Find the mass of the merry go round: conservation of angular momentum?

[n.hirsch1]

Homework Statement

A child exerts a tangential 41.6 N force on the rim of a disk-shaped merry-go-round with a radius of 2.40 m.
If the merry-go-round starts at rest and acquires an angular speed of 0.0850 rev/s in 3.50 s, what is its mass?

Homework Equations

torque = r * F
(I + mr^2) ω / t = torque
I of a solid disk = 1/2 mr^2

The Attempt at a Solution

I found the tangential torque to be 99.84 N/m, and set the momentum equation to it, plugging in the moment of inertia. I got 713.7, and the answer is 227 kg. What am I doing wrong?

 

[Nabeshin]

Two things:
i) Note that the angular speed is given in revolutions/second, not rad/s.

ii) In your second equation, what are you assuming about the placement of the child? Does this assumption make sense?

 

[n.hirsch1]

Once I convert the angular velocity, I can get the correct answer if I multiply it by 2:
2 * [(torque * t) / (angular velocity)] = m*r^2
Why does it work this way and not the other way?

 

[Nabeshin]

The only moment of inertia in question here is just that of the merry go round (i.e 1/2mr^2) with m as the mass of the merry go ground. I think the way you were doing it, you put an extra object of mass m on the rim (so you had an extra +mr^2 term)!