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phy21050

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- Thread starter phy21050
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- #1

phy21050

- #2

Tom Mattson

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I would like to see how you would start this problem before telling you any detailed steps. Here is what you should be thinking about.

1. What shape can the merry-go-round be approximated as?

2. What can the man be approximated as?

3. What law of physics can be used to deterimine the initial and final angular velocities?

For #3, think about what quantity you learned in class lately that involve angular velocities. Is that quantity conserved? If so, can you use it here?

- #3

phy21050

1. the merry go round is like a circle

2. I have no idea what the man is, a mass?

3. and for the 3rd I am not sure

- #4

Tom Mattson

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Originally posted by phy21050

1. the merry go round is like a circle

Yes. Now, you know the rotational inertia of the merry-go-round, and you know the expression for the rotational inertia of a circle (from your book).

2. I have no idea what the man is, a mass?

Yes, he's a mass, but we are also interested in the configuration of the system (that's how you find the rotational inertia). So, try to find the rotational inertia of the (man+merry-go-round). I would treat the man as a point mass for this purpose.

3. and for the 3rd I am not sure

This should be readily apparent from the same chapter in which the problem appears. Try to look for it.

- #5

phy21050

- #6

Tom Mattson

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Here's a hint: The conservation law I alluded to earlier is the

Give it a try, show me how you start the problem, and I will help you from there.

- #7

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The initial angular momentum at the instant the man is about to jump on is:

L(total=L(man)-L(mgr)=Rp(man)-I(mgr)*omega(initial)

Now try to find the angular momentum after he jumps on.

L(total)=I(total)*omega(final)=

(I(mgr)+m(man)*R^2)*omega(final)

You should be able to work through this problem now.

Just remember what you are trying to solve for.

You are trying to solve for the final angular momentum.

*Hint* In order to get the decrease(in angular speed)what has to be equal?

peace out M2k

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