Why Does the Merry-Go-Round Spin at 0.276 rad/sec After the Child Jumps On?

[blue5t1053]

Problem:
In a plyground there is a small merry-go-round of radius 3.91 m and rotational inertia 5.68e+03 kg m^2. A child of mass 199 kg runs at a speed of 3.09 m/sec tangent to the rim of the merry-go-round when it is at rest and then jumps on. Assume no friction in the bearing of the merry-go-round. What is the angular velocity of the merry-go-round and child?

Besides the child be a heavy example, I wasn't able to come up with the right answer of 0.276 rad/sec. What might I be doing wrong?

My Work:
I searched the archives and found two ways of solving the problem.
method \ 1 \ \Rightarrow m*v*r = (I_{merry-go-round} + I_{child}) \omega \ ; solving \ for \ \omega

method \ 2 \ \Rightarrow I_{merry-go-round} * \omega \ + \ m*v*r = (I_{merry-go-round} + I_{child}) \omega^{prime} \ ; \ solving \ for \ \omega^{prime}

What I came up with:

method \ 1 \ \Rightarrow (199 kg \ + (\frac{5680 kg*m^{2}}{3.91m^{2}}))*(3.09 \frac{m}{sec})*(3.91m) = ((5680 kg*m^{2}) + (199 kg * ((3.91m)^{2}))) \omega \ ; solving \ for \ \omega

\omega \ = \ 2.87

method \ 2 \ \Rightarrow ((5680 kg*m^{2}) * (\frac{3.09 \frac{m}{sec}}{3.91 m} * 2 \pi)) + (199 kg \ + (\frac{5680 kg*m^{2}}{(3.91m)^{2}})*(3.09 \frac{m}{sec})*(3.91m) = (5680 kg*m^{2} + (199 kg * (3.91m^{2})) \omega^{prime} \ ;
solving \ for \ \omega^{prime}

\omega^{prime} \ = \ 5.52

 

[Doc Al]

This is a conservation of angular momentum problem, so method #1 makes sense. (You could use method #2 by setting \omega = 0--since it starts at rest, but then you're back to method #1.) But you didn't plug in the numbers properly. Hint: What's "m" stand for?

 

[blue5t1053]

Am I to assume that 'm' is total mass of the system (merry-go-round + child) or merry-go-round for interia of the merry-go-round?

 

[Doc Al]

That would be incorrect. What's the total angular momentum of the system initially?

 

[blue5t1053]

L = r *p

L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child

 

[Doc Al]

L = r *p

This is correct.

L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child

But this isn't quite right. What's 371.53 kg the mass of? What does "p" stand for?

Hint: What's the only thing moving initially?

 

[blue5t1053]

I thought 'p' was for momentum of the merry-go-round. The only object moving initially is the child. If I apply the child attached to the merry-go-round initially with the starting speed, if I understand this correctly, would be

L = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}

 

[Doc Al]

You forgot the speed of the child. (Yes, "p" stands for momentum, not just mass.)

 

[blue5t1053]

I had a lapse of thought there. So I shouldn't include the mass of the merry-go-round?

 

[Doc Al]

Until the kid jumps on it, the merry-go-round isn't moving and thus has no angular momentum. Only the kid has angular momentum (with respect to the center of the merry-go-round).

 

[blue5t1053]

So if I revisit method one...

L_{child} = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}

L_{merry-go-round} = (3.91 m) * (371.531 kg) * (0 \frac{m}{sec}) = 0 \frac{kg*m^{2}}{sec}

I_{child} = (199 kg) * (3.91 m)^{2} = 3042.33 kg*m^{2}

method \ 1 \ \Rightarrow 2404.3 \frac{kg*m^{2}}{sec} = (5680 kg*m^{2} + 3042.33kg*m^{2} ) \omega \ ; solving \ for \ \omega \ = 0.2756 \frac{rad}{sec} = \ 0.276 \frac{rad}{sec}

Thanks!

 

[Doc Al]

Looks good!