# Merry-Go-Round Velocity

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:

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

$$\omega \ = \ 2.87$$

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

$$\omega^{prime} \ = \ 5.52$$

## Answers and Replies

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Doc Al
Mentor
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?

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

$$L = r *p$$

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

Doc Al
Mentor
$$L = r *p$$
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?

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

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

Last edited:
Doc Al
Mentor
You forgot the speed of the child. (Yes, "p" stands for momentum, not just mass.)

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

Doc Al
Mentor
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).

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
Mentor
Looks good!