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Physics 14 Online
OpenStudy (anonymous):

please help ):

OpenStudy (anonymous):

OpenStudy (anonymous):

This is a tricky one. First, we need to calculate the moment of inertia of the merry go rough without the child present. Are you familiar with this concept?

OpenStudy (anonymous):

perhaps, moment of inertia of child would be : I= \[mr ^{2}\] yes?

OpenStudy (anonymous):

Correct!

OpenStudy (anonymous):

We know the mass and radius of the merry go round. Therefore, we can calculate the moment of inertia.

OpenStudy (anonymous):

Let's treat the merry go round as thing disk. This has a moment of inertia of \[I = { mr^2 \over 2}\]

OpenStudy (anonymous):

46*1.5=66

OpenStudy (anonymous):

as a thin*

OpenStudy (anonymous):

We get. \[I = {135 * 1.5^2 \over 2}\]

OpenStudy (anonymous):

Once the girl jumps onto the merry go round, she changes the moment of inertia because she has mass. By the parallel axis theorem, the two moments of inertia will add. So, this means that once the girl lands on the merry go round, our moment of inertia changes. We now have. \[I = {135 \cdot 1.5^2 \over 2} + 46 \cdot 1.5^2\]

OpenStudy (anonymous):

By conservation of momentum, the momentum of the girl BEFORE she lands on the merry go round must equal the momentum AFTER she lands on the merry go round.

OpenStudy (anonymous):

Her angular momentum BEFORE impact is defined has\[L = r \times mv\]

OpenStudy (anonymous):

The angular momentum AFTER she lands must equal this. \[r \times mv = I \omega\]

OpenStudy (anonymous):

where \(I \omega\) is the angular momentum of the girl and the merry go round.

OpenStudy (anonymous):

We know what I is from earlier.

OpenStudy (anonymous):

We can solve for \(\omega\) which is the angular velocity.

OpenStudy (anonymous):

above you said Her angular momentum BEFORE impact is defined has L=r×mv the m here is her mass or the merrygorounds mass or both masses added together?

OpenStudy (anonymous):

Her mass. She has a tangential velocity of 9 m/s at a radius of 1.5 meters from the center of the merry go round.

OpenStudy (anonymous):

I'm also unclear later when I say that this equals \(I \omega\). I here is the moment of inertia of both the girl and the merry go round. Or \[r \times m_{girl} v = \left [ {m_{mrg} r^2 \over 2} + m_{girl} r^2 \right ] \omega\]

OpenStudy (anonymous):

With a little algebra, this equation becomes more manageable.

OpenStudy (anonymous):

I hope this made sense.

OpenStudy (anonymous):

im 21 years old and never has anything ever made so much sense to me or explained as clearly as this was. you deserve 20787175451278 medals. thanks..! and yes. I got the correct answer.

OpenStudy (anonymous):

Awesome! I glad you made sense of it and learned something! I think this calls for a celebratory beer! Cheers and happy studying!

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