Ask Me Help Desk - Pre-AP Physics.Rotational mechanics.angular momentum

Quote:

Originally Posted by tesia

I'm totally lost. We were given the equation that angular momentum = rotational inertia * angular velocity. I get that some what. Then I look at my homework. First of all, my teacher gives us notes she makes. But our homework isn't through her, she doesn't even always look at the questions we get. We get and enter our homework online from a state college. so the homework and notes don't always match up right. The notes will make it seem easy but then the college online stuff is extremely hard. My question is....

A playground merry-go-round of radius
1.83 m has a moment of inertia 251 kg · m2
and is rotating at 13.9 rev/min about a fric-
tionless vertical axle. Facing the axle, a
30.4 kg child hops onto the merry-go-round,
and sits down on the edge.
What is the new angular speed of the
merry-go-round?
Answer in units of rev/min.

I'm already lost. Mass is not in my equation and neither is radius...??? I wrote down my numbers and labeled what they are, then was stumped.

Let's see. Let me recollect my physics. Let the states before and after the kid hops on the merry go round as, say, i and f, respectively. Representing initially and finally.
The conservation of angular momentum is
$L_{i}=I_{i}{\omega}_{i}=L_{f}=L_{f}{\omega}_{f}$

Now, solve for ${\omega}_{f}$ , the final angular speed.

${\omega}_{f}=\frac{I_{i}{\omega}_{i}}{I_{i}+ml^{2} }$

$=\frac{(251 \;\ kg\cdot m^{2})(87.34 \;\ rad/min)}{(251 \;\ kg\cdot m^{2})+(30.4 \;\ kg)(1.83)^{2}}=62.134 \;\ rad/min$

Now, I will let you convert back to rev.min.