onto it

[Tina20]

Homework Statement

A merry-go-round is a common piece of playground equipment. A 2.85 m diameter merry-go-round with a mass of 212.0 kg is spinning at 22.5 rpm. John runs tangent to the merry-go-round at 5.76 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 33.7 kg. What is the merry-go-round's angular velocity, in rad/s, after John jumps on?

Homework Equations

Iiwi = Ifwf

where i = initial, f = final
I = moment of inertia
w = angular velocity

The Attempt at a Solution

Ijwj = Ifwf
Ijohn*wjohn = (Ijohn + Imgr)wf

I know that momentum of John = Iw = mvr, so

mvr = (Ijohn + Imgr)wf

Ijohn = mr^2 where r is the radius of the merry go round
Imgr --> I don;t know how to calculate the moment of inertia of the merry go round...is it mr^2 or do I assume that the merry go round is like a disk with its axis about the center being 1/2MR^2?

Then I need to solve for wf...I did that, but unfortunately I got the wrong answer. Is my equation set up incorrectly? Or it may be that my moment of inertia for the merry go round is incorrect.

 

[Delphi51]

Looks good. I would use the 1/2MR^2 since no better value is given.
A little over 3 for angular velocity?

 

[Tina20]

Yeah, that's what I got but the computer says its wrong :( Any ideas why?

 

[Delphi51]

Don't know. Might be worth comparing our answers.
Since the question is pretty casual about moment of inertia, I wonder if the person's mass is supposed to be simply added to the mass of the merry-go-round, assuming the whole thing is a uniform disk?

 

[Tina20]

Ok, so I tried adding John's weight and the weight of the merry-go-round and inputted that as the weight for the moment of inertia final. Below is my calculation. It is still wrong. I think I need to incorporate the initial momentum of the merry-go round and john's initial momentum as well?

Ijwj = (Imgr+j)(wf)
mvr = 1/2MR^2(wf)
(33.7)(5.76)(1.425) = 1/2(212+33.5)(1.425)^2(wf)
276.609 = 249.259wf
1.109 = wf

The above answer is wrong. Would this seem like an ideal equation to solve the question:

Iiwi = Ifwf
Ijwj + (Imgr)(wmgr) = (Imgr + j)wf
mvr + (1/2MR^2)w = 1/2(Mmgr + Mj)R^2 *wf

 

[Delphi51]

Iiwi = Ifwf
Ijwj + (Imgr)(wmgr) = (Imgr + j)wf

276.6 + 507 = (.5*M*r² + m*r²)wf
783.6 = (.5*212 + 33.7)r²*wf
783.6 = 139.7*1.425²*wf
2.76 = wf

The first time through I got 3.12 due to using .5*106 instead of .5*212; just one of those slips!

 

[Tina20]

Thank you so much, it worked!

Two questions left for me to solve today :)