OK this one has me totally stumped. My teacher attempted to explain it in class and it made me even more confused.

Given:A child of mass m stands at rest near the rim of a stationary merry-go-round of radius R and moment of inertia I. The child now begins to walk around the circumference of the merry-go-round with a tangential speed v with respect to the merry-go-round's surface.
Questions:What is the child's speed with respect to the ground?
Express your answer in terms of some or all of the variables m, R, I, and v.
Next it asks me to check the result in the limits I to 0.

I am LOST! Help?

Thanks.

Use conservation of angular mometum. Let \omega_m = rotational velocity of the merry-go-round, and \omega_c be the rotational velocity of the child. Then:


I \omega_m \ + \ mR^2 \omega_c = 0


The velocity of the child relative to the merry-go-round is v, so:


v = R(\omega _c - \omega_m)


Rearrange, and shove back into the first equation:


\omega_m = \omega _c - v/R \\
I (\omega _c - \frac v R) + mR^2 \omega_c = 0 \\
\omega_c = \frac {Iv} {R(I+mR^2) \\
\omega_c = \frac {v/R} {(1 + \frac {mR^2} I )}


And his speed with respect to the ground is R \omega_c :


R\omega_c = \frac {v} {(1 + \frac {mR^2} I )}